Linear Algebra

Compute Power of Matrix If Eigenvalues and Eigenvectors Are Given

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

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
\[\mathbf{w}=\begin{bmatrix}
7 \\
2 \\
-3
\end{bmatrix}.\]

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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
\[A^5\mathbf{w}=\begin{bmatrix}
92 \\
-2 \\
-96
\end{bmatrix}.\]

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