Task is to find all $3 \times 3$ matrices X, $x_{ij} \in R$, such that $X^2+E=0$
I used suggestions from this question, though I stuck anyway.
$X^2=Y=-E$
Then $det(Y-\lambda E)=0$, which results $\lambda=-1$ and that gives only trivial solutions for eigenvectors.
So, would you be so kind to point a direction for me?
There is no real $3 \times 3$ matrix $X$ satisfying the equation $X^2 + E = 0$. This point may be argued as follows: since the characteristic polynomial of $X$ is of (odd) degree $3$, it has at least one real root and hence $X$ has at least one real eigenvalue $\rho$. Then there is a non-zero vector $v$ with $Xv = \rho v$. Now $X^2 + E = 0$ implies $(X^2 + E)v = 0$, or $X^2 v + E v = 0$; but
$X^2 v = X(X v) = X(\rho v) = \rho X v = \rho^2 v, \tag{1}$
so that
$0 = (X^2 + E) v = (\rho^2 + 1) v, \tag{2}$
which since $v \ne 0$ forces $\rho^2 + 1 = 0$. But no real $\rho$ can satisfy this equation. Thus such an $X$ cannot exist.
This demonstration easily generalizes to the case of $X$ $n \times n$ for $n$ odd.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
