block diagonal matrix | Abstract Algebra

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A matrix is nilpotent iff the trace of every power of the matrix is zero

Posted: March 25, 2024 in Basic Algebra, Matrices
Tags: block diagonal matrix, Fitting's Lemma, nilpotent matrix, trace of a matrix

If $A$ is a square matrix with entries from a field of characteristic zero such that the trace of $A^m$ is zero for all positive integers $m,$ then $A$ is nilpotent. This is a very well-known result in linear algebra and there are at least two well-known proofs of that (see here for example) that use the Vandermonde determinant and Newton identities. In this post, I’m going to give a different proof. Since I have not seen this proof anywhere, I think I can claim it’s mine! 🙂

Let $F$ be a field, and let $M_n(F)$ be the ring of $n \times n$ matrices with entries from $F.$ For $A \in M_n(F),$ let $\text{tr}(A)$ denote the trace of $A.$ If $A$ is nilpotent, then clearly $A^m$ is also nilpotent for all positive integers $m$ and hence $\text{tr}(A^m)=0.$ The well-known fact we are now going to prove is that the converse is also true if the characteristic of $F$ is zero.

Theorem. Let $F$ be a field of characteristic zero, and let $A \in M_n(F)$ such that $\text{tr}(A^m)=0$ for all integers $m \ge 1.$ Then $A$ is nilpotent.

Proof (Y. Sharifi). As we showed here as an application of Fitting’s lemma, there exists an integer $0 \le k \le n,$ an invertible matrix $Y \in M_k(F)$ and a nilpotent matrix $Z \in M_{n-k}(F)$ such that $A$ is similar to a block diagonal matrix $B=\begin{pmatrix}Y & 0 \\ 0 & Z\end{pmatrix}.$ If $k=0,$ then $B,$ hence $A,$ is nilpotent and we are done. We now suppose that $k \ge 1$ and show that this case is impossible. For any positive integer $m,$ we have

$0=\text{tr}(A^m)=\text{tr}(B^m)=\text{tr}(Y^m)+\text{tr}(Z^m)=\text{tr}(Y^m). \ \ \ \ \ \ \ \ (*)$

Let $p(t)=t^k+c_{k-1}t^{k-1}+ \cdots + c_1t+c_0, \ c_i \in F,$ be the characteristic polynomial of $Y.$ Note that since $Y$ is invertible, $c_0 \ne 0.$ By Cayley-Hamilton, $Y^k+c_{k-1}Y^{k-1} + \cdots + c_1Y+c_0I=0,$ where $I$ is the $k \times k$ identity matrix. Thus, taking the trace of both sides and using $(*)$ , gives $kc_0=0,$ which is not possible since $c_0 \ne 0$ and $F$ has characteristic zero. $\ \Box$

Remark 1. The Theorem also holds true if $F$ has positive characteristic $p > n.$ To see that, look at the last sentence in the proof of the Theorem again. We got $kc_0=0, \ c_0 \ne 0,$ which implies $k1_F=0$ and so $p \mid k,$ which is not possible since $k \le n < p.$

Remark 2. The Theorem does not necessarily hold if $F$ has positive characteristic $p \le n.$ For example, choose $A \in M_2(\mathbb{Z}_2)$ to be the identity matrix. Then $\text{tr}(A^m)=0$ for all positive integers $m$ but $A$ is not nilpotent.

The Theorem has many applications; let me give you one of them here.

The following problem was posted on the Art of Problem Solving a few weeks ago; you can see the problem and my solution (post #4) here. The proposer assumes that matrices have real entries but that is not necessary; the entries can come from any field of characteristic zero $F.$ For any $A,B \in M_n(F),$ we denote by $[A,B]$ the additive commutator of $A,B,$ i.e. $[A,B]=AB-BA.$ Clearly $[\ , \ ]$ is bilinear and $\text{tr}([A,B])=0.$

Problem (V. Brayman). Let $F$ be a field of characteristic zero, and let $\{A,B_k, \ k \ge 1\} \subset M_n(F)$ be such that $[B_i,B_j]=A^{j-i}$ for all $j > i \ge 1.$ Prove that $A=0.$

Solution (Y. Sharifi). First notice that for any positive integer $k,$

$\text{tr}(A^k)=\text{tr}([B_1,B_{k+1}])=0,$

and so, by the Theorem, $A$ is nilpotent. Let $m$ be the smallest positive integer such that $A^m=0.$
If $m=1,$ we are done. Suppose now that $m \ge 2.$ Since $\dim_F M_n(F)=n^2 < \infty,$ the set $\{B_k, \ k \ge 1\}$ is $F$ -linearly dependent and so there exist an integer $p \ge 2$ and $c_i \in F$ such that $B_p=\sum_{i=1}^{p-1}c_iB_i.$ But then

$\displaystyle \begin{aligned}A^{m-1}=[B_p,B_{m+p-1}]=[\sum_{i=1}^{p-1}c_iB_i,B_{m+p-1}]=\sum_{i=1}^{p-1}c_i[B_i,B_{m+p-1}]=\sum_{i=1}^{p-1}c_iA^{m+p-1-i}=0,\end{aligned}$