# 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)

## 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{13}\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.