TSWJ The Scientific World Journal 1537-744X Hindawi Publishing Corporation 640350 10.1155/2013/640350 640350 Research Article A Note on Decomposing a Square Matrix as Sum of Two Square Nilpotent Matrices over an Arbitrary Field http://orcid.org/0000-0001-6844-2321 Song Xiaofei 1 Zheng Baodong 1 Cao Chongguang 2 Badawi A. Bracken P. 1 Department of Mathematics Harbin Institute of Technology Harbin 150001 China hit.edu.cn 2 School of Mathematical Sciences Heilongjiang University Harbin 150080 China hlju.edu.cn 2013 11 11 2013 2013 31 08 2013 01 10 2013 2013 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 αL, there exist idempotent matrices P1 and P2 over L such that X=αP1αP2.

1. Introduction

Botha (see ) 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 ) 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  with the result obtained in . However, the relation between these two facts has not been formally discussed yet (more details in ).

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]={zZ+1zs} for some sZ+. 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 AMm,n(K). E denotes a vector space over K and dim(E) is the dimension of E. XMn(K) is called   s2N in Mn(K) if there exist square nilpotent N1 and N2Mn(K) such that X=N1+N2, while X is called an (α,β) composite in Mn(K) if there exist idempotent P1 and P2Mn(K) such that X=αP1+βP2, where α, βK{0} (Definition 1 in ); 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 αL (when car(K)=2, we still use ±P for the meaning of (α,α) composites).

For XMn(K), on the one hand, we will prove that X is s2N in Mn(K) implies X is ±P; that is, the form provided by Botha satisfies the condition as in ; on the other hand, we will also prove that X is ±P implies X is s2N in Mn(K); that is, we can derive the form provided in  from the results obtained in . In fact, the following theorem is the main result of this paper.

Theorem 1 (main theorem).

Suppose K is an arbitrary field and XMn(K); then X is s2N in Mn(K) if and only if X is ±P.

In Section 2, we will state some related theorems and notations from  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 XMn(K) and XiMni(K), we denote by X=X1Xs the following matrix with i=1sni=n: (1)(X1X2Xs).

Notation 1 (Notation 2 in [<xref ref-type="bibr" rid="B2">2</xref>]).

Let XMn(K), λK¯ and kZ+; we denote by

jk(X,λ) the number of blocks of size k for the eigenvalue λ in the Jordan reduction of X;

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 [<xref ref-type="bibr" rid="B2">2</xref>]).

Two sequences (uk)k  1 and (vk)k  1 are side to be intertwined if for  all  kZ+, vkuk+1, and ukvk+1.

Notation 2 (Notation 4 in [<xref ref-type="bibr" rid="B2">2</xref>]).

Given a monic polynomial, P=xn-an-1xn-1--a1x-a0, denote the following C(P) by its companion matrix: (2)C(P)=(000a01000a10100a210an-2001an-1).

Theorem 3 (Theorem 1 in [<xref ref-type="bibr" rid="B2">2</xref>]).

Assume car (K)2 and let XMn(K). Then X is an (α,-α) composite if and only if all the following conditions hold.

The sequences (nk(X,α))k  1 and (nk(X,-α))k  1 are intertwined;

for  all  λK¯{0,α,-α} and for all kZ+, jk(X,λ)=jk(X,-λ).

Theorem 4 (Theorem 5 in [<xref ref-type="bibr" rid="B2">2</xref>]).

Assume car(K)=2 and let XMn(K). Then X is an (α,-α) composite if and only if for every λK¯{0,α}, all blocks in the Jordan reduction of X with respect to λ have an even size.

Suppose XMn(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, βL{0} with α±β. By Theorem 3, the following statements are true:

for  all  λK¯{0,α,-α} and for all kZ+, jk(X,λ)=jk(X,-λ);

for  all  λK¯{0,β,-β}  and  for  all  kZ+, jk(X,λ)=jk(X,-λ).

so for  all  λK¯{0} and for  all  kZ+, jk(X,λ)=jk(X,-λ).

On the other hand, note that for nonzero αK¯ with car(K)2, the sequences (nk(X,α))k  1 and (nk(X,-α))k  1 are intertwined if for  all  kZ+, jk(X,α)=jk(X,-α). Then for  all  λK¯{0}, kZ+, jk(X,λ)=jk(X,-λ) implies that for every algebraic extension L of K and arbitrary nonzero αL, 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 XMn(K). Then X is ±P if and only if for  all   λK¯{0}   for  all   kZ+, jk(X,λ)=jk(X,-λ).

Similarly, we can derive the following corollary from Theorem 4.

Corollary 6.

Assume car(K)=2 and let XMn(K). Then X isP± if and only if for every λK¯{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   s2N.

Lemma 8.

Every nilpotent NMn(K) is s2N.

Proof.

For arbitrary field K, let NMn(K) is nilpotent; then N is similar to N1N2Ns, where for every i[s], NiMri(K), i=1sri=n, and both the characteristic polynomial and minimal polynomial of Ni are xri. Furthermore, Ni is similar to C(xri) as follows: (3)(000100010)ri×ri. That is, C(xri)=E2,1+E3,2++Eri,ri-1Mri(K).

When ri is even, C(xri)=j=1ri/2E2j,2j-1+j=1ri/2-1E2j+1,2j; when ri is odd, C(xri)=j=1(ri-1)/2E2j,2j-1+j=1(ri-1)/2E2j+1,2j. Note that both E2j,2j-1 and E2j+1,2j are square nilpotent matrices then C(xri) is s2N, and Ni is s2N follows. Hence N is s2N.

