Showing there exists an $n\times n$ matrix that solves a given polynomial iff $n$ is even

Question

Prove or disprove: There exists a real $n\times n$-matrix $A$ satisfying: $$ A^2+2A+5I_n=0 $$

if and only if $n$ is even.

If $n=2$ this is quite easy, we can just compute the companion matrix. However, I'm not quite sure how to show this for other even numbers. If we consider blockmatrices, with the 2x2 companion matrix in the bottom right, it doesn't work. Also the $n$ x $n$ companion matrix with $c_0, ..., c_{n-3} = 0, c_{n-2} = 5, c_{n-1}=2$ doesn't seem to work. I can't seem to find a counterexample for odd matrices either since the roots of the polynomial are complex.
How else can one approach this problem using Cayley-Hamilton theorem and minimal polynomials?

Answer 1

If $n=2k$, then consider, say, the block matrix $$\begin{bmatrix}M&0&0&\ldots&0\\0&M&0&\ldots&0\\0&0&M&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&M\end{bmatrix},$$where each $0$ is actually the $2\times2$ null matrix and $M=\left[\begin{smallmatrix}0&1\\-5&-2\end{smallmatrix}\right]$; the whole matrix is a $(2k)\times(2k)$ matrix.

If $n$ is odd, $A$ has some real eigenvalue $\lambda$. But then you would have $\lambda^2+2\lambda+5=0$, which is impossible.

Answer 2

$$A^2+2A+5I_n=0\iff\left(\frac{A+I_n}2\right)^2=-I_n,$$ so you just have to prove that $\exists B\in M_n(\Bbb R), B^2=-I_n$ iff $n$ is even.

$\implies:$ if $B^2=-I_n$ then $0\le(\det(B))^2=\det(-I_n)=(-1)^n$ hence $n$ is even.

$\Longleftarrow:$ if $n$ is even, $n=2m$, then $$\begin{pmatrix}0&-I_m\\I_m&0\end{pmatrix}^2=-I_n.$$