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The commuting triples of matrices
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The commuting triples of matrices

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Content THE COMMUTING TRIPLES OF MATRICES
by
Yongho Han
A Dissertation Presented to the
FACULTY OF THE GRADUATE SCHOOL
UNIVERSITY OF SOUTHERN CALIFORNIA
In Partial Ful¯llment of the
Requirements for the Degree
DOCTOR OF PHILOSOPHY
(MATHEMATICS)
August 2007
Copyright 2007 Yongho Han
Table of Contents
Abstract iii
1 Introduction 1
1.1 Historical Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 4+2+1 case 6
3 3+3+1 case 23
4 3+2+2 case 45
5 3+2+1+1 case 82
6 Conclusions 110
Reference List 111
ii
Abstract
The irreducibility of the variety C(m;n) commuting m-tuples of n£n matrices has
been studied by several authors. It is known that this variety is irreducible if m=2;n=
1;2;3andreducibleifm;n¸4. TheirreducibilityofC(3;n)wasaskedbyGersteinhaber.
Until now we know that C(3;n) is reducible for n¸32, but irreducible for n·4.
If the characteristic of an arbitrary algebraically closed ¯eld is zero, then we have
more speci¯c answer for this question. In this case C(3;n) is reducible for n ¸ 30 and
irreducible for n·6.
This dissertation is primarily concerned with the study of the variety C(3;7). We
show that this variety is also irreducible. Guralnick and Omladi· c have conjectured that
it is reducible for n>7.
iii
Chapter 1
Introduction
1.1 Historical Background
LetF be an algebraically closed ¯eld. Let
C(m;n)=f(A
1
;A
2
;¢¢¢;A
m
)jA
i
A
j
=A
j
A
i
;A
i
2M
n
(F)g.
This set may be viewed as a variety inF
mn
2
de¯ned by
m(m¡1)
2
n
2
quadratic equations.
It is not always easy to determine whether this variety is irreducible or to compute its
dimension[1]. In 1955 Motzkin and Taussky[5] proved the irreducibility of the variety
C(2;n) and found its dimension.
Theorem 1.1. (Motzkin-Taussky) The variety C(2;n) is irreducible and has dimension
n
2
+n.
If we take any commuting pairs A and B in M
n
(F), then we can consider a matrix
T 2 M
n
2(F) whose columns are the matrices A
i
B
j
, 0 · i;j < n. Let A be an algebra
generated by A and B. Then the dimension condition of algebra A is equivalent to the
rank condition of matrix T. Moreover the rank condition is a closed condition. Let
U = f(A;D) 2 C(2;n)jD : regularg. Then we can check that U is a dense subset of
C(2;n) and is a subset ofA. So we have the following famous result.
Theorem 1.2. Suppose A, B in M
n
(F) with AB =BA. Let A be the algebra generated
by A and B. Then dim(A)·n.
By similar reason of the above we have the following lemma.
1
Lemma 1.3. If C(m;n) is irreducible, then any commutative subalgebra of M
n
(F) gen-
erated by m elements has at most dimension n.
Proof. Let (A
1
;A
2
;¢¢¢;A
m
)2C(m;n). LetA be an algebra generated by the above ele-
ment. Let V =f(A
1
;A
2
;¢¢¢;A
m
)2C(m;n)jdim(A)·ng andU =f(D
1
;D
2
;¢¢¢;D
m
)2
C(m;n)jD
m
:regularg. Then U is an open subset of C(m;n) and is in V. Since C(m;n)
is irreducible and V is a closed subset, C(m;n)=V.
IfwetakefourunitelementsE
13
,E
14
,E
23
,E
24
inM
4
(F). Thentheseelementsarein
C(4;4) and the dimension of algebra generated by these four elements is 5. Therefore we
know thatC(4;4) is reducible. By considering block diagonal matrices, it follows that for
any n¸4, C(4;n) is reducible. This implies that C(m;n) is not irreducible for m;n¸4.
Since C(m;1) is equal toF
m
, obviously C(m;1) is irreducible.
Theorem1.4. ThematrixvarietyC(m;2)isirreducibleandhasadimension4+2(m¡1).
Proof. Since any 2£2 matrix is either a regular matrix or a scalar matrix, it follows that
the set of commuting m-tuples where the ¯rst term is regular is dense in C(m;2). Thus
as previous, C(m;2) is irreducible and has a dimension 4+2(m¡1).
Theorem 1.5. The matrix variety C(m;3) is irreducible.
Proof. It follows from [2].
The above results can be summarized as follows.
The matrix variety C(m;n) is
8
<
:
irreducible; if m=2, n=1;2;3;
reducible; if m;n¸4.
1.2 Introduction
Now we consider the matrix variety C(3;n). It was asked in [1] for which values of n
is this variety irreducible. In 1990 Guralnick [2] had proved by a dimensional argument
that C(3;n) is reducible for n¸32, but irreducible for n·3. Using the results from [6],
Guralnick and Sethuraman [3] had proved that C(3;n) is irreducible when n=4.
If the characteristic ofF is zero, then we have more speci¯c answer for this question.
2
From now on we assume that charF = 0. Holbrook and Omladi· c [4] gave the answer to
this question. In that paper [4] they focused on the problem of approximating 3-tuples
of commuting n£ n matrices over F by commuting m-tuples of generic matrices, i.e.
matrices with distinct eigenvalues. Let G(3;n) be the subset of C(3;n) consisting of
triples (A;B;C) such that A, B, and C are all generic. They asked whether G(3;n)
equals C(3;n). Here the overline means the closure of G(3;n) with respect to the Zariski
topology on F
3n
2
. The problem is equivalent to that of whether the variety of C(3;n)
is irreducible. They had proved that C(3;n) is reducible for n ¸ 30, but irreducible
for n = 5. Recently using this perturbation technique Omladi· c [7] gave the answer to
this problem for the case n = 6; i.e., He proved that the variety C(3;6) is irreducible if
charF=0.
In this paper we consider the a±ne variety C(3;7) and prove that C(3;7) is also
irreducible. Actually, the proofs of the theorems given in section 2.1, 2.2, 2.3, and 2.4
work also if the characteristic of the ¯eld F is nonzero. So it seems likely that some
variations of the proofs will give the results in all characteristic. We have been informed
that Omladi· c has independently obtained the same result.
1.3 Preliminaries
The following theorems summarized some known results.
Theorem 1.6. If the radical of the algebra generated by matrices (A;B;C) has square
zero, then the triple belongs to G(3;n).
Proof. It follows from Theorem 1 of [7].
Theorem 1.7. If in a 3-dimensional linear space L of nilpotent commuting matrices
there is a matrix of maximal possible rank with only one nonzero Jordan block, then any
basis of this space belongs to G(3;n).
Proof. It follows from Theorem 2 of [7].
3
Theorem 1.8. If in a 3-dimensional linear space L of nilpotent commuting matrices
there is a matrix with only two Jordan blocks, then any basis of this space belongs to
G(3;n).
Proof. It follows from Corollary 10 of [6].
At the ¯rst we give a result to be used in the sequel.
Lemma 1.9. Assume that C(3;m) is irreducible for m < n. If (A;B;C) 2 C(3;n)
and all three commute with an element that has at least two distinct eigenvalues, then
(A;B;C)2G(3;n).
Proof. Thisimplieswecanassumethatallthreematricesareblockdiagonalwithatleast
two blocks of sizes m and n¡m with 0 < m < n. So by the assumption, (A;B;C) 2
G(3;m)£G(3;n¡m) and rather clearly G(3;m)£G(3;n¡m)½G(3;n).
Lemma 1.10. Suppose that (A;B;C)2G(3;n) and the algebra generated by (A
0
;B
0
;C
0
)
is contained in the algebra generated by (A;B;C). Then (A
0
;B
0
;C
0
)2G(3;n).
Proof. We can ¯nd polynomials a;b and c in 3 variables such that A
0
= a(A;B;C),
B
0
=b(A;B;C), and C
0
=c(A;B;C). Consider the map from C(3;n) to itself that sends
(X;Y;Z) ! (a(X;Y;Z);b(X;Y;Z);c(X;Y;Z)). This is a morphism of varieties. Note
that it sends G(3;n) to G(3;n) (since it sends diagonalizable triples to diagonalizable
triples) and so the same for G(3;n), whence the result.
In particular, this shows that the property of being in G(3;n) depends only on the
algebra generated by the triple.
Corollary 1.11. Suppose that (A;B;C)2C(3;n) and X is a polynomial in A;B. Then
(A;B;C)2G(3;n) if and only if (A;B;C +X) is.
Lemma 1.12. Let J be a ¯nite dimensional nil algebra generated by A
1
;:::;A
m
. Let I
be the ideal generated by A
i
A
j
with 1·i;j·m. If B
1
;:::;B
m
2I, then J is generated
as an algebra by the A
i
+B
i
;1·i·m.
Proof. Let R be the algebra generated by J and an identity element. Then J is the
Jacobson radical of R. Now apply Nakayama's lemma.
4
In particular, this implies the following.
Corollary1.13. Let(A;B;C)2C(3;n)withA;B andC nilpotent. LetX;Y andZ bein
thealgebrageneratedbywordsoflengthtwoinA;B andC (i.e.,A
2
;B
2
;C
2
;AB;AC;BC).
Then (A;B;C)2G(3;n) if and only if (A+X;B+Y;C +Z) is in G(3;n).
5
Chapter 2
4+2+1 case
In this chapter we give the result that a linear space of nilpotent commuting matrices
of size 7£7 having a matrix of maximal rank with one Jordan block of order 4 and one
Jordan block of order 2 can be perturbed by the generic triples.
Theorem 2.1. If in a 3-dimensional linear space L of nilpotent commuting matrices of
size 7£7 there is a matrix of maximal possible rank with one Jordan block of order 4 and
one Jordan block of order 2, then any basis of this space belongs to G(3;7).
Proof. We can write A2 L of maximal in some basis as
A=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If B 2 L, then the structure of B is well known. It is nilpotent and we may add to it a
polynomial in A, so that it looks like
6
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 a b c
0 0 0 0 0 a 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 d e 0 f g
0 0 0 d 0 0 0
0 0 0 h 0 i 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let C be a matrix in a L. Then C looks like
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 a
0
b
0
c
0
0 0 0 0 0 a
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 d
0
e
0
0 f
0
g
0
0 0 0 d
0
0 0 0
0 0 0 h
0
0 i
0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If B
4
6= 0 then B has only one Jordan block and consequently we are done by Theorem
1.7. So we may assume that B
4
=0. So we have agdi=0. Now we may assume that at
least one of a;g;d;i is 0. We will consider 15 separate cases.
Case 1. Assume that there is a matrix B 2 L such that a 6= 0, d 6= 0, and g 6= 0 ,
so that for any B2 L the corresponding entry i = 0. Without loss of generality we may
assume that a = 1. As above we may assume that i
0
= 0, but also a
0
= 0. Since the
algebra generated by A and B contains AB, by Corollary 1.13., we may assume that
b=0 and b
0
=0. The commutative relation of B and C implies that
ch
0
+e
0
=c
0
h, fd
0
+gh
0
=f
0
d+g
0
h, d
0
=f
0
=g
0
=0.
Therefore we have that h
0
=0 and e
0
=c
0
h. Let
7
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡c
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Sincefort6=0thematrixB+tX hasmorethanonepointintheitsspectrum,wecanuse
Lemma1.9. Thereforethetriple(A;B+tX;C+tY)belongstoG(3;7)forallt2F,t6=0.
Case 2. Assume that there is a matrix B 2 L such that a 6= 0, g 6= 0, and i 6= 0 ,
so that for any B2 L the corresponding entry d=0. Without loss of generality we may
assume that a = 1. So we may assume that a
0
= d
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that b = 0 and b
0
= 0. The
commutative relation of B and C implies that
ch
0
+e
0
=c
0
h, ci
0
+f
0
=c
0
i, gh
0
=g
0
h, gi
0
=g
0
i, g
0
=0.
So we have that h
0
= 0, i
0
= 0, e
0
= c
0
h, and f
0
= c
0
i. In this case we may assume that
f
0
6=0. Indeed, if f
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0
1
i
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0
h
i
0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
8
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
f
0
0 0 0 0 0 0
0 f
0
0 0 0 0 0
0 0 f
0
0 0 0 0
0 0 0 f
0
0 0 0
0 0 ¡e
0
0 0 0 0
0 0 0 ¡e
0
0 0 0
0 0 0 0 0 0 f
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
We can easily ¯nd that XC =CX bythe direct computation. So B+tX commutes with
C for all t2F. Moreover for t6=0 the matrix B+tX has more than one point in the its
spectrum. Therefore, by Lemma 1.9., the triple (A;B+tX;C) belongs to G(3;7) for all
t2F, t6=0.
Case 3. Assume that there is a matrix B 2 L such that a 6= 0, d 6= 0, and i 6= 0 ,
so that for any B 2 L the corresponding entry g = 0. Now we may assume that a = 1.
So we may assume that a
0
= g
0
= 0. Since the algebra generated by A and B contains
AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. The commutative relation
of B and C implies that
ch
0
+e
0
=c
0
h, ci
0
+f
0
=c
0
i, fd
0
=f
0
d, id
0
=i
0
d, d
0
=0.
Sowehavethatf
0
=0,i
0
=0,c
0
=0, andch
0
+e
0
=0. Ifh
0
=0, thenwecouldintroduce
the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 ¡c 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Now, we may assume that h
0
6=0. Let
9
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 h
0
0 ¡e
0
0 0 0 0 0 h
0
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Thenwecaneasily¯ndthatXC =CX. SoB+tX commuteswithC forallt2F. Then
for t 6= 0 the matrix B +tX has more than one point in the its spectrum. By Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 4. Assume that there is a matrix B 2 L such that d 6= 0, g 6= 0, and i 6= 0 ,
so that for any B2 L the corresponding entry a=0. Without loss of generality we may
assume that d = 1. So we may assume that a
0
= d
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that e = 0 and e
0
= 0. The
commutative relation of B and C implies that
ch
0
=c
0
h+b
0
, ci
0
=c
0
i, gh
0
=g
0
h+f
0
, gi
0
=g
0
i, i
0
=0.
