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I have to find the eigenvalues and the eigenvectors of the following matrix :

$\begin{equation*} \mathbf{S} = \left( \begin{array} {cc} .0144 & .0117\\ .0117 & .01466 \end{array} \right) \end{equation*}$

I found the eigenvalues $\lambda_1=.026$ and $\lambda_2=.002$. Then I tried to find the eigenvectors.

$$\mathbf{S}\mathbf{x}=\lambda_1\mathbf{x}$$ or

$$.0144x_1+.0117x_2=.026x_1$$ $$.0117x_1+.0146x_2=.026x_2$$

From the first equation , $.0117x_2=.0116x_1$

The result is

for $\lambda_1=.026$, $\begin{equation*} \mathbf{x_1} = \left( \begin{array} {c} .704\\ .710 \end{array} \right) \end{equation*}$

and for $\lambda_2=.002$, $\begin{equation*} \mathbf{x_2} = \left( \begin{array} {c} -.710\\ .704 \end{array} \right) \end{equation*}$

I suppose those are normalizing eigenvectors but don't come up with the results. How can I compute the eigenvector ?

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  • $\begingroup$ Who is "they" in your last sentence? $\endgroup$ – 5xum Sep 13 '15 at 16:36
  • $\begingroup$ @5xum Please see the edit. $\endgroup$ – ABC Sep 13 '15 at 16:37
  • $\begingroup$ If you are working with $0.26$ as the eigenvalue, you will make large mistakes because of numeric errors... $\endgroup$ – 5xum Sep 13 '15 at 16:39
  • $\begingroup$ @5xum It is $.026$ , not $.26$. $\endgroup$ – ABC Sep 13 '15 at 16:41
  • $\begingroup$ OK, but my point stays. The thing with eigenvalues is that if you are calculating them by hand, you need to use exact values, not numeric approximations. $\endgroup$ – 5xum Sep 13 '15 at 16:42
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From your equation (note that the system is underdetermined, so you should get an infinite number of solutions): $$.0117x_2=.0116x_1$$ you can get an eigenvector by setting $x_1$ and solving for $x_2$. If you set $x_1 = 1$ you get: $$x_2 = \frac{.0116}{.0117} = \frac{116}{117}$$ so $$v_1 = \begin{pmatrix} 1 \\ \frac{116}{117} \end{pmatrix}$$ should be an eigenvector for $S$ with eigenvalue $\lambda_1$. You will see that this is not quite right, if you compare the vectors $Av_1$ and $\lambda_1v_1$. This is because you have too few decimals in your $\lambda_1$.

The exact eigenvalues for your given matrix are: $$\lambda_1 = \frac{1453 + 13\sqrt{8101}}{100000}$$ $$\lambda_2 = \frac{1453 - 13\sqrt{8101}}{100000}.$$

If you do the calculations with the exact value for $\lambda_1$, you get the equation $$.0144x_1+.0117x_2=\lambda_1x_1$$ which simplies to: $$.0117x_2 = (\lambda_1 - .0144)x_1.$$ Set $x_1 = 1$ and solve for $x_2$: $$x_2 = \frac{\lambda_1 - .0144}{.0117} = \frac{\frac{1453+13 \sqrt{8101}}{100000} - .0144}{.0177} = \frac{1}{90} \left(1+\sqrt{8101}\right)$$ and so $$v_1 = \begin{pmatrix} 1 \\ \frac{1}{90} \left(1+\sqrt{8101}\right)\end{pmatrix}$$ is your sought after eigenvector.

Note that the difference between the values for $x_2$ calculated by approximate values and exact values is quite large, approximately $.0197$.

A useful trick here is to using fractional numbers all the time, never decimal ones. You can e.g. write S as: $$S = \frac{1}{100000} \begin{pmatrix} 1440 & 1170 \\ 1170 & 1466 \end{pmatrix}.$$

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  • $\begingroup$ But softwares provide the eigenvectors of the problem are: for $\lambda_1=.026$, $\begin{equation*} \mathbf{x_1} = \left( \begin{array} {c} .704\\ .710 \end{array} \right) \end{equation*}$ and for $\lambda_2=.002$, $\begin{equation*} \mathbf{x_2} = \left( \begin{array} {c} -.710\\ .704 \end{array} \right) \end{equation*}$. How all of the softwares assume the same arbitrary value while it comes to mind to take $1$ when doing it by hand? $\endgroup$ – ABC Sep 14 '15 at 11:03
  • $\begingroup$ @ABC, note that if $x$ is an eigenvector, then $\alpha x$ is an eigenvector as well, for all scalar $\alpha \neq 0$. The software has given you a normalized eigenvector. $\endgroup$ – Calle Sep 15 '15 at 17:05

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