TITLE: How do I prove that the trace of a matrix to its $k$th power is equal to the sum of its eigenvalues raised to the $k$th power?
QUESTION [14 upvotes]: Let $A$ be an $n \times n$ matrix with eigenvalues $\lambda_{1},...\lambda_{n}$. How do I prove that tr$(A^k) = \sum_{i=1}^{n}\lambda_{i}^{k}$?

REPLY [10 votes]: $tr(A)=\sum \lambda_i$ 
If $\lambda_i$ is eigenvalue of $A$, then $\lambda_i^k$  is eigenvalue of $A^k$. This mapping preserves multiplicities.

The first one is a classic result, easily deduced from the characteristic polynomial. The second one is a little trickier (if there are repeated eigenvalues).