# Compute the Product $A^{2017}\mathbf{u}$ of a Matrix Power and a Vector

## Problem 114

Let

\[A=\begin{bmatrix}

-1 & 2 \\

0 & -1

\end{bmatrix} \text{ and } \mathbf{u}=\begin{bmatrix}

1\\

0

\end{bmatrix}.\]
Compute $A^{2017}\mathbf{u}$.

(*The Ohio State University, Linear Algebra Exam*)

Add to solve later

Sponsored Links

## Solution.

We first compute $A\mathbf{u}$. We have

\[A\mathbf{u} = \begin{bmatrix}

-1 & 2 \\

0 & -1

\end{bmatrix} \begin{bmatrix}

1\\

0

\end{bmatrix}=

\begin{bmatrix}

-1\\

0

\end{bmatrix}

=-\begin{bmatrix}

1\\

0

\end{bmatrix}=-\mathbf{u}.\]
Thus we have

\[A\mathbf{u}=-\mathbf{u}.\]

Then using this result repeatedly we have

\begin{align*}

A^2\mathbf{u}=A(A\mathbf{u})=A(-\mathbf{u})=-A\mathbf{u}=-(-\mathbf{u})=\mathbf{u}.

\end{align*}

Thus we have

\[A^2\mathbf{u}=\mathbf{u}.\]

Next, we compute

\[A^3\mathbf{u}=A(A^2\mathbf{u})=A(\mathbf{u})=-\mathbf{u}.\] Now you see the pattern and obtain

\begin{align*}

A^n\mathbf{u}=\begin{cases}

-\mathbf{u} & \text{ if } n \text{ is odd}\\

\mathbf{u} & \text{ if } n \text{ is even}.

\end{cases} \tag{*}

\end{align*}

(To prove this pattern, use mathematical induction. See below.)

Thus we conclude that

\[A^{2017}\mathbf{u}=-\mathbf{u}=\begin{bmatrix}

-1\\

0

\end{bmatrix}\]
since $2017$ is odd.

### The proof of the formula.

For completeness, we prove the pattern (*) we see above.

We use mathematical induction.

When $n=1$, we computed $A\mathbf{u}=-\mathbf{u}$ and the base case is true.

Suppose that(*) is true for $n=k$.

Then we have

\begin{align*}

A^{k+1}\mathbf{u}=A(A^{k}\mathbf{u}) &=

\begin{cases}

A(-\mathbf{u}) &\text{ if } k \text{ is odd}\\

A(\mathbf{u}) & \text{ if } k \text{ is even}

\end{cases} \\

&=\begin{cases}

-A\mathbf{u} &\text{ if } k \text{ is odd}\\

A\mathbf{u} & \text{ if } k \text{ is even}

\end{cases} \\

& =\begin{cases}

\mathbf{u} &\text{ if } k \text{ is odd}\\

-\mathbf{u} & \text{ if } k \text{ is even}

\end{cases}

& =\begin{cases}

\mathbf{u} &\text{ if } k+1 \text{ is even}\\

-\mathbf{u} & \text{ if } k+1 \text{ is odd}

\end{cases}

\end{align*}

Here the second equality follows from the induction hypothesis.

This proves (*) for $n=k+1$ and this completes the induction step and (*) is true for any positive integer $n$.

Add to solve later

Sponsored Links

## 1 Response

[…] $A^{2017}mathbf{u}$. This is one of the exam problems at the Ohio State University. Check out the solutions of this problem […]