Prove that if $I-A$ is invertible then $I-A^p$ is invertible in $\mathrm{Mat}_n(\mathbb{Z}_p)$. How do you prove that if $I-A$ invertible then $I-A^p$ is invertible for  $A \in \mathrm{Mat}_n(\mathbb{Z}_p)$, where $p$ is prime?
 A: Given the tag (finite-fields), I assume that $\mathbb Z_p$ is the finite field with $p$ elements. Then we can argue as follows.
Consider the polynomial ring $\mathbb Z_p[X,Y]$ with indeterminates $X$ and $Y.$ There, we have
$$
(X+Y)^p = \sum_{j=0}^p{p\choose j}X^jY^{p-j}.
$$
Note (or check) that, for $1 \leq j \leq p-1,$ the binomial coefficient $p\choose j$ is a multiple of $p,$ and hence
$$
{p\choose j} = 0\quad in\ \mathbb Z_p\quad for\ 1 \leq j \leq p-1.
$$
This implies
$$
(X+Y)^p = X^p + Y^p\quad in\ \mathbb Z_p[X,Y].
$$
Also, the matrices $I$ and $-A$ commute (obviously): $I(-A) = (-A)I.$ Thus, we can replace the indeterminates $X$ and $Y$ with the matrices $I$ and $-A$ (technically speaking, we're using a universal property of polynomial rings), and conclude that
$$
(I-A)^p = I^p + (-A)^p = I-(A^p) \qquad\qquad\qquad\qquad\qquad\qquad\qquad (*)
$$
(observe (or check) that $(-A)^p = -(A^p)$ is true for any $p$).
From $(*),$ we see that, if $I-A$ is invertible, then $I - (A^p) = (I-A)^p$ is invertible, since a product of invertible matrices is invertible.
A: The eigenvalues of $I-A^p$ are of the form $1-\lambda^p$ where $\lambda$ ranges over the eigenvalues of $A$, so $I-A^p$ has 0 as am eigenvalue if and only if $1-\lambda^p=0$ for some eigenvalue $\lambda$ of $A$. But this is equivalent to $\lambda^p=1$, which is equivalent to $\lambda=1$ (since the Frobenius map $x \mapsto x^p$ is injective), and this in turn is equivalent to $I-A$ having 0 as an eigenvalue.
