Looking for an identity for characteristic polynomial of a matrix to the power of n

Question

Is there any identity to help finding the characteristic polynomial of a power of a matrix in terms of its own characteristic polynomial?

Let $A\in \mathbb{Z}^{m\times m}$, we have $$p(t)=\text{det}(A-t I)$$ If we define $$q(t)=\text{det}(A^n-t I)\,\,,n\in\mathbb{N}$$ Is there any identity relating $p(t)$ to $q(t)$?

If there is no general relation between $p(t)$ and $q(t)$, is there any relation to compute $q(1)$ from $p(1)$?

Answer 1

Now if $A$ is any martix, then there is an invertable matrix $G$, such that $GAG^{-1}$ is upper-triangular. Since we have $det(GAG^{-1})=det(A)$, $det(GAG^{-1}-\lambda I)=det(G(A-\lambda I)G^{-1})=det(A-\lambda I)$, and $GA^nG^{-1}=(GAG^{-1})^n$, we may assume that $A$ is upper triangular. Now note that if $A$ is upper-triangular, write $A=D+N$ for $D$ the diagonal and $N$ a strictly upper-triangular matrix. Then we have that $A^n=(D+N)^n=\sum_i D^iN^{n-i}\binom{n}{i}=D^n+N'$, where $N'$ is strictly upper triangular. We have that $p_{A^n}(\lambda)=det(A^n-\lambda I)=det(D^n-\lambda I)$ since the strictly upper triangular part doesn't affect the determinant. We also have $p_A(\lambda)=det(D-\lambda I)$. We thus can see, by the formula for the determinant of a diagonal matrix, that if $p_A(\lambda)=\prod_i (\lambda-y_i)$ then $p_A(\lambda)=\prod_i (\lambda-y_i^n)$.

Answer 2

If $$A=\pmatrix{r&0\cr0&r/(r-1)\cr}$$ then $p(1)=1$ but (using $n=2$) $q(1)=(2r^2+r-1)/(r-1)$ if I've done my arithmetic correctly. This shows there is no hope of computing $q(1)$ if all you know is $p(1)$.