So we have that c
0
=0, g
0
=0, ch
0
=b
0
, and gh
0
=f
0
. If f
0
=0, then we could introduce
the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0
c
g
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0
1
g
0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that f
0
6=0. Let
10
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
f
0
0 0 0 ¡b
0
0 0
0 f
0
0 0 0 ¡b
0
0
0 0 f
0
0 0 0 0
0 0 0 f
0
0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 f
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then we can easily ¯nd that XC = CX. So B +tX commutes with C for all t 2 F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. By
Lemma 1.9, the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 5. Assume that there is a matrix B 2 L such that a 6= 0, d 6= 0. Then at
least one of g and i is zero. If exactly one of g and i is zero, we are done by Case 1. and
Case 3. So we may assume that there is a matrix B 2 L such that a6= 0 and d6= 0, so
that for any B2 L the corresponding entry g =i=0. Without loss of generality we may
assume that a=1. So we may assume that a
0
=g
0
=i
0
=0. Since the algebra generated
by A and B contains AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. The
commutative relation of B and C implies that
ch
0
+e
0
=c
0
h, d
0
=0, f
0
=0.
If h
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 ¡c 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that h
0
6=0. Let
11
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 h
0
0 ¡e
0
0 0 0 0 0 h
0
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then we can easily ¯nd that XC = CX. So B +tX commutes with C for all t 2 F.
Moreover for t6=0 the matrix B+tX has more than one point in the its spectrum. We
are done by Lemma 1.9.
Case 6. Assume that there is a matrix B 2 L such that a 6= 0, i 6= 0. Then at
least one of d and g is zero. If only one of d and g is zero, we are done by Case 2. and 3.
So we may assume that there is a matrix B 2 L such that a6= 0 and i6= 0, so that for
any B2 L the corresponding entry d=g =0. Without loss of generality we may assume
that a = 1. So we may assume that a
0
= g
0
= d
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that b = 0 and b
0
= 0. The
commutative relation of B and C implies that
ch
0
+e
0
=c
0
h, ci
0
+f
0
=c
0
i.
If c
0
=0 and i
0
6=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 h 0 i 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
6=0 and i
0
=0, then we could introduce the matrix
12
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 ¡c 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
=0 and i
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 h 0 i¡c 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In all of these cases the matrix Z clearly commutes with both A andB, so that it su±ces
to prove our case for all triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that
c
0
6=0 and i
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
x 0 0 0 0 0 0
0 x 0 0 0 0 0
0 0 x 0 0 0 0
0 0 0 x 0 0 0
0 0 y 0 0 0 z
0 0 0 y 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 w
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=c
0
i
0
, y =¡c
0
h
0
, z =¡c
02
i, and w =¡c
02
i
0
.
Then a straightforward computation reveals that B + tX commutes with C + tY for
all t 2F. Since for t 6= 0 the matrix B +tX has more than one point in the its spec-
trum,byLemma1.9.,thetriple(A;B+tX;C+tY)belongstoG(3;7)forallt2F,t6=0.
13
Case 7. Assume that there is a matrix B 2 L such that d 6= 0, i 6= 0. Then at
least one of a and g is zero. If only one of a and g is zero, we are done by Case 3. and 4.
So we may assume that there is a matrix B 2 L such that d6= 0 and i6= 0, so that for
any B2 L the corresponding entry a=g =0. Without loss of generality we may assume
that d = 1. So we may assume that a
0
= g
0
= d
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that e = 0 and e
0
= 0. The
commutative relation of B and C implies that
ch
0
=b
0
+c
0
h, ci
0
=c
0
i, f
0
=0, i
0
=0.
Therefore c
0
=0 and ch
0
=b
0
. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡i
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 h
0
b 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 h
0
0 0 h
0
f 0
0 0 0 h
0
0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straight computation reveals that B+tX commutes with C +tY for all t2F. Since
for t 6= 0 the matrix B +tX has more than one point in the its spectrum, we can use
Lemma 1.9. So the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 8. Assume that there is a matrix B 2 L such that d 6= 0, g 6= 0. Then at
least one of a and i is zero. If only one of a and i is zero, we are done by Case 1. and 4.
So we may assume that there is a matrix B 2 L such that d6= 0 and g 6= 0, so that for
any B2 L the corresponding entry a=i=0. Without loss of generality we may assume
that d = 1. So we may assume that a
0
= i
0
= d
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that e = 0 and e
0
= 0. The
commutative relation of B and C implies that
ch
0
=b
0
+c
0
h, gh
0
=f
0
+g
0
h.
14
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 ¡h 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 c
0
0 0
0 0 0 0 0 c
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 g
0
0 0
0 0 0 0 0 g
0
0
0 0 0 0 0 gh
0
g
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 g
0
h
0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 9. Assume that there is a matrix B 2 L such that g 6= 0, i 6= 0. Then at
least one of a and d is zero. If only one of a and d is zero, we are done by Case 2. and
4. So we may assume that there is a matrix B 2 L such that g 6= 0 and i6= 0, so that
for any B 2 L the corresponding entry a = d = 0. Without loss of generality we may
assume that g = 1. So we may assume that a
0
= d
0
= g
0
= 0. The commutative relation
of B and C implies that
ch
0
=c
0
h, h
0
=0, i
0
=0, ci
0
=c
0
i.
Thus c
0
=0 and ch
0
=0. So the matrix C looks like
15
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 b
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 e
0
0 f
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Therefore there is a projection commuting with A and C.
Case 10. Assume that there is a matrix B 2 L such that a 6= 0, g 6= 0. Then at
least one of d and i is zero. If only one of d and i is zero, we are done by Case 1. and 2.
So we may assume that there is a matrix B 2 L such that a6= 0 and g 6= 0, so that for
any B2 L the corresponding entry d=i=0. Without loss of generality we may assume
that a = 1. So we may assume that d
0
= i
0
= a
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that b = 0 and b
0
= 0. The
commutative relation of B and C implies that
ch
0
+e
0
=c
0
h, f
0
=0, g
0
=0, gh
0
=0.
Thus h
0
=0 and e
0
=c
0
h. Introduce matrices X and Y,
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡c
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A direct computation reveals that B+tX commutes with C+tY for all t2F. Moreover
for t6=0 the matrix B+tX has more than one point in the its spectrum. So, by Lemma
1.9, the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
16
Case 11. Assume that there is a matrix B 2 L such that a 6= 0. Then at least one
of d, i, and g is zero. If at least one of d, i, and g is nonzero, we are done by the above
Cases. So we may assume that there is a matrix B2 L such that a6= 0, so that for any
B2 L the corresponding entry d=i=g =0. Without loss of generality we may assume
that a = 1. So we may assume that d
0
= i
0
= g
0
= 0 and a
0
= 0. The commutative
relation of B and C implies that
ch
0
+e
0
=c
0
h, f
0
=0.
If c
0
=0 and h
0
6=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 h 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
6=0 and h
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 ¡c 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
=0 and h
0
=0, then we could introduce the matrix
17
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 h¡c 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In all of these cases the matrix Z clearly commutes with both A andB, so that it su±ces
to prove our case for all triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that
c
0
6=0 and h
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 h
0
0 ¡e
0
0 0 0 0 0 h
0
0
0 0 0 0 0 x 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, where x=¡
b
0
h
0
c
0
.
A straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 12. Assume that there is a matrix B 2 L such that g 6= 0. Then at least one
of a, d, and i is zero. If at least one of a, d, and i is nonzero, we are done by the above
Cases. So we may assume that there is a matrix B2 L such that g6= 0, so that for any
B2 L the corresponding entry a=d=i=0. Without loss of generality we may assume
that g = 1. So we may assume that a
0
= d
0
= i
0
= 0 and g
0
= 0. The commutative
relation of B and C implies that
ch
0
=c
0
h, h
0
=0.
18
So we have c
0
h=0. If c
0
=0 then we are done as we can ¯nd a projection P commuting
with A and C. If c
0
6=0 then h=0. If f
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that f
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
c
0
0 0 0 0 0 0
0 c
0
0 0 0 0 0
0 0 c
0
0 0 0 0
0 0 0 c
0
0 0 0
0 0 x 0 0 0 0
0 0 0 x 0 0 0
0 0 0 y 0 b
0
c
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, where x=¡
e
0
c
0
f
0
and y =
e
0
b
0
f
0
.
Then we can easily ¯nd that XC =CX. So B+tX commutes with C for all t2F.
Thus for t6= 0 the matrix B +tX has more than one point in the its spectrum. So, by
Lemma 1.9, the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 13. Assume that there is a matrix B 2 L such that d 6= 0. Then at least one
of a, g, and i is zero. If at least one of a, g, and i is nonzero, we are done by the above
Cases. So we may assume that there is a matrix B2 L such that d6= 0, so that for any
B2 L the corresponding entry a=g =i=0. Without loss of generality we may assume
that d = 1. So we may assume that a
0
= g
0
= i
0
= 0 and d
0
= 0. The commutative
relation of B and C implies that
ch
0
=b
0
+c
0
h, f
0
=0.
19
If c
0
=0 and h
0
6=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 ¡h 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
6=0 and h
0
=0 , then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 c 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If c
0
=h
0
=0 , then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 c¡h 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In all of these three cases the matrix Z clearly commutes with both A and B, so that it
su±ces to prove our case for all triples (A;B;C+tZ) for t2F, t6=0. So we may assume
that c
0
6=0 and h
0
6=0. Let
20
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 h
0
0 ¡e
0
0 0 0 0 0 h
0
0
0 0 0 0 0 x 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, where x=¡
b
0
h
0
c
0
.
By a straight computation we have XC = CX. Since XC = CX, B +tX commutes
with C for all t2F. Moreover, for t6=0 the matrix B+tX has more than one point in
the its spectrum. Thus, by Lemma 1.9., the triple (A;B+tX;C) belongs to G(3;7) for
all t2F, t6=0.
Case 14. Assume that there is a matrix B 2 L such that i 6= 0. Then at least one
of a, d, and g is zero. If at least one of a, d, and g is nonzero, we are done by the above
Cases. So we may assume that there is a matrix B 2 L such that i6= 0, so that for any
B2 L the corresponding entry a=d=g =0. Without loss of generality we may assume
that i = 1. So we may assume that a
0
= g
0
= d
0
= 0 and i
0
= 0. The commutativity
condition of B and C implies that
ch
0
=c
0
h, c
0
=0.
So we have ch
0
=0. If h
0
=0 then we are done as we can ¯nd a projection P commuting
with A and C. If h
0
6=0 then c=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b
0
0 0
0 0 0 0 0 b
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 e
0
0 f
0
0 0
0 0 0 e
0
0 f
0
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
21
Then we can ¯nd that XC =CX. So B+tX commutes with C for all t2F. Moreover
we may assume that f
0
6=0. Indeed, if f
0
=0, then we can introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with A and B, so that it su±ces to prove our case for all triples
(A;B;C+tZ) for t2F, t6=0. So for t6=0 the matrix B+tX has more than one point
in the its spectrum. Thus, by Lemma 1.9., the triple (A;B +tX;C) belongs to G(3;7)
for all t2F, t6=0.
Case 15. Assume that there is a matrix B 2 L such that a, d, g, and i are zero.
If at least one of a
0
, d
0
, g
0
, and i
0
is not zero, then we can change a role of B and C. So
we are done by above Cases. Thus we may assume that a
0
=g
0
=d
0
=0 and i
0
=0. The
commutativity condition of B and C implies that
ch
0
=c
0
h.
It means that (c;h) and (c
0
;h
0
) are linearly dependent, so that we may choose one of
them, say the ¯rst one, to be zero. It follows that the projection
P =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
commutes with A and B, and we are done.
22
Chapter 3
3+3+1 case
In this chapter we give the result that a linear space of nilpotent commuting matrices
of size 7£7 having a matrix of maximal rank with two Jordan blocks of order 3 can be
perturbed by the generic triples.
Theorem 3.1. If in a 3-dimensional linear space L of nilpotent commuting matrices of
size 7£7 there is a matrix of maximal possible rank with two Jordan block of order 3,
then any basis of this space belongs to G(3;7).
Proof. We can write A2 L of maximal in some basis as
A=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If B 2 L, then the structure of B is well known. It is nilpotent and we may add to it a
polynomial in A, so that it looks like
23
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a b c d
0 0 0 0 a b 0
0 0 0 0 0 a 0
e f g 0 h i j
0 e f 0 0 h 0
0 0 e 0 0 0 0
0 0 k 0 0 l 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let C be a matrix in a L. Then C looks like
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a
0
b
0
c
0
d
0
0 0 0 0 a
0
b
0
0
0 0 0 0 0 a
0
0
e
0
f
0
g
0
0 h
0
i
0
j
0
0 e
0
f
0
0 0 h
0
0
0 0 e
0
0 0 0 0
0 0 k
0
0 0 l
0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Since the matrix B is nilpotent, we ¯nd a relation ae = 0. Now we may assume that at
least one of a and e is 0. We will consider 2 separate cases.
Case 1. Assume that there is a matrix B 2 L such that a6= 0, so that for any B 2 L
the corresponding entry e = 0. Without loss of generality we may assume that a = 1.
So we may assume that a
0
= 0 and e
0
= 0. If B
4
6= 0 then B has only one Jordan block
and consequently we are done by Theorem 1.7. So we may assume that B
4
= 0. Since
B
4
=0, wehavethatf =0andjk =0. BycommutativityconditionofB andC wehave
f
0
=0. Nowwematassumethatatleastoneofj andkiszero. Sothereare3possibilities.
