If $A^4=4A^2$ then $m_A(x)=x^2-4$ and if it isn't diagonalaziable over $\mathbb R$ then $0$ is an eigenvalue 
Given $A_{n\times n} \in \mathbb R$ such that $A^4=4A^2$ then

*

*if $A$ is invertible and isn't of the form $cI, c\in \mathbb R$ then $m_A(x)=x^2-4$.


*if $A$ isn't diagonalizable over  $\mathbb R$ then $0$ is an eigenvalue of $A$.

From the given we get $x^2(x-2)(x+2)=0$ zero $A$ so the eigenvalues of $A$ are from $\{0,2,-2\}$ (btw, are all of them have to be from this set?).

*

*since $A$ is invertible then it's hard to find a counter example that won't allow to lose the $x^2$ from $x^2(x-2)(x+2)=0$ , is there a theorem that say that if $A$ is invertible then $A^2\neq0$?


*here we can tell that the equation $|A-xI|=0$ doesn't have a real solution since it isn't diagonalizable over  $\mathbb R$, but $0\in \mathbb R$ so how can this be?
 A: The minimal polynomial of $A$ divides $x^4 - 4x^2 = x^2(x-2)(x+2)$ and the characteristic polynomial and the minimal polynomial have the same irreducible factors (possibly with different multiplicities), so a priori, you can only tell that the possible eigenvalues are $0,\pm 2$ and not necessarily all must occur (for example, the matrix $A = cI$ where $c \in \{ 0, 2, -2 \}$ satisfies the polynomial equation but has $c$ as the only eigenvalue).


*

*If $A$ is invertible, then $0$ can't be an eigenvalue of $A$. Thus, by the result mentioned above, the only possibilities for the minimal polynomial of $A$ are $x-2$, $x + 2$ and $x^2 - 4$. Since $A$ is not a scalar matrix, we must have $m_A(x) = x^2 - 4$.

*If $0$ is not an eigenvalue of $A$, then the minimal polynomial of $A$ must divide $(x-2)(x+2)$ and thus splits into distinct linear factors which implies that $A$ is diagonalizable.

A: You are given that $A^2(A^2-4)=0$. Now


*

*If $A$ is invertible, multiply by $A^{-2}$ to get $A^2-4=0$; if in addition it is given that $A$ is not a scalar matrix, its minimal polynomial has degree${}>1$, so must be $X^2-4$.

*Here we remark that $X^2(X^2-4)=X^2(X+2)(X-2)$ splits over$~\Bbb R$. If $A$ were annihilated by a spilt polynomial without repeated factors, then $A$ would be diagonalisable. Since it is given that this is not the case, and $X$ is the only repeated factor in the annihilating polynomial $X^2(X^2-4)$, it must be that $X^2$ divides the minimal polynomial, which in particular implies that $0$ is an eigenvalue of $A$ (and even that $0$ is a double root of the minimal polynomial).
