Hindawi Publishing Corporation
The Scientific World Journal
Volume 2013, Article ID 640350, 4 pages
http://dx.doi.org/10.1155/2013/640350
Research Article
A Note on Decomposing a Square Matrix as Sum of Two Square
Nilpotent Matrices over an Arbitrary Field
Xiaofei Song,1 Baodong Zheng,1 and Chongguang Cao2
1
2
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
School of Mathematical Sciences, Heilongjiang University, Harbin 150080, China
Correspondence should be addressed to Baodong Zheng; [email protected]
Received 31 August 2013; Accepted 1 October 2013
Academic Editors: A. Badawi and P. Bracken
Copyright © 2013 Xiaofei Song et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let K be an arbitrary field and X a square matrix over K. Then X is sum of two square nilpotent matrices over K if and only if,
for every algebraic extension L of K and arbitrary nonzero 𝛼 ∈ 𝐿, there exist idempotent matrices 𝑃1 and 𝑃2 over L such that
𝑋 = 𝛼𝑃1 − 𝛼𝑃2 .
1. Introduction
Botha (see [1]) proved that a square matrix 𝐴 over a field 𝐾 is a
sum of two nilpotent matrices over 𝐾 if and only if 𝐴 is similar
to a particular form. In an early paper, Pazzis (see [2]) gave
necessary and sufficient conditions in which a matrix can be
decomposed as a linear combination of two idempotents with
given nonzero coefficients. The goal of this paper is to build a
bridge that connects the result obtained in [1] with the result
obtained in [2]. However, the relation between these two facts
has not been formally discussed yet (more details in [3–9]).
If there is no statement, the meanings of notations mentioned in this paragraph hold all over the paper. 𝐾 denotes
an arbitrary field, 𝐾 is its algebraic closure, 𝐿 is an arbitrary
algebraic extension of 𝐾, and car(𝐾) is the characteristic of 𝐾.
𝑍+ denotes the set of all positive integers, [𝑠] = {𝑧 ∈ 𝑍+ | 1 ≤
𝑧 ≤ 𝑠} for some 𝑠 ∈ 𝑍+ . 𝑀𝑚,𝑛 (𝐾) denotes the space consisting
of all 𝑚 × 𝑛 matrices over 𝐾; 𝑀𝑛 (𝐾) = 𝑀𝑛,𝑛 (𝐾). 𝑟(𝐴) is the
rank of 𝐴 ∈ 𝑀𝑚,𝑛 (𝐾). 𝐸 denotes a vector space over 𝐾 and
dim(𝐸) is the dimension of 𝐸. 𝑋 ∈ 𝑀𝑛 (𝐾) is called 2𝑠 𝑁 in
𝑀𝑛 (𝐾) if there exist square nilpotent 𝑁1 and 𝑁2 ∈ 𝑀𝑛 (𝐾)
such that 𝑋 = 𝑁1 + 𝑁2 , while 𝑋 is called an (𝛼, 𝛽) composite
in 𝑀𝑛 (𝐾) if there exist idempotent 𝑃1 and 𝑃2 ∈ 𝑀𝑛 (𝐾) such
that 𝑋 = 𝛼𝑃1 + 𝛽𝑃2 , where 𝛼, 𝛽 ∈ 𝐾 \ {0} (Definition 1 in
[2]); in particular, 𝑋 is called ± 𝑃 if 𝑋 is an (𝛼, −𝛼) composite
in 𝑀𝑛 (𝐿) for every algebraic extension 𝐿 of 𝐾 and arbitrary
nonzero 𝛼 ∈ 𝐿 (when car(𝐾) = 2, we still use ± 𝑃 for the
meaning of (𝛼, 𝛼) composites).
For 𝑋 ∈ 𝑀𝑛 (𝐾), on the one hand, we will prove that 𝑋 is
in 𝑀𝑛 (𝐾) implies 𝑋 is ± 𝑃; that is, the form provided by
Botha satisfies the condition as in [2]; on the other hand, we
will also prove that 𝑋 is ± 𝑃 implies 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾); that
is, we can derive the form provided in [1] from the results
obtained in [2]. In fact, the following theorem is the main
result of this paper.