(i) Assume that there is a matrix B 2 L such that j 6= 0, so that for any B 2 L
the corresponding entry k = 0. Then we may assume that k
0
= 0. The commutative
relation of B and C implies that
g
0
=h
0
=j
0
=0, dl
0
+i
0
=b
0
h+d
0
l, jl
0
=0.
24
So we have l
0
=0 and i
0
=b
0
h+d
0
l. If b
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 h 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that b
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 x y 0 0
0 0 0 0 x y 0
0 0 0 0 0 x 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 z 0 0
0 0 0 0 0 z 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=b
0
, y =bb
0
¡c
0
, and z =b
02
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B2 L such that k6=0, so that for any B2 L the
corresponding entry j =0. Then we may assume that j
0
=0. The commutative relation
of B and C implies that
d
0
=g
0
=h
0
=k
0
=0, dl
0
+i
0
=b
0
h.
Let
25
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 b 0 0
0 0 0 0 1 b 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 b
0
c
0
0
0 0 0 0 0 b
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
By the straightforward computation we ¯nd that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0thematrixB+tX hasmorethanonepointintheitsspectrum.
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
(iii) Assume that there is a matrix B2 L such that j and k are zero. If at least one
of j
0
and k
0
is nonzero, then we can change a role of B and C. So we are done by above
Cases. Thus we may assume that j
0
=k
0
=0. Then B and C look like
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 1 b c d
0 0 0 0 1 b 0
0 0 0 0 0 1 0
0 0 g 0 h i 0
0 0 0 0 0 h 0
0 0 0 0 0 0 0
0 0 0 0 0 l 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b
0
c
0
d
0
0 0 0 0 0 b
0
0
0 0 0 0 0 0 0
0 0 g
0
0 h
0
i
0
0
0 0 0 0 0 h
0
0
0 0 0 0 0 0 0
0 0 0 0 0 l
0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
The commutative relation of B and C implies that g
0
=0, h
0
=0, dl
0
+i
0
=b
0
h+d
0
l. If
l
0
=0, then we could introduce the matrix
26
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 ¡d 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that l
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 l
0
bl
0
0 d
0
l
0 0 0 0 l
0
bl
0
0
0 0 0 0 0 l
0
0
0 0 0 0 0 0 l
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 b
0
l
0
c
0
l
0
d
0
l
0
0 0 0 0 0 b
0
l
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 2. Assume that there is a matrix B 2 L such that e6= 0, so that for any B 2 L
the corresponding entry a = 0. Without loss of generality we may assume that e = 1.
So we may assume that a
0
= 0 and e
0
= 0. If B
4
6= 0 then B has only one Jordan block
and consequently we are done by Theorem 1.7. So we may assume that B
4
= 0. Since
B
4
=0, we have that b=0 and dl =0. By commutativity condition of B and C we have
b
0
=0. Nowwemayassumethatatleastoneofdandliszero. Sothereare3possibilities.
(i) Assume that there is a matrix B2 L such that d6= 0, so that for any B2 L the
corresponding entry l = 0. Then we may assume that l
0
= 0. The commutative relation
of B and C implies that c
0
=d
0
=h
0
=k
0
=0, hf
0
=j
0
k+i
0
. Let
27
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 ¡f 0 0
0 0 0 0 0 ¡f 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 ¡f
0
¡g
0
0
0 0 0 0 0 ¡f
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A direct computation reveals that B+tX commutes with C+tY for all t2F. Moreover
for t6=0 the matrix B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B2 L such that l6= 0, so that for any B2 L the
corresponding entry d=0. Then we may assume that d
0
=0. The commutative relation
of B and C implies that
c
0
=j
0
=h
0
=l
0
=0, hf
0
+jk
0
=i
0
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 f 0 0
0 0 0 0 1 f 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 f
0
g
0
0
0 0 0 0 0 f
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii) Assume that there is a matrix B 2 L such that d and l are zero. If at least
one of d
0
and l
0
is nonzero, then we can change a role of B and C. So we are done by
28
above Cases. Thus we may assume that d
0
=l
0
=0. The commutative relation of B and
C implies that
c
0
=h
0
=0, hf
0
+jk
0
=j
0
k+i
0
.
If j
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 ¡k 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that j
0
6= 0, say j
0
= 1. Now
we may assume that j =0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 f 0 0
0 0 0 0 1 f 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 f
0
g
0
0
0 0 0 0 0 f
0
0
0 0 0 0 0 0 0
0 0 0 0 0 k
0
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A direct computation reveals that B+tX commutes with C+tY for all t2F. Moreover
for t6=0 the matrix B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 3. Assume that there is a matrix B 2 L such that a and e are zero. If at
least one of a
0
and e
0
is nonzero, then we can change a role of B and C. So we are done
by Case 1. and Case 2. Thus we may assume that a
0
= e
0
= 0. The commutativity
relation of B and C implies that
29
bf
0
+dk
0
=b
0
f +d
0
k, bh
0
+dl
0
=b
0
h+d
0
l,
hf
0
+jk
0
=h
0
f +j
0
k, fb
0
+jl
0
=f
0
b+j
0
l.
Now we will try to consider 2 cases.
(i) Assume that (k;l) and (k
0
;l
0
) are linearly independent. Without loss of gener-
ality we may assume that (k;l) = (1;0) and (k
0
;l
0
) = (0;1). By the commutativity
relation, we have a relation b(j
0
h
0
)¡f(dh
0
)=b
0
(j
0
h)¡f
0
(dh). Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 x
0
0 0 y
0
0 0 0 0 x
0
0 0
0 0 0 0 0 x
0
0
z
0
0 0 0 0 0 w
0
0 z
0
0 0 0 0 0
0 0 z
0
0 0 0 0
0 0 0 0 0 0 ®
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 x 0 0 y
0 0 0 0 x 0 0
0 0 0 0 0 x 0
z 0 0 0 0 0 w
0 z 0 0 0 0 0
0 0 z 0 0 0 0
0 0 0 0 0 0 ®
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where ®
0
= ¡h
0
d, ® = ¡hj
0
, x
0
= hd, x = h
0
d, z
0
= hj
0
, z = h
0
j
0
, y
0
= xi¡ x
0
i
0
,
y = cz¡gx¡c
0
z
0
+g
0
x
0
, w
0
= g
0
x
0
¡c
0
z
0
+cz¡gx, and w = iz¡i
0
z
0
. A direct com-
putation reveals that B+tX commutes with C +tY for all t2F. If at least one of h
0
d
and hj
0
is nonzero then one of B +tX and C +tY has more than one point in the its
spectrum. So, by Lemma 1.9., the triple (A;B +tX;C +tY) belongs to G(3;7) for all
t2F, t6=0. Now we will consider the case that h
0
d=0 and hj
0
=0.
(1) Assume that d = h = h
0
= j
0
= 0. By the commutativity relation, we have
that j =bf
0
¡fb
0
and d
0
=j. If f
0
6=0, then we can take X and Y as follows.
30
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ® 0 0 x
0 0 0 0 ® 0 0
0 0 0 0 0 ® 0
0 0 0 0 0 0 y
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 f
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ¯ 0 0 y
0 0 0 0 ¯ 0 0
0 0 0 0 0 ¯ 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where ® =¡f, ¯ =¡f
0
, x = fi
0
¡f
0
i, and y = f
0
g¡fg
0
. By a straight computation
we can ¯nd that B +tX commutes with C +tY for all t 2F. Moreover for t 6= 0 the
matrix B +tX has more than one point in the its spectrum. So, by Lemma 1.9., the
triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
If f
0
= 0. By commutativity condition of B and C we have j = d
0
=¡fb
0
. Then we
can take X and Y as follows.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 0 0 ¡c
0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡g
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
if j =0 and f 6=0.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 g 0 0 x
0 0 0 0 g 0 0
0 0 0 0 0 g 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡g
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 g
0
0 0 0
0 0 0 0 g
0
0 0
0 0 0 0 0 g
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=g
0
i¡gi
0
, if j =0 and f =0.
31
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 y
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 b
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ® 0 0 z
0 0 0 0 ® 0 0
0 0 0 0 0 ® 0
¯ 0 0 0 0 0 w
0 ¯ 0 0 0 0 0
0 0 ¯ 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where ®=¡b, ¯ =¡f, x=¡bi, y =gb¡fc, z =bg¡cf, and w =¡fi, if j6=0.
IneachcaseastraightforwardcomputationrevealsthatB+tX commuteswithC+tY
for all t2F. In the second case we may assume that g
0
6= 0. In the third case we also
assume that b6= 0. Moreover for t6= 0 the matrix B +tX has more than one point in
the its spectrum. So, by Lemma 1.9., the triple (A;B +tX;C +tY) belongs to G(3;7)
for all t2F, t6=0.
(2) Assume that d = h = h
0
= 0, and j
0
6= 0. By the commutativity condition
we have j
0
=hf
0
¡h
0
f. But this relation means j
0
=0. So this case can not occur.
(3) Assume that d = h
0
= j
0
= 0, and h 6= 0. Then by the commutativity con-
dition we have
j =d
0
=b
0
=f
0
=0.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
1 0 0 0 0 0 ¡c
0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡g
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
32
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(4) Assume that h = h
0
= j
0
= 0, and d 6= 0. By the commutativity condition
we have d=hb
0
¡h
0
b. But this relation means d=0. So this case can not occur.
(5) Assume that d = h = j
0
= 0, and h
0
6= 0. The commutativity relation of B
and C implies that
j =f =b=d
0
=0.
If c=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. Now we may assume that c6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 ¡ci
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 c
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 x 0 0 0
0 0 0 0 x 0 0
0 0 0 0 0 x 0
y 0 0 0 0 0 z
0 y 0 0 0 0 0
0 0 y 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡c, y =¡g, and z =¡gi.
33
A direct computation reveals that B+tX commutes with C+tY for all t2F. Moreover
for t6=0 the matrix B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(6) Assume that h = h
0
= 0, d 6= 0, and j
0
6= 0. By the commutativity condi-
tion we have d=hb
0
¡h
0
b. But this relation means d=0. So this case can not occur.
(7) Assume that h
0
= j
0
= 0, d 6= 0, and h 6= 0. The commutative relation of B
and C implies that
j =d
0
=¡b
0
f, d=b
0
h.
If b=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. Now we may assume that b6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 y
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 b
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 z 0 0 y
0 0 0 0 z 0 0
0 0 0 0 0 z 0
® 0 0 ¯ 0 0 °
0 ® 0 0 ¯ 0 0
0 0 ® 0 0 ¯ 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=ch¡bi, y =gb¡cf, z =¡b, ®=¡f, ¯ =¡h, and ° =gh¡if.
34
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0thematrixC+tY hasmorethanonepointintheitsspectrum.
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
(8) Assume that d = h = 0, j
0
6= 0, and h
0
6= 0. The commutativity relation of
B and C implies that
b=0, j =d
0
=¡b
0
f, j
0
=¡h
0
f.
If f
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that f
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
x 0 0 y 0 0 z
0 x 0 0 y 0 0
0 0 x 0 0 y 0
0 0 0 0 0 0 w
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 b
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ® 0 0 w
0 0 0 0 ® 0 0
0 0 0 0 0 ® 0
0 0 0 ¯ 0 0 °
0 0 0 0 ¯ 0 0
0 0 0 0 0 ¯ 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡
j
0
f
0
, y =
j
f
0
, z =¡b
0
i+ch
0
+
c
0
j
0
f
0
¡
i
0
j
f
0
, w =gb
0
+
g
0
j
f
0
®=¡b
0
, ¯ =¡h
0
, and
° =gh
0
+
j
0
g
0
f
0
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0thematrixC+tY hasmorethanonepointintheitsspectrum.
35
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
(9) Assume that d = j
0
= 0, h 6= 0, and h
0
6= 0. The commutativity relation of
B and C implies that
bh
0
=b
0
h, j =d
0
=bf
0
¡b
0
f, hf
0
=h
0
f.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 x 0 0 y
0 0 0 0 x 0 0
0 0 0 0 0 x 0
0 0 0 0 0 0 z
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 h
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 w 0 0 z
0 0 0 0 w 0 0
0 0 0 0 0 w 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡h, y =hi
0
¡h
0
i, z =gh
0
¡g
0
h, w =¡h
0
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0thematrixB+tX hasmorethanonepointintheitsspectrum.
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that (k;l) and (k
0
;l
0
) are linearly dependent. Without loss of gener-
ality we may assume that (k
0
;l
0
)=(0;0). We will consider 4 possibilities.
(1) Assume that d
0
= j
0
= 0. Then there is a projection P commuting with A
and C.
(2) Assume that d
0
6= 0, and j
0
= 0. Without loss of generality we may assume
thatd
0
=1. Sowemayassumethatd=0. ThecommutativerelationofB andC implies
that
bf
0
=b
0
f +k, bh
0
=b
0
h+l, hf
0
=h
0
f, bf
0
=b
0
f.
36
So we have k =0. If j =0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B +tZ;C) for t 2F, t 6= 0. Now we may assume that j 6= 0. Now we will
consider 2 cases.
(2-a) Assume that h
0
6=0. Then we can ¯nd X and Y as follows.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 h 0 0 0
0 0 0 0 h 0 0
0 0 0 0 0 h 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 x 0 0 y z
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 h
0
0 0 0
0 0 0 0 h
0
0 0
0 0 0 0 0 h
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 w 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=hg
0
¡h
0
g, y =hi
0
¡h
0
i, z =¡jh
0
, w =
hg
0
¡h
0
g
j
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6= 0 the matrix B+tX has more than one point in the its spectrum. So,
by Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(2-b) Assume that h
0
=0. If at least one of f and f
0
is nonzero. Let
37
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 f 0 0 0
0 0 0 0 f 0 0
0 0 0 0 0 f 0
0 0 0 0 0 x 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡f
0
0 0 0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 y 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=cf
0
¡c
0
f, y =
gf
0
¡g
0
f
j
.
