# A Note on Decomposing a Square Matrix as Sum of Two Square Nilpotent Matrices over an Arbitrary Field

Authors: Xiaofei Song; Baodong Zheng; Chongguang Cao
Journal: The Scientific World Journal
 (2013)

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## Abstract

LetK 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.


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## Body

## 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 s2N 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).ForX∈Mn(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 [2]; 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 [1] from the results obtained in [2]. In fact, the following theorem is the main result of this paper.Theorem 1 (main theorem).
SupposeK is an arbitrary field and X∈Mn(K); then X is s2N in Mn(K) if and only if X is ±P.In Section2, 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

SupposeX∈Mn(K) and Xi∈Mni(K), we denote by X=X1⊕⋯⊕Xs the following matrix with ∑i=1s‍ni=n:
(1)(X1X2⋱Xs).Notation 1 (Notation 2 in [2]).
LetX∈Mn(K), λ∈K¯ and k∈Z+; we denote by(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 forallk∈Z+, vk≥uk+1,  and uk≥vk+1.Notation 2 (Notation 4 in [2]).
Given a monic polynomial,P=xn-an-1xn-1-⋯-a1x-a0, denote the following C(P) by its companion matrix:
(2)C(P)=(00⋯⋯0a0100⋯0a1010⋱0a2⋮⋱⋱⋱⋱⋮⋮⋯⋱10an-20⋯⋯01an-1).Theorem 3 (Theorem 1 in [2]).
Assumecar(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)
forallλ∈K¯∖{0,α,-α} and for all k∈Z+, jk(X,λ)=jk(X,-λ).Theorem 4 (Theorem 5 in [2]).
Assumecar(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.SupposeX∈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)
forallλ∈K¯∈{0,α,-α} and for all k∈Z+, jk(X,λ)=jk(X,-λ);(2)
forallλ∈K¯∈{0,β,-β}andforallk∈Z+, jk(X,λ)=jk(X,-λ). so forallλ∈K¯∖{0} and forallk∈Z+, jk(X,λ)=jk(X,-λ).On the other hand, note that for nonzeroα∈K¯ with car(K)≠2, the sequences (nk(X,α))k≥1 and (nk(X,-α))k≥1 are intertwined if forallk∈Z+, jk(X,α)=jk(X,-α). Then forallλ∈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.
Assumecar(K)≠2 and let X∈Mn(K). Then X is ±P if and only if forallλ∈K¯∖{0}forallk∈Z+, jk(X,λ)=jk(X,-λ).Similarly, we can derive the following corollary from Theorem4.Corollary 6.
Assumecar(K)=2 and let X∈Mn(K). Then X isP± 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 isP±.In fact, arbitrary nilpotent is not onlyP± but also s2N.Lemma 8.
Every nilpotentN∈Mn(K) is s2N.Proof.
For arbitrary fieldK, let N∈Mn(K) is nilpotent; then N is similar to N1⊕N2⊕⋯⊕Ns, where for every i∈[s], Ni∈Mri(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)(00⋯010⋯0⋮⋱⋱⋮0⋯10)ri×ri.
That is, C(xri)=E2,1+E3,2+⋯+Eri,ri-1∈Mri(K).
Whenri 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 X∈Mn(K) is s2N 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.
IfX is nonsingular, then X is P±.
SinceX=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 x∈Mn,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.
BecauseN1 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 α∈L, 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.
IfX 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 thatY is s2N. Without loss of generality, we assume X=Y⊕N in the following proof since s2N holds under similarity transformations.
LetN1=(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 thatN is nilpotent, assume its index is r; that is, Nr-1≠0 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.
SoN1 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.
SinceY is P± by Step 1 and N is P± by Corollary 7, it is true that X is P±.
±
P
→
s
2
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 s2N.Step 3.
Supposecar(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.
LetQ1(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.
SupposeS 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).
Whencar(K)=2, X is (α,α) composite for arbitrary nonzero α∈K, we can similarly prove that X is s2N in Mn(K) replacing -α with α in the previous proof.Step 4.
Supposecar(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∑i=1s‍ri=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 s2N.
SinceXi is similar to C((x2-αi2)ri) as follows:
(14)(00⋯⋯0a010⋯⋯00010⋯0a2⋮⋱⋱⋱⋮⋮⋮⋱⋱10a2ri-20⋯⋯010),
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).
Whencar(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 s2N in Mn(K).



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