2
𝑠𝑁
Theorem 1 (main theorem). Suppose 𝐾 is an arbitrary field
and 𝑋 ∈ 𝑀𝑛 (𝐾); then 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾) if and only if 𝑋 is ± 𝑃.
In Section 2, we will state some related theorems and
notations from [2] and we will give some necessary corollaries. The proof of Theorem 1 will be carried out in Section 3.
2. More Notations and Necessary Corollaries
Suppose 𝑋 ∈ 𝑀𝑛 (𝐾) and 𝑋𝑖 ∈ 𝑀𝑛𝑖 (𝐾), we denote by 𝑋 =
𝑋1 ⊕ ⋅ ⋅ ⋅ ⊕ 𝑋𝑠 the following matrix with ∑𝑠𝑖=1 𝑛𝑖 = 𝑛:
𝑋1
(
𝑋2
d
).
(1)
𝑋𝑠
Notation 1 (Notation 2 in [2]). Let 𝑋 ∈ 𝑀𝑛 (𝐾), 𝜆 ∈ 𝐾 and
𝑘 ∈ 𝑍+ ; we denote by
2
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(1) 𝑗𝑘 (𝑋, 𝜆) the number of blocks of size 𝑘 for the
eigenvalue 𝜆 in the Jordan reduction of 𝑋;
(2) 𝑛𝑘 (𝑋, 𝜆) the number of blocks of size greater or equal
to 𝑘 for the eigenvalue 𝜆 in the Jordan reduction of 𝑋.
Definition 2 (Definition 3 in [2]). Two sequences (𝑢𝑘 )𝑘 ≥1 and
(V𝑘 )𝑘 ≥1 are side to be intertwined if for all 𝑘 ∈ 𝑍+ , V𝑘 ≥ 𝑢𝑘+1 ,
and 𝑢𝑘 ≥ V𝑘+1 .
Notation 2 (Notation 4 in [2]). Given a monic polynomial,
𝑃 = 𝑥𝑛 − 𝑎𝑛−1 𝑥𝑛−1 − ⋅ ⋅ ⋅ − 𝑎1 𝑥 − 𝑎0 , denote the following 𝐶(𝑃)
by its companion matrix:
0
1
(0
.
𝐶 (𝑃) = (
( ..
..
.
(0
0 ⋅⋅⋅ ⋅⋅⋅ 0
0 0 ⋅⋅⋅ 0
1 0 d 0
𝑎0
𝑎1
𝑎2 )
. ).
d d d d .. )
⋅⋅⋅ d 1
⋅⋅⋅ ⋅⋅⋅ 0
(2)
Corollary 6. Assume car(𝐾) = 2 and let 𝑋 ∈ 𝑀𝑛 (𝐾). Then
𝑋 is ± 𝑃 if and only if for every 𝜆 ∈ 𝐾 \ {0}, all blocks in the
Jordan reduction of 𝑋 with respect to 𝜆 have an even size.
Naturally, we derive the following corollary from the
above two corollaries.
Corollary 7. Every nilpotent is ± 𝑃.
In fact, arbitrary nilpotent is not only ± 𝑃 but also 2𝑠 𝑁.
Lemma 8. Every nilpotent 𝑁 ∈ 𝑀𝑛 (𝐾) is 2𝑠 𝑁.
Proof. For arbitrary field 𝐾, let 𝑁 ∈ 𝑀𝑛 (𝐾) is nilpotent; then
𝑁 is similar to 𝑁1 ⊕ 𝑁2 ⊕ ⋅ ⋅ ⋅ ⊕ 𝑁𝑠 , where for every 𝑖 ∈ [𝑠], 𝑁𝑖 ∈
𝑀𝑟𝑖 (𝐾), ∑𝑠𝑖=1 𝑟𝑖 = 𝑛, and both the characteristic polynomial
and minimal polynomial of 𝑁𝑖 are 𝑥𝑟𝑖 . Furthermore, 𝑁𝑖 is
similar to 𝐶(𝑥𝑟𝑖 ) as follows:
0 𝑎𝑛−2
1 𝑎𝑛−1 )
0 0 ⋅⋅⋅
1 0 ⋅⋅⋅
( ..