A straightforward computation reveals that B+tX commutes with C+tY for all t2F.
Moreover for t6=0 at least one of B+tX and C+tY has more than one point in the its
spectrum. So, by Lemma 1.9., the triple (A;B +tX;C +tY) belongs to G(3;7) for all
t2F, t6=0.
If f =f
0
=0. If b=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. Now we may assume that b6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 b 0 0 0
0 0 0 0 b 0 0
0 0 0 0 0 b 0
0 0 0 0 0 x 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡b
0
0 0 0 0 0 0
0 ¡b
0
0 0 0 0 0
0 0 ¡b
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 y 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
38
where x=cb
0
¡c
0
b, y =
gb
0
¡g
0
b
j
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(3) Assume that j
0
6= 0, and d
0
= 0. Without loss of generality we may assume
thatj
0
=1. Sowemayassumethatj =0. The commutativerelationof B andC implies
that
bf
0
=b
0
f, bh
0
=b
0
h, hf
0
=h
0
f +k, b
0
f =bf
0
+l.
So we have l =0. If d=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B +tZ;C) for t 2F, t 6= 0. Now we may assume that d 6= 0. Now we will
consider 2 cases.
(3-a) Assume that h
0
6=0. Then we can ¯nd X and Y as follows.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
h 0 0 0 0 0 0
0 h 0 0 0 0 0
0 0 h 0 0 0 0
0 0 x 0 0 y z
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
h
0
0 0 0 0 0 0
0 h
0
0 0 0 0 0
0 0 h
0
0 0 0 0
0 0 w 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
39
where x=ih
0
¡i
0
h, y =hc
0
¡h
0
c, z =¡dh
0
, w =
hc
0
¡h
0
c
d
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0thematrixB+tX hasmorethanonepointintheitsspectrum.
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
(3-b) Assume that h
0
=0. If at least one of f and f
0
is nonzero. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
f 0 0 0 0 0 0
0 f 0 0 0 0 0
0 0 f 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 x 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 ¡f
0
0 0 0
0 0 0 0 ¡f
0
0 0
0 0 0 0 0 ¡f
0
0
0 0 0 0 0 y 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=gf
0
¡g
0
f, y =
cf
0
¡c
0
f
d
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6=0 at least one of B+tX and C+tY has more than one point in
the its spectrum. So, by Lemma 1.9., the triple (A;B +tX;C +tY) belongs to G(3;7)
for all t2F, t6=0. If f =f
0
=0. If b=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. Now we may assume that b6=0. Let
40
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
b 0 0 0 0 0 0
0 b 0 0 0 0 0
0 0 b 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 x 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 ¡b
0
0 0 0
0 0 0 0 ¡b
0
0 0
0 0 0 0 0 ¡b
0
0
0 0 0 0 0 y 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=gb
0
¡g
0
b, y =
cb
0
¡c
0
b
d
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(4) Assume that j
0
6= 0, and d
0
6= 0. Without loss of generality we may assume
thatj
0
=1. Sowemayassumethatj =0. The commutativerelationof B andC implies
that
bf
0
= b
0
f +kd
0
; (3.1)
bh
0
= b
0
h+d
0
l; (3.2)
hf
0
= h
0
f +k; (3.3)
fb
0
= f
0
b+l: (3.4)
By the equation (3.1) and (3.4) we have
l+kd
0
=0: (3.5)
Now we will consider 2 cases.
(4-a) Assume that l 6= 0. Then due to equation (3.5) we have k 6= 0. If d = 0,
then we could introduce the matrix
41
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. Now we may assume that d6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
x 0 0 y 0 0 0
0 x 0 0 y 0 0
0 0 x 0 0 y 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 z 0 0 w 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ® 0 0 0
0 0 0 0 ® 0 0
0 0 0 0 0 ® 0
0 0 0 ¯ 0 0 0
0 0 0 0 ¯ 0 0
0 0 0 0 0 ¯ 0
0 0 0 0 0 ° 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=h¡
b
d
0
, ¯ =¡h
0
+
b
0
d
0
, y =b¡d
0
h, ®=b
0
¡d
0
h
0
, z =¡g¯¡g
0
x,
w =g®¡g
0
y, and ° =
¡c¯¡c
0
x+®i¡yi
0
+wd
0
d
.
Then a direct computation reveals that B+tX commutes with C +tY for all t2F. If
x=¯ =0,thenb=d
0
handb
0
=d
0
h
0
. Usingtheequation(3.2),wehaved
0
l =bh
0
¡b
0
h=
(d
0
h)h
0
¡(d
0
h
0
)h=0. Since d
0
l6=0, it is a contradiction. Therefore at least one of x and
¯ is nonzero. Thus for t6=0 at least one of B+tX and C+tY has more than one point
in the its spectrum. So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7)
for all t2F, t6=0.
(4-b) Assume that l = 0. Then due to equation (3.5) we have k = 0. Then B
and C look like
42
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b c d
0 0 0 0 0 b 0
0 0 0 0 0 0 0
0 f g 0 h i 0
0 0 f 0 0 h 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b
0
c
0
d
0
0 0 0 0 0 b
0
0
0 0 0 0 0 0 0
0 f
0
g
0
0 h
0
i
0
1
0 0 f
0
0 0 h
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In this case we may assume that d6= 0, say d = 1. Then we may assume that d
0
= 0. If
at least one of h
0
and h is nonzero. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 h 0 0 0
0 0 0 0 h 0 0
0 0 0 0 0 h 0
h 0 0 0 0 0 0
0 h 0 0 0 0 0
0 0 h 0 0 0 0
0 0 x 0 0 y ¡h
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 z 0 0 0
0 0 0 0 z 0 0
0 0 0 0 0 z 0
z 0 0 0 0 0 0
0 z 0 0 0 0 0
0 0 z 0 0 0 0
0 0 y 0 0 x ¡h
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=ih
0
¡i
0
h, y =gh
0
¡ch
0
¡g
0
h+c
0
h, and z =h
0
.
Then a direct computation reveals that B+tX commutes with C+tY. Moreover for
t6= 0 at least one of B +tX and C +tY has more than one point in the its spectrum.
So, by Lemma 1.9., the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0.
If h=h
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 b 0 0 0
0 0 0 0 b 0 0
0 0 0 0 0 b 0
f 0 0 0 0 0 0
0 f 0 0 0 0 0
0 0 f 0 0 0 0
0 0 x 0 0 0 ¡f
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 y 0 0 0
0 0 0 0 y 0 0
0 0 0 0 0 y 0
z 0 0 0 0 0 0
0 z 0 0 0 0 0
0 0 z 0 0 0 0
0 0 0 0 0 w ¡b
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
43
where x = if
0
¡i
0
f, y = b
0
, z = f
0
, and w = b
0
i¡bi
0
, if at least one of b and f
0
is not
zero. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 ¡i 0 0 c 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
¡1 0 0 0 0 0 0
0 ¡1 0 0 0 0 0
0 0 ¡1 0 0 0 0
0 0 c 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
if b=f
0
=0.
In both cases a straightforward computation reveals that B+tX commutes with C+tY.
In the ¯rst case for t6=0 at least one of B+tX and C+tY has more than one point in
theitsspectrum. So, byLemma1.9., thetriple(A;B+tX;C+tY)belongstoG(3;7)for
all t2F, t6=0. In the second case B+tX has more than one point in the its spectrum.
So the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
44
Chapter 4
3+2+2 case
In this chapter we give the result that a linear space of nilpotent commuting matrices
of size 7£7 having a matrix of maximal rank with one Jordan block of order 3 and two
Jordan blocks of order 2 can be perturbed by the generic triples.
Theorem 4.1. If in a 3-dimensional linear space L of nilpotent commuting matrices of
size 7£7 there is a matrix of maximal possible rank with one Jordan block of order 3 and
two Jordan blocks of order 2, then any basis of this space belongs to G(3;7).
Proof. We can write A2 L of maximal in some basis as
A=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If B 2 L, then the structure of B is well known. It is nilpotent and we may add to it a
polynomial in A, so that it looks like
45
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a b c d
0 0 0 0 a 0 c
0 0 0 0 0 0 0
0 e f 0 g h i
0 0 e 0 0 0 h
0 j k l m 0 n
0 0 j 0 l 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let C be a matrix in a L. Then C looks like
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a
0
b
0
c
0
d
0
0 0 0 0 a
0
0 c
0
0 0 0 0 0 0 0
0 e
0
f
0
0 g
0
h
0
i
0
0 0 e
0
0 0 0 h
0
0 j
0
k
0
l
0
m
0
0 n
0
0 0 j
0
0 l
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Since B is nilpotent, hl = 0. Now we may assume that at least one of h and l is 0. We
will consider 2 separate cases.
Case 1. Assume that there is a matrix B 2 L such that l 6= 0, so that for any
B2 L the corresponding entry h=0. In this case we may assume that l =1. So we may
assume that l
0
=0 and h
0
=0. Since the algebra generated by A and B contains AB, by
Corollary 1.13., we may assume that m=0 and m
0
=0. By the rank condition of B, we
have ce=0. So we will consider 3 subcases.
(i) Assume that there is a matrix B 2 L such that e 6= 0, so that for any B 2 L
46
the corresponding entry c=0. By assumption we may assume that c
0
=0. The commu-
tative relation of B and C implies that a
0
= e
0
= i
0
= 0. Since rank of B is less than or
equal to 4, we have the following equation:
det
2
6
6
6
6
6
6
6
6
6
4
0 0 a b d
e f 0 g i
0 e 0 0 0
j k 1 0 n
0 j 0 1 0
3
7
7
7
7
7
7
7
7
7
5
=e(e(d¡an)+aij)=0: (4.1)
We will consider 2 cases.
(i)-(1) Assume that a = 0. Then, by (4.1), we have d = 0. The commutativity
relation of B and C implies that
d
0
=b
0
=0, ij
0
=g
0
e, f
0
+j
0
n=jn
0
, g
0
=n
0
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 e 0 1 0 0 n
0 0 e 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(i)-(2) Assume that a6=0. The commutative relation of B and C implies that
af
0
+dj
0
=b
0
e+d
0
j, ag
0
=d
0
, ij
0
=g
0
e, f
0
+j
0
n=jn
0
,g
0
=n
0
+aj
0
.
If j
0
=0, then we could introduce the matrix
47
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 x 0 y
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 z 0 w 0 0
0 0 0 0 0 0 0
0 1 0 0 0 0 v
0 0 1 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
d¡a
2
j¡an
e
, y =
ai
e
, z =
ij
e
¡aj¡n, w =
i
e
, v =
i
e
¡a
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that j
0
6= 0, say j
0
= 1. Then
we may assume that j =0. So we have af
0
+d=b
0
e and f
0
+n=0. From (4.1) we have
d¡an=0. Therefore we have b
0
=0. Therefore C looks like
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 d
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f
0
0 g
0
0 0
0 0 0 0 0 0 0
0 1 k
0
0 0 0 n
0
0 0 1 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 a 0 0 d
0 0 0 0 a 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
48
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B2 L such that c6=0, so that for any B2 L the
corresponding entry e = 0. By assumption we may assume that e
0
= 0. The commuta-
tivity relation of B and C implies that c
0
= i
0
= j
0
= 0. Since rank of B is less than or
equal to 4, we have the following equation:
det
2
6
6
6
6
6
6
6
6
6
4
0 a b c d
0 0 a 0 c
f 0 g 0 i
k 1 0 0 n
j 0 1 0 0
3
7
7
7
7
7
7
7
7
7
5
=¡(c(f¡gj)+aij)=0: (4.2)
We will consider 2 cases.
(ii)-(1) Assume that j = 0. Then, by (4.2), we have f = 0. The commutativity
relation of B and C implies that
f
0
=k
0
=0, ag
0
=a
0
g+d
0
, cn
0
=a
0
i, g
0
=n
0
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 a 0
0 0 0 0 0 0 a
0 0 0 0 0 0 0
0 0 0 0 0 0 g
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
49
(ii)-(2) Assume that j6=0. The commutativity relation of B and C implies that
af
0
+ck
0
=a
0
f +d
0
j, ag
0
=a
0
g+d
0
, cn
0
=a
0
i, f
0
=jn
0
,ja
0
+g
0
=n
0
.
If a
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 1 0 0 x
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 y 0 z 0 0
0 0 0 0 0 0 0
0 0 w 0 0 0 v
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
ai
c
¡aj¡g, y =
ij
c
, z =
i
c
¡j, w =
f¡aj
2
¡gj
c
, v =
i
c
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that a
0
6= 0, say a
0
= 1. Then
we may assume that a=0. So we have ck
0
=f +d
0
j and d
0
+g =0. From (4.2) we have
f¡gj =0. Therefore we have k
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 f
0 0 0 0 0 0 0
0 0 0 0 0 j 0
0 0 0 0 0 0 j
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii) Assume that there is a matrix B 2 L such that c and e are zero. If at least
50
one of c
0
and e
0
is nonzero, then we can change a role of B and C. So we are done by (i)
and (ii). Thus we may assume that c
0
= e
0
= 0. Since rank of B is less than or equal to
4, we have the following condition:
det
2
6
6
6
6
6
6
6
6
6
4
0 0 a b d
0 0 0 a 0
0 f 0 g i
j k 1 0 n
0 j 0 1 0
3
7
7
7
7
7
7
7
7
7
5
=a
2
ij
2
=0.
So at least one of a, i and j is zero. Now we will consider 7 possibilities.
(iii)-(1) Assume that there is a matrix B2 L such that i6=0 and j6=0, so that for
any B 2 L the corresponding entry a = 0. By assumption we may assume that a
0
= 0.
