Contents
Two Dimensional Example
Let
$$
A = \left( \begin{array}{cc}
1 & 2\\
4 & 3
\end{array} \right)
$$
and choose
$$
x_0 = \left( \begin{array}{c}
1\\
0
\end{array} \right)
$$
Then
$$
x_0 = \left( \begin{array}{c}
1\\
0
\end{array} \right),
Ax_0 = \left( \begin{array}{c}
1\\
4
\end{array}
\right),
A^2 x_0 = \left( \begin{array}{c}
9\\
16
\end{array} \right)
$$
and hence there exist $c_i$ such that
$$
c_1 \left( \begin{array}{c}
1\\
0
\end{array} \right) +
c_2 \left( \begin{array}{c}
1\\
4
\end{array}
\right) +
c_3 \left( \begin{array}{c}
9\\
16
\end{array} \right) = 0
$$
Now solving the simultaneous equations
$$c_1 + c_2 + 9 c_3 = 0$$
$$0c_1 + 4c_2 +16c_3 = 0$$
we obtain for instance $c_3=1$, $c_2=-4$ and $c_1=-5$.
Hence
$$
A^2 x_0 -4A x_0 -5 x_0 = -5\left( \begin{array}{c}
1\\
0
\end{array}\right) -4\left( \begin{array}{c}
1\\
4
\end{array} \right) + \left( \begin{array}{c}
9\\
16
\end{array} \right) = 0
$$
and so
$$
(A-5I)(A+I)x_0 = (A^2 -4A -5 I )x_0 = 0
$$
Therefore letting
$$
v_1 = (A+I)x_0 = \left( \begin{array}{c}
2\\
4
\end{array} \right)
$$
we see that
$$
(A-5I)v_1=0 \Rightarrow Av_1=5v_1
$$
Also letting
$$
v_2 = (A-5I)x_0 = \left( \begin{array}{c}
-4\\
4
\end{array} \right)
$$
we see that
$$
(A+I)v_2 = 0 \Rightarrow Av_2 = -v_2
$$
For
$$
A = \left( \begin{array}{cc}
1 & 2\\
4 & 3
\end{array} \right)
$$
then
$$
\left( \begin{array}{c}
2\\
4
\end{array} \right)
$$
is an eigenvector with eigenvalue $5$ and
$$
\left( \begin{array}{c}
-4\\
4
\end{array} \right)
$$
is an eigenvector with eigenvalue $-1$.
Three Dimensional Example
Now we find the eigenvalues and eigenvectors of the matrix
$$A=\left(\begin{array}{ccc}
3&-1&1\\
2&0&-2\\
3&-3&1
\end{array}\right)$$
Since the second column appears to be the simplest we choose our initial vector to be
$$e_2=\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)$$
and then we calculate
$$Ae_2=\left(\begin{array}{c}
-1\\
0\\
-3
\end{array}\right),
A^{2}e_2=\left(\begin{array}{c}
-6\\
4\\
-6
\end{array}\right),
A^{3}e_2=\left(\begin{array}{c}
-28\\
0\\
-36
\end{array}\right).
$$
There are no obvious dependence relations between three of the vectors $e_2$, $Ae_2$, $A^{2}e_2$ and $A^{3}e_2$.
However we are certain that there is a dependence relation between all four:
$$
a\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right)+
b\left(\begin{array}{c}
-1\\
0\\
-3
\end{array}\right)+
c\left(\begin{array}{c}
-6\\
4\\
-6
\end{array}\right)+
d\left(\begin{array}{c}
-28\\
0\\
-36
\end{array}\right)
=0
$$
and so we solve the system:
$$-b-6c-28d=0$$
$$a+4c=0$$
$$-3b-6c-36d=0$$
for which one solution is $d=1$, $c=-4$, $b=-4$ and $a=16$.
Therefore
$$\left(\alpha^{3}-4\alpha^{2}-4\alpha+16I\right)e_2=0$$
and we notice that this cubic has a factor of $(\alpha -2I)$ so
$$\left(\alpha - 2I\right)\left(\alpha^{2} -2\alpha -8I\right)e_2=0$$
$$\left(\alpha - 2I\right)\left(\alpha +2I)(\alpha -4I\right)e_2=0$$
This means that $\alpha$ has an eigenvalue $\lambda_1 = 2$ with eigenvector
$$(\alpha^{2}-2\alpha-8I)e_2=\left(\begin{array}{c}
-4\\
-4\\
0
\end{array}\right)$$
an eigenvalue $\lambda_2=-2$ with eigenvector
$$(\alpha^{2}-6\alpha+8I)e_2=\left(\begin{array}{c}
0\\
12\\
12
\end{array}\right)$$
and an eigenvalue $\lambda_3=4$ with eigenvector
$$(\alpha^{2}-4I)e_2=\left(\begin{array}{c}
-6\\
0\\
-6
\end{array}\right)$$
Example of Obtaining a Subset of the Eigenvalues Quickly
Let $A$ be the matrix
$$\left(\begin{array}{ccc}
1 & -2 & 1\\
-4&-1&2.5\\
0&0&1.5
\end{array}\right)$$
If we choose the initial vector to be
$$e_1=\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)$$
then
$$Ae_1= \left(\begin{array}{c}
1\\
-4\\
0
\end{array}\right)\text{ and }
A^{2}e_1=\left(\begin{array}{c}
9\\
0\\
0
\end{array}\right)$$
and so immediately we see that
$$\left(A^{2}-9I\right)e_1 = 0$$
$$\left(A-3I\right)\left(A+3I\right)e_1=0$$
Therefore
$$(A+3I)e_1 = \left(\begin{array}{c}
4\\
-4\\
0
\end{array}\right)$$
is an eigenvector with eigenvalue $3$ and
$$(A-3I)e_1 = \left(\begin{array}{c}
-2\\
-4\\
0
\end{array}\right)$$
is an eigenvector with eigenvalue $-3$.