Existence of a real $4 \times 4$ matrix satisfying the equation $x^2+1=0$.

Question

Does there exist a real $4 \times 4$ matrix satisfying the equation $x^2+1=0$?

I have tried using determinants to see if I can arrive at some sort of contradiction, but it doesn't quite help. Also, I have tried putting entries ($a,b,c,d \ldots$ etc.) in the matrix to see if it gets me anywhere, and it doesn't. And, companion matrices don't help either because the polynomial is of degree $2$ while the matrix is a real $4 \times 4$ matrix.

Is there some sort of approach that I am definitely not aware of at this point?

Answer 1

You can try with $$ A= \begin{pmatrix}0&-1\\1&0 \end{pmatrix}$$ (so $A^2+1=0$) and then $$M= \begin{pmatrix}0&A\\A&0 \end{pmatrix}$$

Answer 2

This works:

$$\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}^2=-I_4$$