Let $A$ be $n\times n$ matrix and let $\lambda_1, \lambda_2, \dots, \lambda_n$ be all the eigenvalues of $A$. (Some of them may be the same.)

For each positive integer $k$, prove that $\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k$ are all the eigenvalues of $A^k$.

Proof.

By the triangularization (or Jordan canonical form), there exists a nonsingular matrix $S$ such that
\[S^{-1}AS=\begin{bmatrix}
\lambda_1 & * & * & * &*\\
0 &\lambda_2 & * & * &*\\
\vdots & \cdots & \ddots & \cdots& \vdots \\
0 & 0 & 0 & \lambda_{n-1} & *\\
0 & 0 & 0 & 0& \lambda_n
\end{bmatrix}.\] Here the right matrix is an upper triangular matrix whose diagonal entries are eigenvalues of $A$.

Then we have
\begin{align*}
S^{-1}A^k S=(S^{-1}AS)^k=\begin{bmatrix}
\lambda_1^k & * & * & * &*\\
0 &\lambda_2^k & * & * &*\\
\vdots & \cdots & \ddots & \cdots& \vdots \\
0 & 0 & 0 & \lambda_{n-1}^k & *\\
0 & 0 & 0 & 0& \lambda_n^k
\end{bmatrix}.
\end{align*}

The characteristic polynomial of the matrix $A^k$ is given by
\begin{align*}
p(t)&=\det(A^k-tI)\\
&=\det(S^{-1})\det(A^k-tI)\det(S)\\
&=\det(S^{-1}(A^k-tI)S)\\
&=\det(S^{-1}A^kS-tI)\\[6pt] &=\begin{vmatrix}
\lambda_1^k-t & * & * & * &*\\
0 &\lambda_2^k-t & * & * &*\\
\vdots & \cdots & \ddots & \cdots& \vdots \\
0 & 0 & 0 & \lambda_{n-1}^k-t & *\\
0 & 0 & 0 & 0& \lambda_n^k-t
\end{vmatrix}\\[6pt] &=\prod_{i=1}^n(\lambda_i^k-t).
\end{align*}

Since the roots of the characteristic polynomial are all the eigenvalues, we see that $\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k$ are all the eigenvalues of $A^k$.

The post Find All the Eigenvalues of $A^k$ from Eigenvalues of $A$ first appeared on Problems in Mathematics.