. d d
0 ⋅⋅⋅ 1
Theorem 3 (Theorem 1 in [2]). Assume car(𝐾) ≠ 2 and let
𝑋 ∈ 𝑀𝑛 (𝐾). Then 𝑋 is an (𝛼, −𝛼) composite if and only if all
the following conditions hold.
(1) The sequences (𝑛𝑘 (𝑋, 𝛼))𝑘 ≥1 and (𝑛𝑘 (𝑋, −𝛼))𝑘 ≥1 are
intertwined;
(2) for all 𝜆 ∈ 𝐾 \ {0, 𝛼, −𝛼} and for all 𝑘 ∈ 𝑍+ , 𝑗𝑘 (𝑋, 𝜆) =
𝑗𝑘 (𝑋, −𝜆).
Theorem 4 (Theorem 5 in [2]). Assume car(𝐾) = 2 and let
𝑋 ∈ 𝑀𝑛 (𝐾). Then 𝑋 is an (𝛼, −𝛼) composite if and only if for
every 𝜆 ∈ 𝐾\{0, 𝛼}, all blocks in the Jordan reduction of 𝑋 with
respect to 𝜆 have an even size.
Suppose 𝑋 ∈ 𝑀𝑛 (𝑘) is ± 𝑃, where car(𝐾) ≠ 2. Then 𝑋 is
(𝛼, −𝛼) composite and (𝛽, −𝛽) composite in 𝑀𝑛 (𝐿) for some
algebraic extension 𝐿 of 𝐾, where 𝛼, 𝛽 ∈ 𝐿 \ {0} with 𝛼 ≠ ± 𝛽.
By Theorem 3, the following statements are true:
(1) for all 𝜆 ∈ 𝐾 ∈ {0, 𝛼, −𝛼} and for all 𝑘 ∈ 𝑍+ ,
𝑗𝑘 (𝑋, 𝜆) = 𝑗𝑘 (𝑋, −𝜆);
(2) for all 𝜆 ∈ 𝐾 ∈ {0, 𝛽, −𝛽} and for all 𝑘 ∈ 𝑍+ ,
𝑗𝑘 (𝑋, 𝜆) = 𝑗𝑘 (𝑋, −𝜆).
so for all 𝜆 ∈ 𝐾\{0} and for all 𝑘 ∈ 𝑍+ , 𝑗𝑘 (𝑋, 𝜆) = 𝑗𝑘 (𝑋, −𝜆).
On the other hand, note that for nonzero 𝛼 ∈ 𝐾 with
car(𝐾) ≠ 2, the sequences (𝑛𝑘 (𝑋, 𝛼))𝑘 ≥1 and (𝑛𝑘 (𝑋, −𝛼))𝑘 ≥1
are intertwined if for all 𝑘 ∈ 𝑍+ , 𝑗𝑘 (𝑋, 𝛼) = 𝑗𝑘 (𝑋, −𝛼). Then
for all 𝜆 ∈ 𝐾 \ {0}, 𝑘 ∈ 𝑍+ , 𝑗𝑘 (𝑋, 𝜆) = 𝑗𝑘 (𝑋, −𝜆) implies that
for every algebraic extension 𝐿 of 𝐾 and arbitrary nonzero
𝛼 ∈ 𝐿, 𝑋 is an (𝛼, −𝛼) composite in 𝑀𝑛 (𝐿); that is, 𝑋 is ± 𝑃.
Therefore the following corollary is true.
Corollary 5. Assume car(𝐾) ≠ 2 and let 𝑋 ∈ 𝑀𝑛 (𝐾). Then
𝑋 is ± 𝑃 if and only if for all 𝜆 ∈ 𝐾 \ {0} for all 𝑘 ∈ 𝑍+ ,
𝑗𝑘 (𝑋, 𝜆) = 𝑗𝑘 (𝑋, −𝜆).
Similarly, we can derive the following corollary from
Theorem 4.
0
0
.. )
.
0
.
