Generalization of Gracia's Results
Volume
1081-3810
Generalization of Gracia's Results
Jun Liao Hubei Institute 0 1 2 3 4
0 1 2 3 4
Heguo Liu 0 1 2 3 4
0 Xingzhong Xu Hubei University
1 Hubei University
2 Zuohui Wu Hubei University
3 Yulei Wang Henan University of Technology
4 Jun Liao, Heguo Liu , Yulei Wang, Zuohui Wu, and Xingzhong Xu
Follow this and additional works at: https://repository.uwyo.edu/ela
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http://math.technion.ac.il/iic/ela
GENERALIZATION OF GRACIA’S RESULTS∗
JUN LIAO† , HEGUO LIU†, YULEI WANG‡ , ZUOHUI WU†, AND XINGZHONG XU†
AMS subject classifications. 15A24, 15A27.
1. Introduction. The notation used in this paper is standard, see [
2, 3
] for
example. Let C be the complex field. Suppose A ∈ Mm×m(C) and B ∈ Mn×n(C).
Let αAB be a linear transformation of Mm×n(C) defined by
αAB(X ) = AX − X B, for X ∈ Mm×n(C).
If A = B, then we will write αA instead of αAA for brevity. In the case of no confusion,
we write α = αAB for short.
The well known dimension formula of the kernel kerαA is due to Frobenius [3,
Theorem VII.1]. Then Gracia has obtained the dimension formulas of kerα2A and
kerα3A in [
1
].
It is obvious that the kernels of the liner transformations α2 and α3 are the
solutions of the matrix equations A(AX − X B) − (AX − X B)B = 0 and A[A(AX −
X B) − (AX − X B)B] − [A(AX − X B) − (AX − X B)B]B = 0, respectively. In
this paper, we obtain the dimensions of kerα2 and kerα3, which generalizes Gracia’s
results.
∗Received by the editors on September 18, 2014. Accepted for publication on May 1, 2015.
Handling Editor: Joao Filipe Queiro.
†School of Mathematics and Statistics, Hubei University, Wuhan, 430062, P.R. China
(, , , ).
Supported by the National Natural Science Foundation of China (grant no. 11371124 and
11401186).
‡School of Mathematics, Henan University of Technology, Zhengzhou, 450001, P.R. China
(). Supported by the National Natural Science Foundation of China (grant no.
11301150).
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For convenience, we introduce the following notations. Suppose that the
elementary divisors of A and B are (λ − λ1)p1 , (λ − λ2)p2 , . . . , (λ − λu)pu and (λ − μ1)q1 , (λ −
μ2)q2 , . . . , (λ − μv)qv , respectively. Let En be the unit matrix of size n and let
0 1 0 · · ·
0 0 1 · · ·
Nn = ... ... ... . . .
0 0 0 · · ·
0 0 0 · · ·
0
0
...
1
0
be the square matrix of size n in which all the elements of the first superdiagonal are
1 and all the other elements are 0. Let
and
and
JA =
JB =
λ1Ep1 + Np1
μ1Eq1 + Nq1
0
0
λ2Ep2 + Np2
μ2Eq2 + Nq2
. . .
. . .
λuEpu + Npu
0
0
μvEqv + Nqv
0
F(α2β) = 2min(pα, qβ ) − 1
2min(pα, qβ )
if λα 6= μβ ;
if λα = μβ and pα = qβ ;
if λα = μβ and pα 6= qβ
0
F(α3β) = 3min(pα, qβ ) − 2
3min(pα, qβ ) − 1
3min(pα, qβ )
if λα 6= μβ;
if λα = μβ and pα = qβ ;
if λα = μβ and |pα − qβ | = 1;
if λα = μβ and |pα − qβ | ≥ 2.
be respectively the Jordan normal forms of A and B. For 1 ≤ α ≤ u and 1 ≤ β ≤ v.
Let F(α2β) and F(α3β) be defined by the following:
The main results are the following:
Theorem 1.1. Let α : Mm×n(C) → Mm×n(C) be a linear transformation such
that α(X ) = AX − X B, where A and B are m × m and n × n complex matrices,
respectively. Then the dimension formula for kerα2 is
Theorem 1.2. Let α : Mm×n(C) → Mm×n(C) be a linear transformation such
that α(X) = AX − XB, where A and B are m × m and n × n complex matrices,
respectively. Then the dimension formula for kerα3 is
245
(1.1)
(1.2)
u v
dim(kerα2) = X X F(α2β).
α=1 β=1
u v
dim(kerα3) = X X F(α3β).
α=1 β=1
2. The proof of Theorem 1.1. Before proving the theorem, we first give a
lemma in the following:
Lemma 2.1. For a matrix M ∈ Ml×l(C), let Λ(M ) be the set of its different
eigenvalues. If Λ(A) ∩ Λ(B) = ∅, then kerαk = 0 for k = 1, 2, 3, . . .
Proof. It is well known that Λ(A) ∩ Λ(B) = ∅ if and only if the unique solution
of the matrix equation AX − XB = 0 is X = 0. Thus, when Λ(A) ∩ Λ(B) = ∅, we
can prove by induction on k that if Λ(A) ∩ Λ(B) = ∅, the equality αk(X) = 0 implies
X = 0. In fact, if αk(X) = 0 then α(αk−1(X)) = 0 and Aαk−1(X) − αk−1(X)B = 0.
Since Λ(A) ∩ Λ(B) = ∅, it follows that αk−1(X) = 0. By hypothesis of the induction
the equality αk−1(X) = 0 implies that X = 0. So that, Λ(A) ∩ Λ(B) = ∅ implies that
kerαk = 0 for k = 1, 2, 3, . . .
Proof. It is obvious that there are invertible matrices U and V such that A =
U JAU −1 and B = V JBV −1. Assume X ∈ kerα2, then A(AX −XB)−(AX −XB)B =
0. Hence,
U JAU −1(U JAU −1X − XV JBV −1) = (U JAU −1X − XV JBV −1)V JBV −1.
Thus,
JA(JAU −1XV − U −1XV JB) = (JAU −1XV − U −1XV JB)JB.
Let X = U −1XV . Then, the equation is
JA(JAX − XJB) = (JAX − XJB)JB.
(2.1)
Now we partition X into blocks (Xαβ) where Xαβ = (εik)pα×qβ is a pα × qβ matrix
for 1 ≤ α ≤ u and 1 ≤ β ≤ v. Then we get uv matrix equations from (3):
(λαEpα + Npα )[(λαEpα + Npα )Xαβ − Xαβ(μβ Eqβ + Nqβ )]
= [(λαEpα + Npα )Xαβ − Xαβ(μβEqβ + Nqβ )](μβEqβ + Nqβ ).
http://math.technion.ac.il/iic/ela
246
Write Pα := Npα and Qβ := Nqβ . An easy (...truncated)