The commutativity relation of B and C implies that
d
0
=i
0
=j
0
=0, g
0
=n
0
, f
0
=jn
0
.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 j 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
51
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 0 0
0 g
0
0 0 ¡b
0
0 0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
A direct computation reveals that B+tX commutes with C for all t2F. Moreover for
t6=0 B+tX has more than one point in the its spectrum. So, by Lemma 1.9., the triple
(A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(2) Assume that there is a matrix B 2 L such that a 6= 0 and j 6= 0, so
that for any B 2 L the corresponding entry i = 0. By assumption we may assume that
i
0
=0. The commutativity relation of B and C implies that
dj
0
+af
0
=d
0
j+a
0
f, ag
0
=a
0
g+d
0
, jn
0
=f
0
+j
0
n, aj
0
+n
0
=g
0
+a
0
j.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 ¡a ¡b
0 0 0 0 0 0 ¡a
0 0 0 0 0 0 0
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 ¡1 0
0 0 0 0 0 0 ¡1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 j
0
k
0
0 0 0 ¡aj
0
0 0 j
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡g¡aj.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii)-(3) Assume that there is a matrix B 2 L such that a 6= 0 and i 6= 0, so
52
that for any B 2 L the corresponding entry j = 0. By assumption we may assume that
j
0
=0. The commutativity relation of B and C implies that
a
0
=f
0
=i
0
=0, g
0
=n
0
, ag
0
=d
0
.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 a
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 ¡d
0
0
0 g
0
0 0 ¡b
0
0 ¡d
0
0 0 g
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(4) Assume that there is a matrix B 2 L such that j 6= 0. Then at least
one of a and i is zero. If only one of a and i is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that j6=0, so that for any B2 L the
corresponding entry a = i = 0. So we may assume that a
0
= i
0
= 0. The commutative
relation of B and C implies that
53
d
0
=0, g
0
=n
0
, dj
0
=0, and f
0
+nj
0
=n
0
j.
If j
0
= 0. Now we may assume that g
0
6= 0. Indeed, if g
0
= 0, then we could introduce
the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 j 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 0 0
0 g
0
0 0 ¡b
0
0 0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Ifj
0
6=0,sayj
0
=1. Thenwemayassumethatj =0. Bythecommutativecondition,
we have d=0. Let
54
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 ¡f
0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 ¡b
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii)-(5) Assume that there is a matrix B 2 L such that i 6= 0. Then at least
one of a and j is zero. If only one of a and j is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that i6= 0, so that for any B2 L the
corresponding entry a=j =0. So we may assume that a
0
=j
0
=0. The commutativity
relation of B and C implies that
g
0
=n
0
, d
0
=i
0
=f
0
=0.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
55
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 0 0
0 g
0
0 0 ¡b
0
0 0
0 0 g
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(6) Assume that there is a matrix B 2 L such that a 6= 0. Then at least
one of i and j is zero. If only one of i and j is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that a6=0, so that for any B2 L the
corresponding entry i = j = 0. So we may assume that i
0
= j
0
= 0. The commutativity
relation of B and C implies that
f
0
=0, g
0
=n
0
, a
0
f =0, ag
0
=a
0
g+d
0
.
If a
0
= 0. Now we may assume that g
0
6= 0. Indeed, if g
0
= 0, then we could introduce
the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 a
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
56
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 ¡d
0
0
0 g
0
0 0 ¡b
0
0 ¡d
0
0 0 g
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Ifa
0
6=0,saya
0
=1. Thenwemayassumethata=0. Bythecommutativecondition,
we have f =0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡d
0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 ¡k
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii)-(7) Assume that there is a matrix B 2 L such that a, i, and j are zero. If
at least one of a
0
, i
0
andj
0
is nonzero, then we can change the role of B and C. So we are
done by above Cases. Thus we may assume that a
0
= i
0
= j
0
= 0. The commutativity
relation of B and C implies that d
0
=f
0
=0, g
0
=n
0
.
If g
0
=0, then we could introduce the matrix
57
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 0 0
0 g
0
0 0 ¡b
0
0 0
0 0 g
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 2. Assume that there is a matrix B 2 L such that h 6= 0, so that for any
B 2 L the corresponding entry l = 0. So we may assume that h = 1. Thus we may
assume that h
0
=0 and l
0
=0. Since the algebra generated by A and B contains AB, by
Corollary 1.13., we may assume that i = 0 and i
0
= 0. By the rank condition of B, we
have aj =0. Thus at least one of a and j is zero. Now we will consider 3 subcases.
(i) Assume that there is a matrix B 2 L such that j 6= 0, so that for any B 2 L
58
the corresponding entry a=0. By assumption we may assume that a
0
=0. The commu-
tativity relation of B and C implies that c
0
= m
0
= j
0
= 0. Since rank of B is less than
or equal to 4, we have the following condition:
det
2
6
6
6
6
6
6
6
6
6
4
0 0 b c d
e f g 1 0
0 e 0 0 1
j k m 0 n
0 j 0 0 0
3
7
7
7
7
7
7
7
7
7
5
=j(j(b¡gc)+ecm)=0: (4.3)
We will consider 2 cases.
(i)-(1) Assume that c = 0. Then, by (4.3), we have b = 0. The commutativity
relation of B and C implies that
d
0
=b
0
=0, me
0
=n
0
j, k
0
+e
0
g =eg
0
, g
0
=n
0
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 e 0 0 g 1 0
0 0 e 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(i)-(2) Assume that c6=0. The commutativity relation of B and C implies that
be
0
+ck
0
=b
0
e+d
0
j, cn
0
=b
0
, me
0
=n
0
j, k
0
+e
0
g =eg
0
,n
0
=g
0
+ce
0
.
If e
0
=0, then we could introduce the matrix
59
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 x 0 y
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 0 0 z 0 0
0 0 1 0 0 0 0
0 0 w 0 0 0 v
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
mc
j
, y =
b¡ec
2
¡gc
j
, z =
m
j
¡c, w =
me
j
¡ec¡g, v =
m
j
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that e
0
6= 0, say e
0
= 1. Then
we may assume that e=0. So we have ck
0
+b=d
0
j and k
0
+g =0. From (4.3) we have
b¡gc=0. Therefore we have d
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 b c 0
0 0 0 0 0 0 c
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B+tX commutes with C for all t2F. Moreover
for t6= 0 B +tX has more than one point in the its spectrum. So, by Lemma 1.9., the
triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B2 L such that a6=0, so that for any B2 L the
60
corresponding entry j = 0. By assumption we may assume that j
0
= 0. The commuta-
tivity relation of B and C implies that a
0
=e
0
=m
0
=0. Since rank of B is less than or
equal to 4, we have the following equation:
det
2
6
6
6
6
6
6
6
6
6
4
0 a b c d
0 0 a 0 c
f 0 g 1 0
e 0 0 0 1
k 0 m 0 n
3
7
7
7
7
7
7
7
7
7
5
=a(a(k¡en)+ecm)=0: (4.4)
We will consider 2 cases.
(ii)-(1) Assume that e = 0. Then, by (4.4), we have k = 0. The commutativity
relation of B and C implies that f
0
=k
0
=0, ag
0
=c
0
m, cn
0
=b
0
+c
0
n, g
0
=n
0
.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 c 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 n 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(ii)-(2) Assume that e6=0. The commutativity relation of B and C implies that
af
0
+ck
0
=b
0
e+c
0
k, ag
0
=c
0
m, cn
0
=b
0
+c
0
n, k
0
=eg
0
,ec
0
+n
0
=g
0
.
If c
0
=0, then we could introduce the matrix
61
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 x 1 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 y 0 z 0 0
0 0 0 0 0 0 0
0 0 w 0 0 0 v
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
mc
a
¡ec¡n, y =
k¡e
2
c¡en
a
, z =
m
a
, w =
me
a
, v =
m
a
¡e
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that c
0
6= 0, say c
0
= 1. Then
we may assume that c=0. So we have af
0
=k+b
0
e and b
0
+n=0. From (4.4) we have
k¡en=0. Therefore we have f
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 e 0 0 0
0 0 0 0 e 0 0
0 0 0 0 k 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii) Assume that there is a matrix B 2 L such that a and j are zero. If at least
one of a
0
and j
0
is nonzero, then we can change the role of B and C. So we are done by
(i) and (ii). Thus we may assume that a
0
=j
0
=0. Since rank of B is less than or equal
to 4, we can ¯nd the following equation:
62
det
2
6
6
6
6
6
6
6
6
6
4
0 0 b c d
0 0 0 0 c
e f g 1 0
0 e 0 0 1
j k m 0 n
3
7
7
7
7
7
7
7
7
7
5
=c
2
e
2
m=0.
Thus at least one of c, e and m is zero. So there are 7 possibilities.
(iii)-(1) Assume that there is a matrix B 2 L such that e 6= 0 and m 6= 0, so
that for any B 2 L the corresponding entry c = 0. By assumption we may assume that
c
0
=0. The commutativity relation of B and C implies that
b
0
=e
0
=m
0
=0, g
0
=n
0
, k
0
=eg
0
.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 e 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 0 0 ¡d
0
0
0 g
0
0 0 0 0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
63
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(2) Assume that there is a matrix B 2 L such that c 6= 0 and e 6= 0, so
that for any B2 L the corresponding entry m=0. By assumption we may assume that
m
0
=0. The commutativity relation of B and C implies that
be
0
+ck
0
=b
0
e+c
0
k, cn
0
=c
0
n+b
0
, eg
0
=k
0
+e
0
g, ec
0
+n
0
=g
0
+e
0
c.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ¡c ¡d 0 0
0 0 0 0 ¡c 0 0
0 0 0 0 0 0 0
0 0 0 ¡1 0 0 0
0 0 0 0 ¡1 0 0
0 0 0 0 x 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 e
0
f
0
0 ¡e
0
c 0 0
0 0 e
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡n¡ce.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii)-(3) Assume that there is a matrix B 2 L such that c 6= 0 and m 6= 0, so
that for any B 2 L the corresponding entry e = 0. By assumption we may assume that
e
0
=0. The commutativity relation of B and C implies that
c
0
=k
0
=m
0
=0, g
0
=n
0
, cn
0
=b
0
.
If g
0
=0, then we could introduce the matrix
64
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 c 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 ¡d
0
0
0 g
0
0 0 ¡b
0
0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(4) Assume that there is a matrix B 2 L such that e 6= 0. Then at least
one of c and m is zero. If only one of c and m is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that e6=0, so that for any B2 L the
correspondingentryc=m=0. Sowemayassumethatc
0
=m
0
=0. Thecommutativity
relation of B and C implies that
b
0
=0, g
0
=n
0
, be
0
=0, and k
0
+ge
0
=g
0
e.
If e
0
= 0. Now we may assume that g
0
6= 0. Indeed, if g
0
= 0, then we could introduce
the matrix
65
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 e 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 0 0 ¡d
0
0
0 g
0
0 0 0 0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0. If e
0
6=0, say e
0
=1.
Then we may assume that e=0. By the commutative condition, we have b=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 ¡k
0
1 0
0 0 0 0 0 0 1
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 ¡d
0
0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
66
(iii)-(5) Assume that there is a matrix B 2 L such that m 6= 0. Then at least
one of c and e is zero. If only one of c and e is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that m6=0, so that for any B2 L the
corresponding entry c = e = 0. So we may assume that c
0
= e
0
= 0. The commutative
relation of B and C implies that
g
0
=n
0
, b
0
=k
0
=m
0
=0.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 0 0 ¡d
0
0
0 g
0
0 0 0 0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii)-(6) Assume that there is a matrix B 2 L such that c 6= 0. Then at least
67
one of e and m is zero. If only one of e and m is zero, we are done by above subcases. So
we may assume that there is a matrix B2 L such that c6= 0, so that for any B2 L the
correspondingentrye=m=0. Sowemayassumethate
0
=m
0
=0. Thecommutativity
relation of B and C implies that
k
0
=0, g
0
=n
0
, c
0
k =0, cn
0
=c
0
n+b
0
.
If c
0
= 0. Now we may assume that g
0
6= 0. Indeed, if g
0
= 0, then we could introduce
the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 c 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 ¡b
0
0 ¡d
0
0
0 g
0
0 0 ¡b
0
0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B+tX commutes with C for all t2F. Moreover
for t6= 0 B +tX has more than one point in the its spectrum. So, by Lemma 1.9., the
triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Ifc
0
6=0, sayc
0
=1. Thenwemayassumethatc=0. Bythecommutativecondition,
we have k =0. Let
68
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 ¡b
0
0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f
0
0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(iii)-(7) Assume that there is a matrix B 2 L such that c, e, and m are zero. If
at least one of c
0
, e
0
, and m
0
is nonzero, then we can change the role of B and C. So we
aredonebyaboveCases. Thuswemayassumethatc
0
=e
0
=m
0
=0. Thecommutativity
relation of B and C implies that
b
0
=k
0
=0, g
0
=n
0
.
If g
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that g
0
6=0. Let
69
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
g
0
0 0 0 0 ¡d
0
0
0 g
0
0 0 0 0 ¡d
0
0 0 g
0
0 0 0 0
0 ¡f
0
0 0 0 0 0
0 0 ¡f
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
Case 3. Assume that there is a matrix B 2 L such that h and l are zero. If at
least one of h
0
and l
0
is nonzero, then we can change the role of B and C. So we are
done by Case 1. and Case 2. Thus we may assume that h
0
=l
0
= 0. In this case we will
consider 16 subcases.
