TITLE: Find characteristic polynomial of $\,A^2$ if the characteristic polynomial of $\,A$ is $\,t^4 -t$
QUESTION [6 upvotes]: $A \in M_{4\times4}(\mathbb{R})$. The characteristic polynomial of $A$ is $P_A(t)=t^4-t$. I have to find the characteristic polynomial of $A^2$ and $A^4$. 

So I know that due to the Cayley–Hamilton theorem that $P_A(A) = A^4-A=0$ , therefore $A^4 = A$, Therefore  $P_{A^4}(t)=P_A(t)=t^4-t$.
But what do I do with  $A^2$?
I also know that the eigenvalues of A are $0,1$... how does this help?
Thank you all.

REPLY [3 votes]: For $\mathbb{C}$:
The eigenvalues of $A$ are $0,1,e,e^2$ where $e$ is a primitive root out $1$: $e^3=1$. Hence the eigenvalues of $A^2$ and also of $A^4$ are $0,1,e^2, e^4=e$ i.e. the same. So the characteristic polynomials of $A^2$ and $A^4$ also are $t^4-t$.
For $\mathbb{R}$:
Сharacteristic polynomials of $A^2$ and $A^4$ do not change, i.e. $t^4-t$.
Addendum: Theorem from the Gantmacher's book: If $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ and $g(x)$ a polynomial, then $g(\lambda_1),\ldots,g(\lambda_n)$ are the eigenvalues of $g(A)$ (with the same multiplicities).