Cool simple solution to: $-A^2$ is not the identity matrix this is not a question per se, just a simple cool solution to a potentially difficult question, that I want to share. I liked it.
The question is:
Let $A$ be a $3\times3$ matrix with real values.
Show that $A^2 \neq -I_3$
There are probably many solutions for this problem, but i really thought this one was simple.
Solution is posted down below.
 A: Hint: What can $\operatorname{det}(A)$ be?
A: Solution:
Let's suppose that $A^2 = -I_3$, then $A^2+I_3=0$
That means that $A$ is a root of the polynomial $t^2+1$.
However, since we are dealing with real values only, this polynomial has no roots, and so it is irreducible in $\mathbb R$.
But it is still a polynomial that $A$ is a root of, and so, from the definition of the minimal polynomial (let's call it $m_A$):
$m_A|t^2+1$ but $t^2+1$ is irreducible and so $m_A=t^2+1$
Invoking the Cayley-Hamilton theorem, we know that the characteristic polynomial of $A$ , $p_A$ can be divided by the minimal polynomial, and it has the same roots, so we can say that:
$p_A =(m_A)^r = (t^2+1)^r$, $r \in \mathbb N$
But since $A$ is a 3-by-3 matrix, the characteristic polynomial should be in a degree of 3:
$deg(p_A)=deg((t^2+1)^r) = 2r = 3$ so 3 is an even number. Contradiction!
A: Well, you can go much simpler. You have that the eigenvalues of $-A^2$ are $-\lambda^2$, where $\lambda$ runs through the eigenvalues of $A$. Since $A$ is real odd-dimensional, it has a real eigenvalue $\lambda$ and then $-\lambda^2<0$, which is not possible for the identity matrix.
A: Suppose $A^2=-1$. Then $$(\det A)^2=\det (?)={?^?}\det(?)={?},$$ a contradiction
