The eigenvalues of a power are powers of eigenvalues

Let $A$ be an $n \times n$ matrix over a field $F$. Show that if $\lambda_i$, $1 \leq i \leq n$ (all contained in $F$), are the eigenvalues of $A$, then $\lambda_i^k$ are the eigenvalues of $A^k$.

We begin with a lemma.

Lemma: If $A$ and $B$ are upper triangular matrices with diagonal entries all equal to $a$ and $b$, respectively, then $AB$ is upper triangular with diagonal entries all equal to $ab$. Proof: Certainly $AB$ is upper triangular. (We showed this here.) Recall now that if $A = [a_{i,j}]$, then we have $a_{i,j} = a$ if $j = i$ and $0$ if $j < i$. Similarly if $B = [b_{i,j}]$, then $b_{i,j} = b$ if $j = i$ and $0$ if $j < i$. Now $AB = [\sum_k a_{i,k}b_{k,j}]$. Consider this sum with $j = i$. If $i > k$, then $a_{i,k} = 0$. If $i < k$, then $b_{k,i} = 0$. If $k = i = j$, then $a_{i,k}b_{k,j} = ab$. In particular, the $(i,i)$-entry of $AB$ is $ab$ as desired. $\square$

Note that if $J$ is a Jordan block whose diagonal entries are all $\lambda$, then using the lemma and induction we see that $J^k$ is upper triangular with diagonal entries all equal to $\lambda^k$.

Now suppose $A$ is an arbitrary matrix with eigenvalues $\lambda_i$. By Theorem 23, $A$ is similar to a matrix $J$ in Jordan canonical form. In particular, $A$ and $J$ have the same eigenvalues. If $P^{-1}AP = J$, then $P^{-1}A^kP = (P^{-1}AP)^k = J^k$, so that $A^k$ and $J^k$ are similar, and thus have the same eigenvalues. Now $J = J_1 \oplus \cdots \oplus J_t$ where $J_i$ are Jordan blocks, so that $J^k = J_1^k \oplus \cdots \oplus J_t^k$. The eigenvalues of $J^k$ (diagonal entries) are thus $\lambda_i^k$, so that the eigenvalues of $A^k$ are $\lambda_i^k$ as desired.