$A^2=-I_4$. Find possible values of minimal polynomial and characteristic polynomial Let $A\in\mathbb{R}^{4\times 4}$ satisfy $$A^2=-I_4 .$$

(a) Find possible values of $m_a$ (minimal polynomial) and $p_a$ (characteristic polynomial).
(b) Find an example for A satisfying the condition.

Please help me approach the first question. I can assume (b) would immediately follow.
 A: The minimal polynomial is $m(x)=x^2+1$ because it's irreducible over $\mathbb{R}$. We know $m(A)=0$ and (due to Cayley-Hamilton) $p(A)=0,$ Since $\operatorname{deg}(p)=4,$ the characteristic polynomial is $p(x)=(x^2+1)^2=x^4+2x^2+1.$ Note, that the minimal polynomial and the characteristic polynomial share roots. That is, we cannot add roots to the characteristic polynomial that are not present in the minimal polynomial already.
A: (a) The polynomial $f(x) = x^2 + 1$ vanishes at $A$. Therefore $m_A$ divides $f$. As $f$ is irreducible in $\mathbb R(x)$, $m_A=f$.
$p_A$ is a real polynomial. Its complex roots are the ones of $m_A$. And by Cayley-Hamilton theorem $m_A$ divides $p_A$. Hence $p_A=(x^2+1)^2$ as the degree of $p_A$ is equal to $4$.
(b) Now, consider a non-zero vector $u$. $(e_1 = u, e_2 = A \cdot u)$ is a linear independent family of vectors (see the note at the end of the post). You have $A \cdot e_1 = e_2$ and $A \cdot e_2 = A^2 \cdot u = -u = -e_1$. Therefore, the restriction of $A$ to the plane $(e_1, e_2)$ has for matrix:
$$\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}.$$
Find a vector $e_3$ such that $(e_1, e_2,e_3)$ is linear independent. This is possible as your space dimension is equal to $4$. Then you'll be able to prove that $(e_1, e_2,e_3, A \cdot e_3)$ is also linear independent. The matrix of $A$ in this basis is 
$$\begin{pmatrix}
0 & -1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 0
\end{pmatrix}.$$
And we have proven that $A$ is always similar to such a matrix.
Note: proof that $(u, A \cdot u)$ are linear independent for $u \neq 0$.
Suppose that $\alpha u + \beta Au = 0$ with $\alpha$ and $\beta$ non-zero. Then applying $A$ on both sides of the equality $\alpha Au - \beta u = 0$. Multiply the first equality by $\alpha$, the second one by $\beta$ and substract both resulting equalities. You get $(\alpha^2 + \beta^2) u=0$. As $u$ is supposed to be non-zero, this implies $\alpha = \beta =0$.
A: We have
$$
A^2 = -I\\
A^2 +I = 0
$$
and we direclty read off the polynomial $f(x) = x^2 + 1$, with $f(A) = 0$. Now you just need to figure out how the minimal and characteristic polynomials both relate to any given polynomial where $A$ is a root.
