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

Ohio State University exam problems and solutions in mathematics

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)
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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 $n$.


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1 Response

  1. 01/01/2017

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

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