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

Let $A$ be an $n \times n$ matrix with eigenvalues $\lambda_{1},...\lambda_{n}$. How do I prove that tr$(A^k) = \sum_{i=1}^{n}\lambda_{i}^{k}$?

Answer 1

The trace is similarity-invariant. Supposing the multiplicity of each eigenvalue to be 1. Let $A^k = U^{-1}\Sigma^k U,$ where the matrix $U$ is composed by the eigenvectors of $A$ and $\Sigma$ is the diagonal matrix with the eigenvalues of $A$. So we have: $$\mathrm{tr}(A^k) = \mathrm{tr}(U^{-1}\Sigma^k U) = \mathrm{tr}(U^{-1}(\Sigma^k U)) = \mathrm{tr}((\Sigma^k U)U^{-1}) = \mathrm{tr}(\Sigma^k (U U^{-1})) = \mathrm{tr}(\Sigma^k I) = \mathrm{tr}(\Sigma^k).$$

Thus as $\Sigma = \left(\lambda_i \mathbb{1}_{\{i=j\}} \;;\; (i,j) \in [0,n]^2\right) $, we have $\Sigma^k = \left(\lambda_j^k \mathbb{1}_{\{i=j\}} \;;\; (i,j) \in , [0,n]^2\right) $ and, therefore, $$\mathrm{tr}(\Sigma^k) = \sum_{i=1}^n \lambda_i^k$$

Also, this argument may be extended easily to any matrix A n-by-n using the Jordan form as in wiki

Answer 2

Sum of roots of a polynomial is given by $-\frac{b}{a}$, if the polynomial looks like $a\lambda^N+b\lambda^{N-1}+\dots$. Now, eigenvalues are roots of polynomial $\det(A-\lambda I)$. Try to prove that $-\frac{b}{a}$ in this case is $\mathrm{trace}(A)$. See that $A^kv=A^{k-1}(Av)=A^{k-1}(\lambda_i v)=\lambda_i^k v$ (where $\lambda$ is a eigenvalue). Thus $\lambda_i^k$ are eigenvalues of $A^k$. Thus $\lambda_i^k$ are roots of polynomial $\det (A^k-\lambda I)$. Thus, their sum should equal $\mathrm{trace}(A^k)$.

Answer 3

1. $tr(A)=\sum \lambda_i$
2. 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).