(3)
𝑟𝑖 ×𝑟𝑖
𝑟𝑖
That is, 𝐶(𝑥 ) = 𝐸2,1 + 𝐸3,2 + ⋅ ⋅ ⋅ + 𝐸𝑟𝑖 ,𝑟𝑖 −1 ∈ 𝑀𝑟𝑖 (𝐾).
𝑟 /2
𝑟 /2−1
𝑖
𝑖
𝐸2𝑗,2𝑗−1 + ∑𝑗=1
When 𝑟𝑖 is even, 𝐶(𝑥𝑟𝑖 ) = ∑𝑗=1
𝑟𝑖
(𝑟𝑖 −1)/2
∑𝑗=1
(𝑟𝑖 −1)/2
∑𝑗=1
𝐸2𝑗+1,2𝑗 ;
𝐸2𝑗,2𝑗−1 +
𝐸2𝑗+1,2𝑗 .
when 𝑟𝑖 is odd, 𝐶(𝑥 ) =
Note that both ∑ 𝐸2𝑗,2𝑗−1 and ∑ 𝐸2𝑗+1,2𝑗 are square nilpotent
matrices then 𝐶(𝑥𝑟𝑖 ) is 2𝑠 𝑁, and 𝑁𝑖 is 2𝑠 𝑁 follows. Hence 𝑁 is
2
𝑠 𝑁.
3. Proof of Main Theorem
2
𝑠𝑁
→ ± 𝑃. Suppose 𝑋 ∈ 𝑀𝑛 (𝐾) is 2𝑠 𝑁 in 𝑀𝑛 (𝐾); that is,
there exist square nilpotent matrices 𝑁1 and 𝑁2 ∈ 𝑀𝑛 (𝐾)
such that 𝑋 = 𝑁1 + 𝑁2 . It will take two steps to prove 𝑋 is ± 𝑃.
Step 1. If 𝑋 is nonsingular, then 𝑋 is ± 𝑃.
Since 𝑋 = 𝑁1 + 𝑁2 with 𝑁12 = 𝑁22 = 0, inspect the
eigenspaces of 𝑁1 and 𝑁2 . Note that 𝑁1 and 𝑁2 are square
nilpotent matrices, their ranks satisfy the following inequality
matrices.
𝑟 (𝑁1 ) + 𝑟 (𝑁2 ) ≤ 𝑛,
(4)
where equality holds if and only if 𝑟(𝑁1 ) = 𝑟(𝑁2 ) = 𝑛/2.
At first, 𝑋 is nonsingular implies 0 is not its eigenvalue.
Secondly, if the inequality is strict, then intersection of
eigenspaces of 𝑁1 and 𝑁2 contains nonzero vectors; that is,
there exists nonzero 𝑥 ∈ 𝑀𝑛,1 (𝐾) such that 𝑁1 𝑥 = 𝑁2 𝑥 = 0,
which implies that 0 is one of eigenvalues of 𝑋. This is a
contradiction. Hence, 𝑟(𝑁1 ) + 𝑟(𝑁2 ) = 𝑛; that is, 𝑛 is even
and 𝑁1 and 𝑁2 are similar but not equal.
Because 𝑁1 is square nilpotent with 𝑟(𝑁1 ) = 𝑛/2, we
can choose 𝑛/2 linear independent vectors from the set of its
column vectors which can make up a base of eigenspace of 𝑁1
and denote 𝛽 by the 𝑛 × (𝑛/2) matrix consisting of these 𝑛/2
columns. Correspondingly, we have 𝑛 × (𝑛/2) matrix 𝛾 with
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3
all columns from the set of columns of 𝑁2 . Because 0 is the
only vector in the intersection of eigenspaces of 𝑁1 and 𝑁2 ,
𝑛 × 𝑛 matrix (𝛽 𝛾) is nonsingular.
𝑁12 𝛾 = 0 implies that nonzero column vectors of 𝑁1 𝛾
are eigenvectors of 𝑁1 and 𝑁1 (𝛽 𝛾) = (0 𝑁1 𝛾) implies
𝑟(𝑁1 𝛾) = 𝑛/2. Hence; 𝑁1 𝛾 and 𝛽 are equal under certain
column transformation; that is, there is an invertible matrix
𝑇1 such that 𝑁1 𝛾 = 𝛽𝑇1 . Correspondingly, there is an
invertible matrix 𝑇2 such that 𝑁2 𝛽 = 𝛾𝑇2 .