3. Proof of Main Theorem

s 2 N P ± . Suppose XMn(K) is s2N in Mn(K); that is, there exist square nilpotent matrices N1 and N2Mn(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. (4)r(N1)+r(N2)n, 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 xMn,1(K) such that N1x=N2x=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 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 (βγ) is nonsingular.

N 1 2 γ = 0 implies that nonzero column vectors of N1γ are eigenvectors of N1 and N1(βγ)=(0N1γ) implies r(N1γ)=n/2. Hence; N1γ and β are equal under certain column transformation; that is, there is an invertible matrix T1 such that N1γ=βT1. Correspondingly, there is an invertible matrix T2 such that N2β=γT2.

Let (y1y2) be the inverse of (βγ), where y1 and y2 are (n/2)×n matrices. Naturally, the following equation is true: (5)(y1y2)(βγ)=(y1βy1γy2βy2γ)=(In/20n/20n/2In/2). Now, we carry out the same similarity transformation on N1 and N2 as follows: (6)(y1y2)N1(βγ)=(y1N1βy1N1γy2N1βy2N1γ),(y1y2)N2(βγ)=(y1N2βy1N2γy2N2βy2N2γ). Note that N1γ=βT1 and N2β=γT2, the above three equations imply that N1 is similar to (0n/2T10n/20n/2)Mn(K) and N2 is similar to (0n/20n/2T20n/2)Mn(K).

Hence, X is similar to (0n/2T10n/20n/2)+(0n/20n/2T20n/2). For every algebraic extension L of K and arbitrary nonzero αL, X is also similar to the following matrix: (7)α(In/2α-1T10n/20n/2)-α(In/20n/2-α-1T20n/2) That is, X is P±.

Step 2.

If X is singular and similar to YN, where Y is nonsingular and N is nilpotent. Then X is P±.

At first, we need to prove that Y is s2N. Without loss of generality, we assume X=YN in the following proof since s2N holds under similarity transformations.

Let N1=(n1n2n3n4), 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: (8)n12+n2n3=0,n1n2+n2n4=0n3n1+n4n3=0,n3n2+n42=0.

Since (X-N1)2=N22=0, we get the following equations after replacing n1 with Y-n1 and n4 with N-n4 in the previous equations: (9)(Y-n1)2+n2n3=0,(Y-n1)n2+n2(N-n4)=0,n3(Y-n1)+(N-n4)n3=0,n3n2+(N-n4)2=0.

We can derive the following equations from the 3rd and 4th equations in the above two sets of equations: (10)Yn2+n2N=0,  n3Y+Nn3=0.

Note that N is nilpotent, assume its index is r; that is, Nr-10 and Nr=0. After multiplying the right side of equation Yn2+n2N=0 by Nr-1, we can get Yn2Nr-1=0. Y is nonsingular implies n2Nr-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   s2N.

Since Y is P± by Step 1 and N is P± by Corollary 7, it is true that X is P±.

± P s 2 N . Suppose XMn(K) is P±. If X is similar to YN, 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 Y1Y2, 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   s2N.

Step 3.

Suppose car(K)2 and all eigenvalues of X are not in K; then for arbitrary nonzero αK, X is an (α,-α) composite; that is, there exist idempotent matrices P1 and P2Mn(K) such that X=αP1-αP2.

Let Q1(0) be the eigenspace of P1 with respect to 0, Q1(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 P1S=0 and P2T=T; then (ST) is n×n nonsingular matrix. Let (UV) be its inverse; that is, (11)(UV)(ST)=(USUTVSVT)=(In/20n/20n/2In/2). Then we carry out the same similarity transformation on P1 and P2 as follows: (12)(UV)P1(ST)=(UP1SUP1TVP1SVP1T)=(0n/2UP1T0n/2VP1T),(UV)P2(ST)=(UP2SUP2TVP2SVP2T)=(UP2S0n/2VP2SIn/2), where P1 and P2 are idempotent implies that VP1T and UP2S are idempotent and r(P1)=r(P2)=n/2 implies that VP1T=In/2 and UP2S=0n/2. Hence, X is similar to the following matrix: (13)(0n/2αUP1T0n/20n/2)+(0n/20n/2-αVP2S0n/2). That is, X is s2N in Mn(K).

When car(K)=2, X is (α,α) composite for arbitrary nonzero αK, we can similarly prove that X is s2N 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 kZ+ and arbitrary nonzero αK.

Moreover, X is similar to X1Xs, where both the characteristic polynomial and the minimal polynomial of Xi are [(x-αi)(x+αi)]ri=(x2-αi2)ri with 2i=1sri=n and αiK{0} is one of eigenvalues of X for every i[s]. Without loss of generality, we just need to prove Xi is   s2N.

Since Xi is similar to C((x2-αi2)ri) as follows: (14)(000a010000100a210a2ri-20010), where (x2-αi2)ri=x2ri-a2ri-2x2ri-2--a2x2-a0. We have C((x2-αi2)ri)=E2,1++E2ri,2ri-1+a0E1,2ri+a2E3,2ri++a2ri-2E2ri-1,2ri=(E2,1+E4,3++E2ri,2ri-1)+(E3,2++E2ri-1,2ri-2+a0E1,2ri+a2E3,2ri++a2ri-2E2ri-1,2ri)=N1+N2. Obviously, both N1 and N2 are square nilpotent matrices; that is, Xi is s2N. Hence, X is s2N in Mn(K).

When car(K)=2, all blocks in the Jordan reduction of X with respect to αK{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 s2N in Mn(K).

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