(i) Assume that there is a matrix B 2 L such that a 6= 0, c 6= 0, e 6= 0, and
j6=0. Now we may assume that a=1. Since the algebra generated by A andB contains
AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. Moreover we may assume
that a
0
=0. The commutativity relation of B and C implies that
f
0
+ck
0
=d
0
j, g
0
+cm
0
=0, i
0
+cn
0
=0, g
0
e+i
0
j =0, m
0
e+n
0
j =0,
and c
0
=e
0
=j
0
=0. If d
0
=0 and m
0
6=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 j 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
70
If d
0
6=0 and m
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 ¡c 0
ce
j
0 0 0 0 0 0 0
0 0 0 0 1 0 ¡
e
j
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If d
0
=0 and m
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 j 0 ¡c 0
ce
j
0 0 0 0 0 0 0
0 0 0 0 1 0 ¡
e
j
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In each case the matrix Z clearly commutes with both A and B, so that it su±ces to
prove our case for all triples (A;B;C +tZ) for t 2 F, t 6= 0. So we may assume that
d
0
6= 0 , say d
0
= 1. We can also assume that m
0
6= 0. So we may assume that d = 0. So
B and C look like
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 1 0 c 0
0 0 0 0 1 0 c
0 0 0 0 0 0 0
0 e f 0 g 0 i
0 0 e 0 0 0 0
0 j k 0 m 0 n
0 0 j 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f
0
0 g
0
0 i
0
0 0 0 0 0 0 0
0 0 k
0
0 m
0
0 n
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let
71
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
m
0
0 0 0 0 0 0
0 m
0
0 0 0 0 0
0 0 m
0
0 0 0 0
0 ¡k
0
0 0 0 0 0
0 0 ¡k
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
g
0
k
0
¡f
0
m
0
j
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B 2 L such that a 6= 0, e 6= 0, j 6= 0, and
c = 0. If B
4
6= 0 then B has only one Jordan block and consequently we are done by
Theorem 1.7. So we may assume that B
4
=0. So we have ae=¡cj. So we have a rela-
tionae=0. Sincea6=0ande6=0,itisacontradiction. Thereforethiscasecannotoccur.
(iii) Assume that there is a matrix B 2 L such that a 6= 0, c 6= 0, e 6= 0, and
j =0. By same reason in (ii), it can not occur.
(iv) Assume that there is a matrix B 2 L such that a 6= 0, c 6= 0, j 6= 0, and
e=0. It also can not occur.
(v) Assume that there is a matrix B 2 L such that c 6= 0, e 6= 0, j 6= 0, and
a=0. It also can not occur.
(vi) Assume that there is a matrix B 2 L such that a 6= 0, e 6= 0, c = 0, and
j =0. It also can not occur.
72
(vii) Assume that there is a matrix B 2 L such that a 6= 0, j 6= 0, c = 0, and
e=0. Now we may assume that a=1. Since the algebra generated by A andB contains
AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. Moreover we may assume
that a
0
=0. The commutativity relation of B and C implies that
c
0
=e
0
=g
0
=i
0
=j
0
=n
0
=0, f
0
=d
0
j.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
j 0 0 0 0 0 0
0 j 0 0 0 0 0
0 0 j 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡jd
0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡k
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(viii) Assume that there is a matrix B 2 L such that a 6= 0, c 6= 0, e = 0, and
j =0. Now we may assume that a=1. Since the algebra generated by A andB contains
AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. Moreover we may assume
that a
0
=0. The commutativity relation of B and C implies that
f
0
+ck
0
=c
0
k, g
0
+cm
0
=c
0
m, i
0
+cn
0
=c
0
n, e
0
=j
0
=0.
If c
0
=0, then we could introduce the matrix
73
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 k 0 m 0 n
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples(A;B;C+tZ)fort2F,t6=0. Sowemayassumethatc
0
6=0, sayc
0
=1. Wemay
assume that c = 0. Since the algebra generated by A and C contains AC, by Corollary
1.13., we may assume that d
0
=0 and d=0. Let
®=n
0
i¡mi¡n
2
+ng+f, ¯ =¡m
2
+k
0
¡m
0
n+mn
0
+m
0
g, x=(n¡g)®,
y =¡m®+i¯, z =(n¡g)¯.
If ®6=0, then we can ¯nd X and Y as follows.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡® 0 0 0 0 0 0
0 ¡® 0 0 0 0 0
0 0 ¡® 0 0 0 0
0 x 0 0 0 0 0
0 0 x 0 0 0 0
0 y 0 0 0 0 0
0 0 y 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 y 0 0 0 ® 0
0 0 y 0 0 0 ®
0 z 0 ¯ 0 0 0
0 0 z 0 ¯ 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
If ®=0. Let
74
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 i 0 0 0 0 0
0 0 i 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 i 0 0 0 0 0
0 0 i 0 0 0 0
0 n¡g 0 1 0 0 0
0 0 n¡g 0 1 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
In this case we may assume that i 6= 0. Indeed, if i = 0, then we could introduce the
matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6= 0. Then a straightforward computation reveals that
B +tX commutes with C +tY for all t 2F. Moreover the rank of B +tX is greater
than or equal to 5. We are done by Theorem 1.7. and Theorem 1.8.
(ix) Assume that there is a matrix B 2 L such that e 6= 0, j 6= 0, a = 0, and
c=0. Now we may assume that e=1. Since the algebra generated by A and B contains
AB, by Corollary1.13., wemayassumethat f =0 andf
0
=0. Moreoverwe mayassume
that e
0
=0. The commutativity relation of B and C implies that
b
0
+jd
0
=j
0
d, g
0
+ji
0
=j
0
i, m
0
+jn
0
=j
0
n, a
0
=c
0
=0.
If j
0
=0, then we could introduce the matrix
75
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 d 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 i 0 0
0 0 0 0 0 0 0
0 1 0 0 n 0 0
0 0 1 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t 2 F, t 6= 0. Thus we may assume that j
0
6= 0, say j
0
= 1.
We may assume that j = 0. Since the algebra generated by A and C contains AC, by
Corollary 1.13., we may assume that k
0
=0 and k =0. Let
®=¡b+n
2
¡ng+im¡n
0
m, ¯ =i
2
¡d
0
+i
0
n¡i
0
g¡in
0
, x=(n¡g)®, y =m¯¡i®,
z =(n¡g)¯.
If ®6=0, then we can ¯nd X and Y as follows.
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡® 0 0 x 0 y 0
0 ¡® 0 0 x 0 y
0 0 ¡® 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 y 0 z 0
0 0 0 0 y 0 z
0 0 0 0 0 0 0
0 0 0 0 0 ¯ 0
0 0 0 0 0 0 ¯
0 0 0 ® 0 0 0
0 0 0 0 ® 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
If ®=0. Let
76
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 m 0
0 0 0 0 0 0 m
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 m 0 w 0
0 0 0 0 m 0 w
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where w =n¡g.
In this case we may assume that m6= 0. Indeed, if m = 0, then we could introduce the
matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6= 0. Then a straightforward computation reveals that
B +tX commutes with C +tY for all t 2F. Moreover the rank of B +tX is greater
than or equal to 5. We are done by Theorem 1.7 and Theorem 1.8.
(x) Assume that there is a matrix B 2 L such that e 6= 0, c 6= 0, a = 0, and
j =0. Now we may assume that c=1. Since the algebra generated by A and B contains
AB, by Corollary 1.13., we may assume that d=0 and d
0
=0. Moreover we may assume
that c
0
=0. The commutativity relation of B and C implies that
a
0
=e
0
=h
0
=j
0
=m
0
=n
0
=0, k
0
=b
0
e.
Let
77
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
e 0 0 0 0 0 0
0 e 0 0 0 0 0
0 0 e 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 ¡f
0
0 0
0 0 0 0 0 0 0
0 0 0 0 ¡eb
0
0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straight forward computation reveals that B+tX commutes with C+tY for all
t2F. Moreover for t6= 0 B +tX has more than one point in the its spectrum. So, by
Corollary 1.13., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(xi) Assume that there is a matrix B 2 L such that c 6= 0, j 6= 0, a = 0, and
e=0. It can not occur.
(xii) Assume that there is a matrix B 2 L such that a 6= 0, c = 0, e = 0, and
j =0. Now we may assume that a=1. Since the algebra generated by A andB contains
AB, we may assume that b = 0 and b
0
= 0. Moreover we may assume that a
0
= 0. The
commutativity relation of B and C implies that
f
0
=c
0
k,g
0
=c
0
m, i
0
=c
0
n, e
0
=j
0
=0.
If c
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 k 0 m 0 n
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C + tZ) for t 2 F, t 6= 0. So we may assume that c
0
6= 0, say c
0
= 1.
78
Since the algebra generated by A andC contains AC, by Corollary 1.13., we may assume
that d
0
= 0 and d = 0. Then B and C are exactly same matrices in the subcase (viii).
Therefore we are done by (viii).
(xiii) Assume that there is a matrix B 2 L such that e 6= 0, a = 0, c = 0, and
j =0. Now we may assume that e=1. Since the algebra generated by A and B contains
AB, by Corollary1.13., wemayassumethat f =0 andf
0
=0. Moreoverwe mayassume
that e
0
=0. The commutativity relation of B and C implies that
b
0
=dj
0
, g
0
=hj
0
, m
0
=nj
0
, a
0
=c
0
=0.
If j
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 d 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 i 0 0
0 0 0 0 0 0 0
0 1 0 0 n 0 0
0 0 1 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples(A;B;C+tZ)fort2F,t6=0. Thuswemayassumethatj
0
6=0, sayj
0
=1. Since
the algebra generated by A and C contains AC, by Corollary 1.13., we may assume that
k
0
= 0 and k = 0. Then the matrices B and C are exactly same matrices in the subcase
(ix). Therefore we are done by (ix).
(xiv) Assume that there is a matrix B 2 L such that j 6= 0, a = 0, c = 0, and
e=0. Now we may assume that j =1. Since the algebra generated by A and B contains
AB, by Corollary 1.13., we may assume that k =0 and k
0
=0. Moreover we may assume
that j
0
=0. The commutativity relation of B and C implies that
d
0
=be
0
, i
0
=ge
0
, n
0
=me
0
, a
0
=c
0
=0.
If e
0
=0, then we could introduce the matrix
79
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 b
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 0 0 0 0 g
0 0 1 0 0 0 0
0 0 0 0 0 0 m
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples(A;B;C+tZ)fort2F, t6=0. Thuswemayassumethate
0
6=0, saye
0
=1. Since
the algebra generated by A and C contains AC, by Corollary 1.13., we may assume that
f
0
= 0 and f = 0. If we change the role of B and C, then this case is same as the case
(xiii). So we are done by (xiii).
(xv) Assume that there is a matrix B 2 L such that c 6= 0, a = 0, e = 0, and
j =0. Now we may assume that c=1. Since the algebra generated by A and B contains
AB, by Corollary 1.13., we may assume that d=0 and d
0
=0. Moreover we may assume
that c
0
=0. The commutativity relation of B and C implies that
k
0
=a
0
f,m
0
=a
0
g, n
0
=a
0
i, e
0
=j
0
=0.
If a
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f 0 g 0 i
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that a
0
6= 0, say a
0
= 1. Since
the algebra generated by A and C contains AC, by Corollary 1.13., we may assume that
80
b
0
= 0 and b = 0. If we change the role of B and C, then this case is same as the case
(xii). So we are done by (xii).
(xvi) Assume that there is a matrix B 2 L such that a = c = e = j = 0. If at
least one of a
0
, c
0
, e
0
, and j
0
is nonzero, then we can change the role of B and C. So we
are done by above cases. Thus we may assume that a
0
=c
0
= e
0
=j
0
= 0. If i = 0, then
we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. So we may assume that i6=0. So we may assume
that i
0
=0. Moreover we may assume that g
0
6=0 and n
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
x 0 0 y 0 z 0
0 x 0 0 y 0 z
0 0 x 0 0 0 0
0 v 0 0 0 0 0
0 0 v 0 0 0 0
0 w 0 0 0 0 0
0 0 w 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x = g
0
n
0
, y = m
0
d
0
¡n
0
b
0
, z = ¡d
0
g
0
, v = ¡n
0
f
0
, and w = m
0
f
0
¡k
0
g
0
. Then
a straightforward computation reveals that B + tX commutes with C for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
81
Chapter 5
3+2+1+1 case
In this chapter we give the result that a linear space of nilpotent commuting matrices
of size 7£7 having a matrix of maximal rank with one Jordan block of order 3 and one
Jordan block of order 2 can be perturbed by the generic triples.
Theorem 5.1. If in a 3-dimensional linear space L of nilpotent commuting matrices of
size 7£7 there is a matrix of maximal possible rank with one Jordan block of order 3 and
one Jordan blocks of order 2, then any basis of this space belongs to G(3;7).
Proof. We can write A2 L of maximal in some basis as
A=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 1 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
If B 2 L, then the structure of B is well known. It is nilpotent and we may add to it a
polynomial in A, so that it looks like
82
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a b c d
0 0 0 0 a 0 0
0 0 0 0 0 0 0
0 e f 0 g h i
0 0 e 0 0 0 0
0 0 j 0 k x y
0 0 m 0 n z w
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Since B is nilpotent we may assume that
2
4
x y
z w
3
5
=
2
4
0 l
0 0
3
5
.
Therefore B looks like
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a b c d
0 0 0 0 a 0 0
0 0 0 0 0 0 0
0 e f 0 g h i
0 0 e 0 0 0 0
0 0 j 0 k 0 l
0 0 m 0 n 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Let C be a matrix in a L. Then C looks like
C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 a
0
b
0
c
0
d
0
0 0 0 0 a
0
0 0
0 0 0 0 0 0 0
0 e
0
f
0
0 g
0
h
0
i
0
0 0 e
0
0 0 0 0
0 0 j
0
0 k
0
0 l
0
0 0 m
0
0 n
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Since the rank of B is less than or equal to 3, we have the following condition:
83
det
2
6
6
6
6
6
6
4
0 0 a b
0 0 0 a
e f 0 g
0 e 0 0
3
7
7
7
7
7
7
5
=ea=0.
Thus at least one of a and e is zero. Now we will consider 3 cases.