𝑦
Let ( 𝑦12 ) be the inverse of (𝛽 𝛾), where 𝑦1 and 𝑦2 are
(𝑛/2) × 𝑛 matrices. Naturally, the following equation is true:
0
𝐼
𝑦 𝛽 𝑦1 𝛾
𝑦
) = ( 𝑛/2 𝑛/2 ) .
( 1 ) (𝛽 𝛾) = ( 1
𝑦2
0𝑛/2 𝐼𝑛/2
𝑦2 𝛽 𝑦2 𝛾
(5)
Now, we carry out the same similarity transformation on 𝑁1
and 𝑁2 as follows:
𝑦 𝑁 𝛽 𝑦1 𝑁1 𝛾
𝑦
( 1 ) 𝑁1 (𝛽 𝛾) = ( 1 1
),
𝑦2
𝑦2 𝑁1 𝛽 𝑦2 𝑁1 𝛾
𝑦 𝑁 𝛽 𝑦1 𝑁2 𝛾
𝑦
).
( 1 ) 𝑁2 (𝛽 𝛾) = ( 1 2
𝑦2
𝑦2 𝑁2 𝛽 𝑦2 𝑁2 𝛾
(6)
Note that 𝑁1 𝛾 = 𝛽𝑇1 and 𝑁2 𝛽 = 𝛾𝑇2 , the above three
0
𝑇1
) ∈ 𝑀𝑛 (𝐾) and
equations imply that 𝑁1 is similar to ( 0𝑛/2
𝑛/2 0𝑛/2
𝑁2 is similar to
0
0
( 𝑇𝑛/22 0𝑛/2
𝑛/2
) ∈ 𝑀𝑛 (𝐾).
0
0
𝑇
0
1
) + ( 𝑇𝑛/22 0𝑛/2
). For every
Hence, 𝑋 is similar to ( 0𝑛/2
𝑛/2 0𝑛/2
𝑛/2
algebraic extension 𝐿 of 𝐾 and arbitrary nonzero 𝛼 ∈ 𝐿, 𝑋 is
also similar to the following matrix:
0𝑛/2
𝐼
𝛼−1 𝑇1
𝐼
) − 𝛼 ( 𝑛/2
)
𝛼 ( 𝑛/2
0𝑛/2 0𝑛/2
−𝛼−1 𝑇2 0𝑛/2
(7)
That is, 𝑋 is ± 𝑃.
Step 2. If 𝑋 is singular and similar to 𝑌 ⊕ 𝑁, where 𝑌 is
nonsingular and 𝑁 is nilpotent. Then 𝑋 is ± 𝑃.
At first, we need to prove that 𝑌 is 2𝑠 𝑁. Without loss of
generality, we assume 𝑋 = 𝑌 ⊕ 𝑁 in the following proof
since 2𝑠 𝑁 holds under similarity transformations.
Let 𝑁1 = ( 𝑛𝑛13 𝑛𝑛24 ), where the order of 𝑛1 is the same for 𝑌
and the order of 𝑛4 is the same for 𝑁. Then 𝑁12 = 0 implies
the following equations are true:
𝑛12 + 𝑛2 𝑛3 = 0,
𝑛1 𝑛2 + 𝑛2 𝑛4 = 0
𝑛3 𝑛1 + 𝑛4 𝑛3 = 0,
2
𝑛3 𝑛2 + 𝑛42 = 0.
(8)
𝑁22
= 0, we get the following equations
Since (𝑋 − 𝑁1 ) =
after replacing 𝑛1 with 𝑌−𝑛1 and 𝑛4 with 𝑁−𝑛4 in the previous
equations:
2
(𝑌 − 𝑛1 ) + 𝑛2 𝑛3 = 0,
(𝑌 − 𝑛1 ) 𝑛2 + 𝑛2 (𝑁 − 𝑛4 ) = 0,
𝑛3 (𝑌 − 𝑛1 ) + (𝑁 − 𝑛4 ) 𝑛3 = 0,
2
𝑛3 𝑛2 + (𝑁 − 𝑛4 ) = 0.
