# Find All the Eigenvalues of $A^k$ from Eigenvalues of $A$

## Problem 388

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. Sponsored Links ### More from my site • If Eigenvalues of a Matrix A are Less than 1, then Determinant of I-A is Positive Let A be an n \times n matrix. Suppose that all the eigenvalues \lambda of A are real and satisfy \lambda <1. Then show that the determinant \[ \det(I-A) >0, where $I$ is the $n \times n$ identity matrix. We give two solutions. Solution 1. Let […]
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1. 04/29/2017

[…] Let $lambda_1$ and $lambda_2$ be eigenvalues of $A$. Then we have begin{align*} 3=tr(A)=lambda_1+lambda_2 text{ and }\ 5=tr(A^2)=lambda_1^2+lambda_2^2. end{align*} Here we used two facts. The first one is that the trace of a matrix is the sum of all eigenvalues of the matrix. The second one is that $lambda^2$ is an eigenvalue of $A^2$ if $lambda$ is an eigenvalue of $A$, and these are all the ei…. […]

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