If the square of a matrix is negative definite, then the dimension is even.

Question

I am trying to solve the following problem. I'm studying for an exam.

Let $A$ be a real $n \times n$ matrix such that $A^2$ is negative definite. Prove that $n$ must be even.

The only hint given is that $A^2$ being negative definite means that $\langle A^2x, x \rangle < 0$ for every $x \in \mathbb{R}^n \backslash \{0\}$. It does not say, but I suppose $A$ must also be symmetric. My attempt at a solution is the following. Let $x$ be an arbitrary eigenvector of $A$ with eigenvalue $\lambda$ (negative definite real matrices are diagonalizable). We have that \begin{equation} \langle A^2x, x \rangle = (AAx)^{\top}x = x^{\top}A^2x = \lambda^2 x^{\top}x = \lambda^2 ||x||^2 < 0. \end{equation} This means that $\lambda^2 < 0$ and thus, every eigenvalue of $A$ must be complex. Hence, $n$ must be even. But we also have \begin{equation} ||Ax||^2 = \langle Ax, Ax \rangle = (Ax)^{\top}Ax = x^{\top}A^2x = \langle A^2x, x \rangle < 0, \end{equation} which makes no sense. Where have I gone wrong? Could it be that in this case negative definite does not entail symmetric? Any help is appreciated.

Answer 1

If $A$ is a real symmetric matrix, then as you showed it is impossible for $A^2$ to be negative definite. However, it is possible to have $A^2$ negative definite if $A$ is not symmetric. For example, $$A = \pmatrix{0 & 1\cr -1 & 0\cr}$$ will do it, where $A^2 = -I$.

One way to see the result is that any real $n \times n$ matrix with odd $n$ must have a real eigenvalue (because the non-real ones come in complex-conjugate pairs) and corresponding real eigenvector. If $\lambda$ and $v$ are such an eigenvalue and eigenvector, then $\langle A^2 v, v \rangle = \lambda^2 \|v\|^2 \ge 0$.