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Worked Examples: Distinct Real Eigenvalues

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

$(A-5I)v_1=0 \implies Av_1=5v_1$

Also letting

$v_2 = (A-5I)x_0 = \left( \begin{array}{c} -4\\ 4 \end{array} \right)$

we see that

$(A+I)v_2 = 0 \implies Av_2 = -v_2$

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 $-3$ .