# Compute Power of Matrix If Eigenvalues and Eigenvectors Are Given

## Problem 373

Let $A$ be a $3\times 3$ matrix. Suppose that $A$ has eigenvalues $2$ and $-1$, and suppose that $\mathbf{u}$ and $\mathbf{v}$ are eigenvectors corresponding to $2$ and $-1$, respectively, where
$\mathbf{u}=\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}.$ Then compute $A^5\mathbf{w}$, where
$\mathbf{w}=\begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix}.$

## Solution.

Since $\mathbf{u}$ is an eigenvector corresponding to the eigenvalue $2$, we have
$A\mathbf{u}=2\mathbf{u}.$ Similarly, we have
$A\mathbf{v}=-\mathbf{v}.$ From these, we have
$A^5\mathbf{u}=2^5\mathbf{u} \text{ and } A\mathbf{v}=(-1)^5\mathbf{v}.$

To compute $A^5\mathbf{w}$, we first need to express $\mathbf{w}$ as a linear combination of $\mathbf{u}$ and $\mathbf{v}$. Thus, we need to find scalars $c_1, c_2$ such that
$\mathbf{w}=c_1\mathbf{u}+c_2\mathbf{v}.$ By inspection, we have
$\begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix}=3\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}+2\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix},$ and thus we obtain $c_1=3$ and $c_2=2$,

We compute $A^5\mathbf{w}$ as follows:
\begin{align*}
A^5\mathbf{w}&=A^5(3\mathbf{u}+2\mathbf{v})\\
&=3A^5\mathbf{u}+2A^5\mathbf{v}\\
&=3\cdot 2^5\mathbf{u}+2\cdot (-1)^5\mathbf{v}\\
&=96\mathbf{u}-2\mathbf{v}\6pt] &=96\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}-2\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 92 \\ -2 \\ -96 \end{bmatrix}. \end{align*} Therefore, the result is \[A^5\mathbf{w}=\begin{bmatrix} 92 \\ -2 \\ -96 \end{bmatrix}.

Let $S$ be the subset of $\R^4$ consisting of vectors $\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}$ satisfying...