IMC 2022 · Day 1 · P2

Hint: Consider the eigenvalues of $A$.

Taking the transpose of the matrix equation and substituting we have $$A^T + (A^T)^k = A \ \Longrightarrow \ A + A^k + (A + A^k)^k = A \ \Longrightarrow \ A^k \bigl( I + (I + A^{k-1})^k \bigr) = 0.$$ Hence $p(x) = x^k (1 + (1 + x^{k-1})^k)$ is an annihilating polynomial for $A$. It follows that all eigenvalues of $A$ must occur as roots of $p$ (possibly with different multiplicities). Note that for all $x \in \mathbb{R}$ (this can be seen by considering even/odd cases on $k$), $$(1 + x^{k-1})^k \ge 0,$$ and we conclude that the only eigenvalue of $A$ is 0 with multiplicity $n$.

Thus $A$ is nilpotent, and since $A$ is $n \times n$, $A^n = 0$. It follows $A^k = 0$, and $A = A^T$. Hence $A$ can only be the zero matrix: $A$ is real symmetric and so is orthogonally diagonalizable, and all its eigenvalues are 0.

Remark. It's fairly easy to prove that eigenvalues must occur as roots of any annihilating polynomial. If $\lambda$ is an eigenvalue and $v$ an associated eigenvector, then $f(A) v = f(\lambda) v$. If $f$ annihilates $A$, then $f(\lambda) v = 0$, and since $v \ne 0$, $f(\lambda) = 0$.

If $\lambda$ is an eigenvalue of $A$, then $\lambda + \lambda^k$ is an eigenvalue of $A^T = A + A^k$, thus of $A$ too. Now, if $k$ is odd, then taking $\lambda$ with maximal absolute value we get a contradiction unless all eigenvalues are 0. If $k$ is even, the same contradiction is obtained by comparing the traces of $A^T$ and $A + A^k$.

Hence all eigenvalues are zero and $A$ is nilpotent. The hypothesis that $k \ge n$ ensures $A = A^T$. A nilpotent self-adjoint operator is diagonalizable and is necessarily zero.