(9)
We can derive the following equations from the 3rd and
4th equations in the above two sets of equations:
𝑌𝑛2 + 𝑛2 𝑁 = 0,
𝑛3 𝑌 + 𝑁𝑛3 = 0.
(10)
Note that 𝑁 is nilpotent, assume its index is 𝑟; that is,
𝑁𝑟−1 ≠ 0 and 𝑁𝑟 = 0. After multiplying the right side of
equation 𝑌𝑛2 + 𝑛2 𝑁 = 0 by 𝑁𝑟−1 , we can get 𝑌𝑛2 𝑁𝑟−1 = 0. 𝑌
is nonsingular implies 𝑛2 𝑁𝑟−1 = 0. Repeat the operation, we
eventually get 𝑛2 = 0. Similarly, we can also get 𝑛3 = 0.
So 𝑁1 is quasidiagonal and 𝑁2 is also quasidiagonal
through similar proof; that is, 𝑛1 and 𝑛4 are square nilpotent
same as the corresponding parts of 𝑁2 . Finally, we prove that
𝑌 is 2𝑠 𝑁.
Since 𝑌 is ± 𝑃 by Step 1 and 𝑁 is ± 𝑃 by Corollary 7, it is
true that 𝑋 is ± 𝑃.
2
± 𝑃 → 𝑠 𝑁. Suppose 𝑋 ∈ 𝑀𝑛 (𝐾) is ± 𝑃. If 𝑋 is similar to
𝑌⊕𝑁, where 𝑌 is nonsingular and 𝑁 is nilpotent, then 𝑋 is ± 𝑃
if and only if 𝑌 is ± 𝑃 by Corollaries 5, 6, and 7. Without loss of
generality, we can assume 𝑋 is nonsingular. Furthermore, if
𝑋 is nonsingular and similar to 𝑌1 ⊕ 𝑌2 , where all eigenvalues
of 𝑌1 are not in 𝐾 and all eigenvalues of 𝑌2 are in 𝐾. Then 𝑋 is
± 𝑃 if and only if 𝑌1 is ± 𝑃 and 𝑌2 is ± 𝑃. It will take two steps
to prove 𝑋 is 2𝑠 𝑁.
Step 3. Suppose car(𝐾) ≠ 2 and all eigenvalues of 𝑋 are not
in 𝐾; then for arbitrary nonzero 𝛼 ∈ 𝐾, 𝑋 is an (𝛼, −𝛼)
composite; that is, there exist idempotent matrices 𝑃1 and
𝑃2 ∈ 𝑀𝑛 (𝐾) such that 𝑋 = 𝛼𝑃1 − 𝛼𝑃2 .
Let 𝑄1(0) be the eigenspace of 𝑃1 with respect to 0, 𝑄(1)
1 the
eigenspace of 𝑃1 with respect to 1, let 𝑄2(0) be the eigenspace
of 𝑃2 with respect to 0, and 𝑄2(1) the eigenspace of 𝑃2 with
respect to 1. Both 𝛼 and −𝛼 are not eigenvalues of 𝑋 implies
that 𝑄1(0) ∩ 𝑄2(0) = 𝑄1(1) ∩ 𝑄2(1) = 𝑄1(0) ∩ 𝑄2(1) = 𝑄1(1) ∩ 𝑄2(0) = {0};
then dim(𝑄1(0) ) = dim(𝑄1(1) ) = dim(𝑄2(0) ) = dim(𝑄2(1) ) =
𝑛/2 (otherwise, dim(𝑄1(0) ) ≥ 𝑛/2 implies 𝑄1(0) ∩ 𝑄2(0) ≠ {0} or
𝑄1(0) ∩ 𝑄2(1) ≠ {0}, etc.); that is, 𝑛 is even.
