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.

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