### 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 $-3$.