Case 1. Assume that there is a matrix B 2 L such that a 6= 0, so that for any
B 2 L the corresponding entry e = 0. By assumption we may assume that e
0
= 0. We
may assume that a = 1. So we may assume that a
0
= 0. Since the algebra generated by
A and B contains AB, by Corollary 1.13., we may assume that b=0 and b
0
=0. By the
commutativity condition of B and C we have h
0
=0. Since B
3
=0,
hj+im+clm=hk+in+cln=hl =hlm=hln=0.
So in this case at least one of h and l is zero. Now we will consider 3 possibilities.
(i) Assume that there is a matrix B 2 L such that h 6= 0, so that for any B 2 L
the corresponding entry l = 0. By assumption we may assume that l
0
= 0. Since
rank(B)·3, we have hj =hm=0. So we have j =m=0. The commutativity relation
of B and C implies that
i
0
=0, g
0
+ck
0
+dn
0
=c
0
k+d
0
n, f
0
+cj
0
+dm
0
=0, hk
0
+in
0
=0, hj
0
+im
0
=0.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 c
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverfort6=0rank(B+tX)¸4. WearedonebyTheorem2.1., 3.1., and4.1.
84
(ii) Assume that there is a matrix B 2 L such that l 6= 0, so that for any B 2 L
the corresponding entry h = 0. By assumption we may assume that h
0
= 0. Since
rank(B)·3, we have lm=0. So we have m=0. The commutativity condition implies
that m
0
=0.
(ii)-(1) If there is a matrix B2 L such that n6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 n
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 x 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 y 0 z 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
dn
0
¡d
0
n
l
, y =xj, z =xk.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii)-(2) If there is a matrix B 2 L such that n = 0. Then, by commutativity
condition, we have n
0
=0. If i
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0
1
l
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0
j
l
0
k
l
0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
85
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that i
0
6= 0, say i
0
= 1. Then
we may assume that i=0. By the rank condition, lf =0. So we have f =0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t 6= 0 rank(B +tX) ¸ 4. So, by Theorem 2.1., 3.1., and 4.1., the triple
(A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(iii) Assume that there is a matrix B 2 L such that h = l = 0. If l
0
is not zero,
then we can change the role of B and C. So we are done by above cases. Thus we may
assume that l
0
=0. Since B
3
=0, in=im=0 . So we will consider 2 possibilities.
(iii)-(1) Assume that there is a matrix B 2 L such that i 6= 0. Then we have
m=n=0. Since rank(B)·3, ij =0. So we have j =0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 c
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6=0 rank(B+tX)¸4. Thus, by Theorem 2.1., 3.1., and 4.1., the
triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
86
(iii)-(2) Assume that there is a matrix B 2 L such that i = 0. Then the com-
mutativity condition implies that i
0
= 0. The commutative relation of B and C implies
that
f
0
+cj
0
+dm
0
=c
0
j+d
0
m, g
0
+ck
0
+dn
0
=c
0
k+d
0
n.
If d
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 m 0 n 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that d
0
6= 0, say d
0
= 1. Then
we may assume that d=0. Now the commutativity relation of B and C implies that
f
0
+cj
0
=c
0
j+m, g
0
+ck
0
=c
0
k+n.
If n
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that n
0
6= 0. Finally we also
assume that (j
0
;k
0
) and (m
0
;n
0
) are linearly dependent. If not for each t2F, t6= 0 we
could introduce the matrix
87
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0
j
0
c
t
0
k
0
c
t
0 0
0 0 0 0 0 0 0
0 0 ¡
j
0
t
0 ¡
k
0
t
0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡1 0 0 0 0 0 0
0 ¡1 0 0 0 0 0
0 0 ¡1 0 0 0 0
0
m
0
n
0
0 0 0 c
g
0
n
0
0 0
m
0
n
0
0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 c
0
1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 2. Assume that there is a matrix B 2 L such that e 6= 0, so that for any
B 2 L the corresponding entry a = 0. We may assume that e = 1. So we may assume
that e
0
=0 and a
0
=0. Since the algebra generated by A and B contains AB, by Corol-
lary 1.13., we may assume that f = 0 and f
0
= 0. By the commutative condition of B
and C we have n
0
=0. Since B
3
=0,
ck+dn+clm=hk+in+hlm=ln=cln=hln=0.
Now we will consider 3 possibilities.
(i) Assume that there is a matrix B 2 L such that n 6= 0, so that for any B 2 L
88
the corresponding entry l = 0. By assumption we may assume that l
0
= 0. Since
rank(B)·3, we have dn=cn=0. So we have d=c=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 m
0
0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Sincefort6=0rank(B+tX)¸4, thetriple(A;B+tX;C+tY)belongstoG(3;7)
for all t2F, t6=0.
(ii) Assume that there is a matrix B 2 L such that l 6= 0, so that for any B 2 L
the corresponding entry n = 0. By assumption we may assume that n
0
= 0. Since
rank(B)·3, we havelc=0. So we havec=0. The commutative condition implies that
c
0
=0.
(ii)-(1) If there is a matrix B2 L such that h6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 h 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 x 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 y 0 0
0 0 0 0 0 0 0
0 0 0 0 0 h
0
0
0 0 z 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=zd, y =zi, z =
jh
0
¡j
0
h
l
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. Since for t 6= 0 B +tX has more than one point in the its spectrum, the triple
89
(A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii)-(2) If there is a matrix B 2 L such that h = 0. Then, by commutative con-
dition, we have h
0
= 0. In this case we may assume that k
0
6= 0, say k
0
= 1. Then we
may assume that k =0. By the rank condition, lb=0. So we have b=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Since for t 6= 0 rank(B +tX) ¸ 4, the triple (A;B +tX;C) belongs to G(3;7) for all
t2F, t6=0.
(iii) Assume that there is a matrix B 2 L such that l = n = 0. If l
0
is not zero,
then we can change the role of B and C. So we are done by above cases. Thus we may
assume that l
0
=0. Since B
3
=0, ck =hk =0.
(iii)-(1) Assume that there is a matrix B 2 L such that k 6= 0. Then we have
c=h=0. Since rank(B)·3, ij =0. So we have j =0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 m
0
0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
90
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Since for t6= 0 rank(B+tX)¸ 4, the triple (A;B+tX;C +tY) belongs to G(3;7) for
all t2F, t6=0.
(iii)-(2) Assume that there is a matrix B 2 L such that k = 0. So the commu-
tative condition implies that k
0
= 0. The commutative relation of B and C implies
that
b
0
+c
0
j+d
0
m=cj
0
+dm
0
, g
0
+h
0
j+i
0
m=hj
0
+im
0
.
If m
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 d 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 i 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C+tZ) for t2F, t6=0. So we may assume that m
0
6=0, say m
0
=1. Then
we may assume that m=0. Now the commutativity relation of B and C implies that
b
0
+c
0
j =cj
0
+d, g
0
+h
0
j =hj
0
+i.
If i
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
91
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6= 0. So we may assume that i
0
6= 0. Finally we also
assume that (c
0
;h
0
) and (d
0
;i
0
) are linearly dependent. If not for each t 2F, t 6= 0 we
could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0
c
0
j
t
¡
c
0
t
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0
h
0
j
t
¡
h
0
t
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡1 0 0
d
0
i
0
0 0 0
0 ¡1 0 0
d
0
i
0
0 0
0 0 ¡1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 j 0 0
0 0 0 0
g
0
i
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 j
0
0 0
0 0 0 0 1 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B +tX commutes with C +tY for all t 2 F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
Case 3. Assume that there is a matrix B 2 L such that a = e = 0. If one of
a
0
and e
0
is nonzero, then we can change the role of B and C. So we are done by above
Case 1. and Case 2. So we may assume that a
0
= e
0
= 0. If B
3
6= 0 then B has only
one Jordan block and consequently we are done by Theorem 1.7. So we may assume that
B
3
=0. So we have clm=cln=hlm=hln=0. We will consider 2 subcases.
92
(i) Assume that there is a matrix B 2 L such that l 6= 0. So we may assume
that l = 1 and l
0
= 0. Since B
3
= 0 and l 6= 0, we have cm = cn = hm = hn = 0. We
will consider 4 subcases.
(i)-(1) Assume that there is a matrix B 2 L such that c 6= 0 and h 6= 0, so that
for any B2 L the corresponding entry m=n=0. So we may assume that m
0
=n
0
=0.
The commutative relation of B and C implies that
c
0
=h
0
=j
0
=k
0
=0.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(i)-(2) Assume that there is a matrix B 2 L such that c = 0 and h 6= 0, so that
for any B2 L the corresponding entry m=n=0. So we may assume that m
0
=n
0
=0.
By commutative condition we have c
0
= 0. The commutativity relation of B and C
implies that
h
0
=j
0
=k
0
=0.
Let
93
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
(i)-(3) Assume that there is a matrix B 2 L such that c 6= 0 and h = 0, so that
for any B2 L the corresponding entry m=n=0. So we may assume that m
0
=n
0
=0.
By commutative condition we have h
0
= 0. The commutativity relation of B and C
implies that
c
0
=j
0
=k
0
=m
0
=n
0
=0.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Moreover for t6= 0 B+tX has more than one point in the its spectrum. So, by Lemma
1.9., the triple (A;B+tX;C) belongs to G(3;7) for all t2F, t6=0.
94
(i)-(4) Assume that there is a matrix B 2 L such that c = 0 and h = 0. The
commutativity relation of B and C implies that
c
0
=h
0
=m
0
=n
0
=d
0
m=d
0
n=i
0
m=i
0
n=0.
Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 x
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, where x=m or x=n.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
If at least one of m and n is nonzero, then for t6=0 B+tX has more than one point in
the its spectrum. So, by Lemma 1.9., the triple (A;B+tX;C) belongs to G(3;7) for all
t 2F, t 6= 0. Now we may assume that m = n = 0. In this case we may assume that
d
0
6=0, say d
0
=1. Then we may assume that d=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 i 0
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 ¡j
0
0 ¡k
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. Since for t 6= 0 B +tX has more than one point in the its spectrum, the triple
(A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii) Assume that there is a matrix B 2 L such that l = 0. If l
0
is not zero, then
95
we can change the role of B andC. So we are done by above cases. Thus we may assume
that l
0
=0. The commutative relation of B and C implies that
cj
0
+dm
0
=c
0
j+d
0
m, ck
0
+dn
0
=c
0
k+d
0
n, hj
0
+im
0
=h
0
j+i
0
m, hk
0
+in
0
=h
0
k+i
0
n.
(ii)-(1) Suppose that (m;n) and (m
0
;n
0
) are linearly independent. We may as-
sume that (m;n) = (1;0) and (m
0
;n
0
) = (0;1). By the rank condition of B, we have
k(ci¡dh)=0.
(ii)-(1)-(a) Assume that there is a matrix B2 L such that k =0. Then we have
d
0
=cj
0
¡c
0
j, d=¡ck
0
, i
0
=hj
0
¡h
0
j, i=¡hk
0
.
If c
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 ¡j
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that c
0
6=0. Let
Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 1 0 0 0 0 g
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that C+tY commutes with B for all t2F.
Moreover rank(C +tY)¸4 for t6=0. So we can use theorem 2.1., 3.1., and 4.1. There-
fore the triple (A;B;C +tY) belongs to G(3;7) for all t2F, t6=0.
96
(ii)-(1)-(b) Assume that there is a matrix B2 L such that k6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 ¡h
0
0 0 0 0 0
0 0 ¡h
0
0 0 0 0
0 0 0 0 0 h i
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 y 0 0 0 0 z
0 0 y 0 0 0 0
0 0 0 0 0 h
0
0
0 0 0 0 0 0 h
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=by+b
0
h
0
, y =¡
i
0
+k
0
h
0
k
, z =gy+g
0
h
0
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Soifatleastoneofhandh
0
isnonzero,thenthetriple(A;B+tX;C+tY)belongs
to G(3;7) for all t2F, t6=0. Suppose h=h
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 bc
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ck
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 x
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 c
0
0 ¡c 0 0 y
0 0 c
0
0 ¡c 0 0
0 0 0 0 0 0 z
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=bc
0
, y =fc+gc
0
, z =c
0
k.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. If c 6= 0 then for t 6= 0 C +tY has more than one point in the its spectrum. If
c
0
6= 0 and c = 0, then rank(C +tY) ¸ 4 for t 6= 0. So if at least one of c and c
0
is
nonzero, then the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0. Now
we may assume that c=c
0
=0. Let
97
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 ¡b
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 ¡k
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 ¡f
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreover for t6= 0 C +tY has more than one point in the its spectrum. So, by
Lemma 1.9., the triple (A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii)-(2) Suppose that (m;n) and (m
0
;n
0
) are linearly dependent. We may assume
that (m
0
;n
0
)=(0;0).
(ii)-(2)-(a) Assume that there is a matrix C 2 L such that d
0
= i
0
= 0. Then
there is a projection commuting with A and C.
(ii)-(2)-(b) Assume that there is a matrix C 2 L such that d
0
6= 0 and i
0
= 0.
We may assume that d
0
=1 and d=0. Then B and C looks like
B =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b c 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f 0 g h i
0 0 0 0 0 0 0
0 0 j 0 k 0 0
0 0 m 0 n 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, C =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 b
0
c
0
1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f
0
0 g
0
h
0
0
0 0 0 0 0 0 0
0 0 j
0
0 k
0
0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
The commutative relation of B and C implies that
cj
0
=c
0
j+m, ck
0
=c
0
k+n, hj
0
=h
0
j, hk
0
=h
0
k.