Suppose 𝑆 and 𝑇 are 𝑛 × (𝑛/2) matrices with 𝑟(𝑆) = 𝑟(𝑇) =
𝑛/2 satisfying 𝑃1 𝑆 = 0 and 𝑃2 𝑇 = 𝑇; then (𝑆 𝑇) is 𝑛 × 𝑛
nonsingular matrix. Let ( 𝑈
𝑉 ) be its inverse; that is,
0
𝐼
𝑈
𝑈𝑆 𝑈𝑇
( ) (𝑆 𝑇) = (
) = ( 𝑛/2 𝑛/2 ) .
0𝑛/2 𝐼𝑛/2
𝑉
𝑉𝑆 𝑉𝑇
(11)
Then we carry out the same similarity transformation on 𝑃1
and 𝑃2 as follows:
𝑈𝑃1 𝑇
0
𝑈𝑃 𝑆 𝑈𝑃1 𝑇
𝑈
( ) 𝑃1 (𝑆 𝑇) = ( 1
) = ( 𝑛/2
),
0𝑛/2 𝑉𝑃1 𝑇
𝑉𝑃1 𝑆 𝑉𝑃1 𝑇
𝑉
𝑈𝑃 𝑆 0𝑛/2
𝑈𝑃 𝑆 𝑈𝑃2 𝑇
𝑈
),
)=( 2
( ) 𝑃2 (𝑆 𝑇) = ( 2
𝑉𝑃2 𝑆 𝐼𝑛/2
𝑉𝑃2 𝑆 𝑉𝑃2 𝑇
𝑉
(12)
where 𝑃1 and 𝑃2 are idempotent implies that 𝑉𝑃1 𝑇 and 𝑈𝑃2 𝑆
are idempotent and 𝑟(𝑃1 ) = 𝑟(𝑃2 ) = 𝑛/2 implies that 𝑉𝑃1 𝑇 =
𝐼𝑛/2 and 𝑈𝑃2 𝑆 = 0𝑛/2 . Hence, 𝑋 is similar to the following
matrix:
0
0
𝛼𝑈𝑃1 𝑇
0𝑛/2
) + ( 𝑛/2
).
(13)
( 𝑛/2
0𝑛/2 0𝑛/2
−𝛼𝑉𝑃2 𝑆 0𝑛/2
That is, 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾).
4
The Scientific World Journal
When car(𝐾) = 2, 𝑋 is (𝛼, 𝛼) composite for arbitrary
nonzero 𝛼 ∈ 𝐾, we can similarly prove that 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾)
replacing −𝛼 with 𝛼 in the previous proof.
Step 4. Suppose car(𝐾) ≠ 2 and all eigenvalues of 𝑋 are in 𝐾;
then by Corollary 5, 𝑗𝑘 (𝑋, 𝛼) = 𝑗𝑘 (𝑋, −𝛼) for every 𝑘 ∈ 𝑍+
and arbitrary nonzero 𝛼 ∈ 𝐾.
Moreover, 𝑋 is similar to 𝑋1 ⊕ ⋅ ⋅ ⋅ ⊕ 𝑋𝑠 , where both the
characteristic polynomial and the minimal polynomial of 𝑋𝑖
are [(𝑥 − 𝛼𝑖 )(𝑥 + 𝛼𝑖 )]𝑟𝑖 = (𝑥2 − 𝛼𝑖2 )𝑟𝑖 with 2 ∑𝑠𝑖=1 𝑟𝑖 = 𝑛 and
𝛼𝑖 ∈ 𝐾\{0} is one of eigenvalues of 𝑋 for every 𝑖 ∈ [𝑠]. Without
loss of generality, we just need to prove 𝑋𝑖 is 2𝑠 𝑁.
Since 𝑋𝑖 is similar to 𝐶((𝑥2 − 𝛼𝑖2 )𝑟𝑖 ) as follows:
0
1
0
(
( ..
(.
..
.
0
(
0 ⋅⋅⋅ ⋅⋅⋅ 0
0 ⋅⋅⋅ ⋅⋅⋅ 0
1 0 ⋅⋅⋅ 0
.
d d d ..
