Computing Eigenvector. 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 ?
 A: 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:
$$S = \frac{1}{100000}
\begin{pmatrix}
 1440 & 1170 \\ 1170 & 1466
\end{pmatrix}.$$