In this case we may assume that i6=0. Let
98
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 x 0 0 0
0 0 0 0 x 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 z 0 w 0 ¡yi
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 y 0 0 0
0 0 0 0 y 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where z =xf
0
¡yf, w =xg
0
¡yg, and y =j
0
if x=j, y =k
0
if x=k, or y =h
0
if x=h.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Moreoverifatleastoneofh
0
,j
0
, andk
0
isnonzero, thenfort6=0B+tX hasmore
than one point in the its spectrum. So, by Lemma 1.9., the triple (A;B +tX;C +tY)
belongs to G(3;7) for all t2F, t6=0. Now we may assume that h
0
=j
0
=k
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 1 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
0 0 0 0 ¡b
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 x 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=¡
f
0
i
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. Since for t 6= 0 B +tX has more than one point in the its spectrum, the triple
(A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0.
(ii)-(2)-(c) Assume that there is a matrix C 2 L such that d
0
= 0 and i
0
6= 0.
We may assume that i
0
= 1. So now we may assume that i = 0. By the rank condition
of B we have h(jn¡km)=0. If h=0 and h
0
6=0. If d=0, then we could introduce the
matrix
99
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. So we may assume that d6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ¡c 0 0 0
0 0 0 0 ¡c 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 h
0
1
0 0 x 0 y 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
cf
0
d
, y =
cg
0
d
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. Since for t 6= 0 C +tY has more than one point in the its spectrum, the triple
(A;B+tX;C +tY) belongs to G(3;7) for all t2F, t6=0. If h=0 and h
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 c 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 c
0
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
100
Then a direct computation reveals that B+tX commutes withC+tY for allt2F. So if
at least one of c and c
0
is nonzero, then the triple (A;B+tX;C+tY) belongs to G(3;7)
for all t2F, t6=0. If c=c
0
=0. If k
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 1 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that k
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
k
0
0 0 0 0 0 0
0 k
0
0 0 0 0 0
0 0 k
0
0 0 0 0
0 ¡j
0
0 0 0 0 0
0 0 ¡j
0
0 0 0 0
0 0 0 0 0 0 0
0 0 x 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 y 0 z 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=g
0
j
0
¡f
0
k
0
, y =¡
b
0
j
0
d
, z =¡
b
0
k
0
d
.
ThenadirectcomputationrevealsthatB+tX commuteswithC+tY forallt2F. Since
fort6=0B+tX hasmorethanonepointintheitsspectrum,thetriple(A;B+tX;C+tY)
belongs to G(3;7) for all t2F, t6=0. If h6=0. The commutativity relation of B and C
implies that
cj
0
=c
0
j, ck
0
=c
0
k, hj
0
=h
0
j+m, hk
0
=h
0
k+n.
From the rank condition we have jn¡km=0. So we have jk
0
=j
0
k. Let
101
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 w 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
¡k 0 0 x 0 0 0
0 ¡k 0 0 x 0 0
0 0 ¡k 0 0 0 0
0 j 0 0 0 0 0
0 0 j 0 0 0 0
0 0 0 0 0 0 0
0 0 y 0 z n ¡k
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
ck+dn
h
, y =
xf¡bj
d
, z =
xg¡bk
d
, w =gj¡fk.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. If k6= 0, then for t6= 0 C +tY has more than one point in the its spectrum. So
the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0. Now we may assume
that k =0. From the rank condition of B we have jn=0. So we will consider 3 cases. If
j = 0 and n6= 0. From the commutative condition we have that k
0
6= 0, c = 0, hj
0
=m,
and hk
0
=n. Let
Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 d 0 0 0
0 0 0 0 d 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f 0 g h 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that C+tY commutes with B for all t2F.
Since d6=0 and h6=0, rank(C+tY)¸4 for t6=0. So the triple (A;B;C+tY) belongs
to G(3;7) for all t2F, t6= 0. If j 6= 0 and n = 0. From the commutative condition we
have that k
0
=0, cj
0
=c
0
j, and hj
0
=h
0
j+m. Let
102
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 ® 0 0 0
0 0 0 0 ® 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 y 0 0 0
0 0 0 0 y 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 z 0 w v x
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x = hj, y =¡dhj
0
, z =¡fhj
0
+f
0
hj, w =¡ghj
0
+g
0
hj, v =¡mh, ® =¡dhj.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. Thenfort6=0C+tY hasmorethanonepointintheitsspectrum. So, byLemma
1.9., the triple (A;B +tX;C +tY) belongs to G(3;7) for all t2F, t6= 0. If j = 0 and
n=0. From the commutative condition we have that k
0
=0, cj
0
=0, and hj
0
=m. Let
Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 d 0 0 0
0 0 0 0 d 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 f 0 g h 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that C+tY commutes with B for all t2F.
Since d6=0 and h6=0, rank(C+tY)¸4 for t6=0. So the triple (A;B;C+tY) belongs
to G(3;7) for all t2F, t6=0.
(ii)-(2)-(d) Assume that there is a matrix C 2 L such that d
0
6= 0 and i
0
6= 0.
We may assume that d
0
= 1. So now we may assume that d = 0. By the rank condition
of B we have ci(jn¡km)=0. If i=0, then we could introduce the matrix
103
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. So we may assume that i6=0. If jn¡mk =0 and
h
0
¡i
0
c
0
6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 j x
0 0 0 0 0 m y
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 j
0
0
0 0 0 0 0 0 j
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
ij
0
h
0
¡i
0
c
0
, y =¡
ic
0
j
0
h
0
¡i
0
c
0
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2 F. If j
0
6= 0, then for t 6= 0 C +tY has more than one point in the its spectrum.
So the triple (A;B +tX;C +tY) belongs to G(3;7) for all t 2F, t 6= 0. If j
0
= 0 and
j6= 0, then for t6= 0 B+tX has more than one point in the its spectrum. So the triple
(A;B +tX;C +tY) belongs to G(3;7) for all t 2F, t 6= 0. Now we may assume that
j
0
=j =0. Let
104
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 k x
0 0 0 0 0 n y
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 k
0
0
0 0 0 0 0 0 k
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
ik
0
h
0
¡i
0
c
0
, y =¡
ic
0
k
0
h
0
¡i
0
c
0
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2 F. If k
0
6= 0, then for t 6= 0 C +tY has more than one point in the its spectrum.
So the triple (A;B +tX;C +tY) belongs to G(3;7) for all t2F, t6= 0. If k
0
= 0 and
k6=0, then for t6=0 B+tX has more than one point in the its spectrum. So the triple
(A;B +tX;C +tY) belongs to G(3;7) for all t 2F, t 6= 0. Now we may assume that
k
0
=k =0. If c
0
=0, then we could introduce the matrix
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and B, so that it su±ces to prove our case for all
triples (A;B;C +tZ) for t2F, t6=0. So we may assume that c
0
6=0. Let
105
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 ¡c
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Sincefort6=0B+tX hasmorethanonepointintheitsspectrum,thetriple(A;B+tX;C)
belongs to G(3;7) for all t2F, t6=0. If jn¡mk =0 and h
0
¡i
0
c
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 ¡c
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a direct computation reveals that B + tX commutes with C for all t 2 F. If
c
0
6=0, then for t6=0 B+tX has more than one point in the its spectrum. So the triple
(A;B +tX;C) belongs to G(3;7) for all t2F, t6= 0. Now we may assume that c
0
= 0.
Then, byassumption, wehaveh
0
=0. ThecommutativerelationofB andC impliesthat
cj
0
=m, ck
0
=n, hj
0
=i
0
m, hk
0
=i
0
n.
If c=0, then we could introduce the matrix
106
Z =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 1 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 i
0
0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 j
0
0 k
0
0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
that clearly commutes with both A and C, so that it su±ces to prove our case for all
triples (A;B+tZ;C) for t2F, t6=0. So we may assume that c6=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 n 0
0 0 0 0 0 0 n
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 k
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. If n6=0, then for t6=0 B+tX has more than one point in the its spectrum. So,
by Lemma 1.9 ., the triple (A;B +tX;C +tY) belongs to G(3;7) for all t 2F, t 6= 0.
Now we may assume that n=0. Since n=ck
0
and c6=0, we have k
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 m 0
0 0 0 0 0 0 m
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 j
0
0 0 0 0 0 0 0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. If m 6= 0, then for t 6= 0 B +tX has more than one point in the its spectrum.
107
So the triple (A;B +tX;C +tY) belongs to G(3;7) for all t 2F, t 6= 0. Now we may
assume that m=0. Then j
0
=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 i 0
0 0 0 0 0 0 i
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 x
0 0 0 0 0 0 y
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
i
c
, y =
i
0
c¡h
c
.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t 2F. Since for t 6= 0 B +tX has more than one point in the its spectrum, the triple
(A;B+tX;C+tY)belongstoG(3;7)forallt2F,t6=0. Ifjn¡mk6=0andh
0
¡i
0
c
0
6=0.
Since ic(jn¡mk)=0 and i6=0, we have c=0. Let
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 x z
0 0 0 0 0 w y
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
, Y =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 j 0
0 0 0 0 0 0 j
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
,
where x=
hj
h
0
¡i
0
c
0
, y =¡c
0
z, z =
ij
h
0
¡i
0
c
0
, w =¡c
0
x.
Then a straightforward computation reveals that B+tX commutes with C +tY for all
t2F. If j 6= 0, then for t6= 0 C +tY has more than one point in the its spectrum. So
the triple (A;B+tX;C+tY) belongs to G(3;7) for all t2F, t6=0. If j =0, then by the
commutative condition of B and C we have m = 0. So jn¡mk = 0. It can not occur.
If jn¡mk6=0 and h
0
¡i
0
c
0
=0. Let
108
X =
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 1
0 0 0 0 0 0 ¡c
0
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
.
Then a straightforward computation reveals that B+tX commutes with C for all t2F.
Ifc
0
6=0, thenfort6=0B+tX hasmorethanonepointintheitsspectrum. Sothetriple
(A;B+tX;C) belongs to G(3;7) for all t2F, t6=0. If c
0
=0, then by the commutative
condition of B and C we have m=n=0. So jn¡mk =0. It can not occur.
109
Chapter 6
Conclusions
Nowwearriveatthemainresult. Weneedtoconsideronlythecaseofa3dimensional
linear space L of nilpotent commuting matrices of 7£7. The case of a single nonzero
Jordan block is [7], the case of square zero is [7], and the case of at most two Jordan
blocks is [6]. So this leaves only the case of Jordan blocks of orders 4+2+1, 3+3+1,
3+2+2, or 3+2+1+1 which are handled in the previous 4 chapters. So we can have
the following main result.
Theorem 6.1. Any triple of nilpotent commuting triples of 7£7 over an algebraically
closed ¯eld of characteristic zero belongs to G(3;7).
Corollary 6.2. If A, B, and C are three commuting n£n matrices with n<8, then the
algebra they generate has at most dimension n.
Proof. LetA=A(A;B;C) be the algebra generated by A, B, and C. Then dimA·r is
just a closed condition. ThusV =f(A;B;C)2C(3;n)jdimA·ng is a closed subvariety.
On the other hand it contains G(3;n) which is dense in C(3;n). Thus it is the whole
variety.
110
Reference List
[1] M. Gerstenhaber. On dominance and varieties of commuting matrices. Ann. Math.,
73:324{348, 1961.
[2] R. Guralnick. A note on commuting pairs of matrices. Linear Multilinear Algebra,
31:71{75, 1992.
[3] R.GuralnickandB.Sethurman. Commutingpairsandtriplesofmatricesandrelated
varieties. Linear Algebra Appl., 310:139{148, 2000.
[4] J. Holbrook and M. Omladi· c. Approximating commuting operators. Linear Algebra
Appl., 327:131{149, 2001.
[5] T. Motzkin and O. Taussky. Pairs of matrices with property L. Trans. Amer. Math.
Soc., 80:387{401, 1955.
[6] M. Neubauer and B. Sethuraman. Commuting pairs in the centralizers of 2-regular
matrices. J. Algebra, 214:174{181, 1999.
[7] M. Omladi· c. A variety of commuting triples. Linear Algebra Appl., 383:233{245,
2004.
111 
Abstract (if available)
Abstract The irreducibility of the variety C(m,n) commuting m-tuples of n X n matrices has been studied by several authors. It is known that  this variety is irreducible if m=2, n=1,2,3 and reducible if m,n>=4. The irreducibility of C(3,n) was asked by Gersteinhaber. Until now we know that C(3,n) is reducible for n>=32, but irreducible for n<=4. -- If the characteristic of an arbitrary algebraically closed field is zero, then we have more specific answer for this question. In this case C(3,n) is reducible for n>=30 and irreducible for n<=6. -- This dissertation is primarily concerned with the study of the variety C(3,7). We show that this variety is also irreducible. Guralnick and Omladi\v{c} have conjectured that it is reducible for n > 7 
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Asset Metadata
Creator Han, Yongho (author) 
Core Title The commuting triples of matrices 
School College of Letters, Arts and Sciences 
Degree Doctor of Philosophy 
Degree Program Mathematics 
Publication Date 06/05/2007 
Defense Date 04/02/2007 
Publisher University of Southern California (original), University of Southern California. Libraries (digital) 
Tag approximation by generic matrices,irreducible variety of commuting triples,OAI-PMH Harvest 
Language English
Advisor Guralnick, Robert (committee chair), Huang, Ming-Deh (committee member), Montgomery, Susan (committee member) 
Creator Email yhh@usc.edu 
Permanent Link (DOI) https://doi.org/10.25549/usctheses-m510 
Unique identifier UC1447531 
Identifier etd-Han-20070605 (filename),usctheses-m40 (legacy collection record id),usctheses-c127-492923 (legacy record id),usctheses-m510 (legacy record id) 
Legacy Identifier etd-Han-20070605.pdf 
Dmrecord 492923 
Document Type Dissertation 
Rights Han, Yongho 
Type texts
Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection) 
Repository Name Libraries, University of Southern California
Repository Location Los Angeles, California
Repository Email cisadmin@lib.usc.edu
Tags
approximation by generic matrices
irreducible variety of commuting triples