# Prove that the Length $\|A^n\mathbf{v}\|$ is As Small As We Like.

## Problem 381

Consider the matrix

\[A=\begin{bmatrix}

3/2 & 2\\

-1& -3/2

\end{bmatrix} \in M_{2\times 2}(\R).\]

**(a)** Find the eigenvalues and corresponding eigenvectors of $A$.

**(b)** Show that for $\mathbf{v}=\begin{bmatrix}

1 \\

0

\end{bmatrix}\in \R^2$, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.

(University of California, Berkeley, Linear Algebra Final Exam Problem)

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## Proof.

### (a) Find the eigenvalues and corresponding eigenvectors of $A$.

To find the eigenvalues of $A$, we compute the characteristic polynomial $p(t)$ of $A$.

We have

\begin{align*}

p(t)&=\det(A-tI)\\

&=\begin{vmatrix}

3/2-t & 2\\

-1& -3/2-t

\end{vmatrix}\\

&=t^2-1/4.

\end{align*}

Since the eigenvalues are roots of the characteristic polynomials, the eigenvalues of $A$ are $\pm 1/2$.

Next we find the eigenvectors corresponding to eigenvalue $1/2$.

These are the solutions of $(A-\frac{1}{2}I)\mathbf{x}=\mathbf{0}$.

We have

\begin{align*}

A-\frac{1}{2}I=\begin{bmatrix}

1 & 2\\

-1& -2

\end{bmatrix}

\xrightarrow{R_2+R_1}

\begin{bmatrix}

1 & 2\\

0& 0

\end{bmatrix}.

\end{align*}

Thus, the solution $\mathbf{x}$ satisfies $x_1=-2x_2$, and the eigenvectors are

\[\mathbf{x}=x_2\begin{bmatrix}

-2 \\

1

\end{bmatrix},\]
where $x_2$ is a nonzero scalar.

Similarly, we find the eigenvectors corresponding to the eigenvalue $-1/2$.

We solve $(A+\frac{1}{2}I)\mathbf{x}=\mathbf{0}$.

We have

\begin{align*}

A+\frac{1}{2}I=\begin{bmatrix}

2 & 2\\

-1& -1

\end{bmatrix}

\xrightarrow[\text{then } R_2+R_1]{\frac{1}{2}R_1}

\begin{bmatrix}

1 & 1\\

0& 0

\end{bmatrix}.

\end{align*}

Thus, we have $x_1=-x_2$, and the eigenvectors are

\[\mathbf{x}=x_2\begin{bmatrix}

-1 \\

1

\end{bmatrix},\]
where $x_2$ is a nonzero scalar.

### (b) We can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.

We express the vector $\mathbf{v}=\begin{bmatrix}

1 \\

0

\end{bmatrix}$ as a linear combination of eigenvectors $\begin{bmatrix}

-2 \\

1

\end{bmatrix}$ and $\begin{bmatrix}

-1 \\

1

\end{bmatrix}$ corresponding to eigenvalues $1/2$ and $-1/2$, respectively.

Let

\[\begin{bmatrix}

1 \\

0

\end{bmatrix}=c_1\begin{bmatrix}

-2 \\

1

\end{bmatrix}+c_2\begin{bmatrix}

-1 \\

1

\end{bmatrix}\]
for some scalars $c_1, c_2$.

Solving this for $c_1, c_2$, we find that $c_1=-1$ and $c_2=1$, and thus we have

\[\begin{bmatrix}

1 \\

0

\end{bmatrix}=-\begin{bmatrix}

-2 \\

1

\end{bmatrix}+\begin{bmatrix}

-1 \\

1

\end{bmatrix}.\]
Then for any positive integer $n$, we have

\begin{align*}

A^n\begin{bmatrix}

1 \\

0

\end{bmatrix}&=-A^n\begin{bmatrix}

-2 \\

1

\end{bmatrix}+A^n\begin{bmatrix}

-1 \\

1

\end{bmatrix}\\

&=-\left(\, \frac{1}{2} \,\right)^n\begin{bmatrix}

-2 \\

1

\end{bmatrix}+\left(\, -\frac{1}{2} \,\right)^n\begin{bmatrix}

-1 \\

1

\end{bmatrix}\\

&=\left(\, \frac{1}{2} \,\right)^n\begin{bmatrix}

2-(-1)^n \\

1+(-1)^n

\end{bmatrix}

\end{align*}

Note that in the second equality we used the following fact: If $A\mathbf{x}=\lambda \mathbf{x}$, then $A^n\mathbf{x}=\lambda^n \mathbf{x}$.

Then the length is

\begin{align*}

\left \| A^n\begin{bmatrix}

1 \\

0

\end{bmatrix}\right \|&=\left(\, \frac{1}{2} \,\right)^n \sqrt{\left(\, 2-(-1)^n \,\right)^2+\left(\, 1+(-1)^n \,\right)^2}\\

& \leq \left(\, \frac{1}{2} \,\right)^n \sqrt{3^2+2^2}\\

&= \sqrt{14}\left(\, \frac{1}{2} \,\right)^n \to 0 \text{ as $n$ tends to infinity}.

\end{align*}

Therefore, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we wish.

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