TITLE: Eigenvalues and power of a matrix
QUESTION [40 upvotes]: Let $A$ be an n×n matrix with eigenvalues $\lambda_i, i=1,2,\dots,n$. Then
$\lambda_1^k,\dots,\lambda_n^k$ are eigenvalues of $A^k$.

I was wondering if $\lambda_1^k,\dots,\lambda_n^k$ are all the
eigenvalues of $A^k$?
Are the algebraic and geometric multiplicities of $\lambda_i^k$ for
$A^k$ same as those of $\lambda_i$ for $A$ respectively?

Thanks!

REPLY [4 votes]: If A is not diagonalizable, there exists nonsingular matrix P such that $$\mathbf{P^{-1}AP}=\mathbf{J}=
\begin{bmatrix}
    J_1 & 0& 0 & \dots  & 0 \\
     0  & J_2 & 0& \dots & 0 \\
    \vdots & \vdots& \vdots & \vdots  &\vdots\\
0&0&0&\dots&J_l
\end{bmatrix}$$
with $$J_i=\begin{bmatrix}
    \lambda_i & 1& 0 & \dots  & 0 \\
     0  & \lambda_i & 1& \dots & 0 \\
    \vdots & \vdots& \vdots & \vdots  &\vdots\\
0&0&\dots&0&\lambda_i
\end{bmatrix}.$$
So $$J_i^k=\begin{bmatrix}
    \lambda^k_i & C_k^1\lambda_i^{k-1}& \dots& \dots  & C_k^{k-n_i-1}\lambda_i^{k-n_i-1} \\
     0  & \lambda^k_i & \dots& \dots & \vdots \\
    \vdots & \vdots& \vdots & \vdots  &\vdots\\
   0  & 0 & \dots& \dots & C_k^1\lambda_i^{k-1}\\
   0&0&\dots&\dots&\lambda^k_i
\end{bmatrix}.$$
Then $\mathbf{J}^k=\mathbf{P}^{-1}\mathbf{A}^k\mathbf{P}$,  so $\mathbf{J}^k=\begin{bmatrix}
    J^k_1 & 0& 0 & \dots  & 0 \\
     0  & J^k_2 & 0& \dots & 0 \\
    \vdots & \vdots& \vdots & \vdots  &\vdots\\
0&0&0&\dots&J^k_l
\end{bmatrix}$ and $\mathbf{A}^k$ have the same eigenvalues.
Since the size of each Jordan block does not change, the algebraic multiplicities do not change.