Trace of matrix in power of $k$ (real, symmetric)

Question

I found that if the matrix is real and symmetric then:

$$ \operatorname{Tr} \left( X^k \right) = \operatorname{Tr} \left(Q\Lambda^kQ^{-1} \right) = \operatorname{Tr} \left( \Lambda^k \right) = \sum_i \lambda_i ^k $$

where $Q\Lambda Q^{-1}$ is an eigendecomposition. I understand all steps but

$$ \operatorname{Tr} \left( Q\Lambda^kQ^{-1} \right)= \operatorname{Tr} \left( \Lambda^k \right)$$

Why is it right?

Answer 1

Because $\;\operatorname{tr}(AB)=\operatorname{tr}(BA)\;$ , so

$$\operatorname{tr}(Q\Lambda^kQ^{-1})=\operatorname{tr}\left(Q(\Lambda^kQ^{-1})\right)=\operatorname{tr}((\Lambda^kQ^{-1})Q)=\operatorname{tr}(\Lambda^k(Q^{-1}Q))=\operatorname{tr}(\Lambda^kI)=\operatorname{tr}(\Lambda^k)$$

Answer 2

Another reason is that the trace is the sum of the eigenvalues, and two similar matrices have the same eigenvalues.