𝑎0
0
𝑎2
)
.. ) ,
. )
(14)
d d 1 0 𝑎2𝑟𝑖 −2
⋅⋅⋅ ⋅⋅⋅ 0 1 0 )
where (𝑥2 − 𝛼𝑖2 )𝑟𝑖 = 𝑥2𝑟𝑖 − 𝑎2𝑟𝑖 −2 𝑥2𝑟𝑖 −2 − ⋅ ⋅ ⋅ − 𝑎2 𝑥2 − 𝑎0 . We
have 𝐶((𝑥2 − 𝛼𝑖2 )𝑟𝑖 ) = 𝐸2,1 + ⋅ ⋅ ⋅ + 𝐸2𝑟𝑖 ,2𝑟𝑖 −1 + 𝑎0 𝐸1,2𝑟𝑖 + 𝑎2 𝐸3,2𝑟𝑖 +
⋅ ⋅ ⋅ + 𝑎2𝑟𝑖 −2 𝐸2𝑟𝑖 −1,2𝑟𝑖 = (𝐸2,1 + 𝐸4,3 + ⋅ ⋅ ⋅ + 𝐸2𝑟𝑖 ,2𝑟𝑖 −1 ) + (𝐸3,2 + ⋅ ⋅ ⋅ +
𝐸2𝑟𝑖 −1,2𝑟𝑖 −2 + 𝑎0 𝐸1,2𝑟𝑖 + 𝑎2 𝐸3,2𝑟𝑖 + ⋅ ⋅ ⋅ + 𝑎2𝑟𝑖 −2 𝐸2𝑟𝑖 −1,2𝑟𝑖 ) = 𝑁1 + 𝑁2 .
Obviously, both 𝑁1 and 𝑁2 are square nilpotent matrices; that
is, 𝑋𝑖 is 2𝑠 𝑁. Hence, 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾).
When car(𝐾) = 2, all blocks in the Jordan reduction
of 𝑋 with respect to 𝛼 ∈ 𝐾 \ {0} have an even size by
Corollary 6; that is, both the characteristic polynomial and
minimal polynomial of every block with respect to 𝛼 are
(𝑥 + 𝛼)𝑠𝑖 = ((𝑥 + 𝛼)2 )𝑠𝑖 /2 = (𝑥2 + 𝛼2 )𝑠𝑖 /2 , where 𝑠𝑖 is even.
Similarly, we can also prove that 𝑋 is 2𝑠 𝑁 in 𝑀𝑛 (𝐾).
References
[1] J. D. Botha, “Sums of two square-zero matrices over an arbitrary
field,” Linear Algebra and Its Applications, vol. 436, no. 3, pp.
516–524, 2012.
[2] C. D. Pazzis, “On linear combinations of two idempotent matrices over an arbitrary field,” Linear Algebra and Its Applications,
vol. 433, no. 3, pp. 625–636, 2010.
[3] R. E. Hartwig and M. S. Putcha, “When is a matrix a difference
of two idempotents?” Linear and Multilinear Algebra, vol. 26,
no. 4, pp. 267–277, 1990.
[4] J. H. Wang and P. Y. Wu, “Sums of square-zero operators,” Studia
Mathematica, vol. 99, no. 2, pp. 115–127, 1991.
[5] F. R. Gantmacher, The Theory of Matrices, vol. 1, Chelsea, New
York, NY, USA, 1959.
[6] H. Flanders, “Elementary divisors of 𝐴𝐵 and 𝐵𝐴,” Proceedings
of the American Mathematical Society, vol. 2, no. 6, pp. 871–874,
1951.
[7] W. E. Roth, “The equations 𝐴𝑋 − 𝑌𝐵 = 𝐶 and 𝐴𝑋 − 𝑋𝐵 = 𝐶
in matrices,” Proceedings of the American Mathematical Society,
vol. 3, pp. 392–396, 1952.
[8] C. Pearcy and D. Topping, “Sums of small numbers of idempotents,” The Michigan Mathematical Journal, vol. 14, pp. 453–465,
1967.
[9] R. E. Hartwig and M. S. Putcha, “When is a matrix a sum of
idempotents?” Linear and Multilinear Algebra, vol. 26, no. 4, pp.
279–286, 1990.
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