Eigenvector basis I have this matrix:
$$
\begin{pmatrix}
3 &  \sqrt{2} \\
\sqrt{2}  &  2 \\
\end{pmatrix}
$$
I found the eigenvalues ( that are $1$ and $4$) but I find it really difficult to find the bases... ($(cI-A)x=0$ )
I think maybe one is $(1,1)$ but I'm not sure..
 A: The eigenvalues are correct.
To find the eigenvectors, you need to solve systems of linear equations. Namely, to find eigenvectors of $1$, you need to find all vectors $\mathbf{x}$ for which $(A-(1)I)\mathbf{x}=\mathbf{0}$ (this is equivalent to $(I-A)\mathbf{x}=\mathbf{0}$, but it involves changing fewer entries, so it may be less prone to errors).
$$A-I = \left(\begin{array}{cc}
3&\sqrt{2}\\
\sqrt{2}&2\end{array}\right) - \left(\begin{array}{cc}1&0\\0&1\end{array}\right) = \left(\begin{array}{cc}2 & \sqrt{2}\\
\sqrt{2} & 1
\end{array}\right).$$
When is $(A-I)\mathbf{x}=\mathbf{0}$? We need to solve the system
$$\left(\begin{array}{cc}
2 & \sqrt{2}\\
\sqrt{2}&1\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right).$$
At this point, you probably know how to solve systems of linear equations. For example, we can use Gaussian elimination on the matrix:
$$\begin{align*}
\left(\begin{array}{cc}
2 & \sqrt{2}\\
\sqrt{2} & 1
\end{array}\right) &\to \left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
\sqrt{2} & 1 \end{array}\right) &\text{(divide first row by }2\text{)}\\
&\to \left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
0 & 1 - \sqrt{2}\left(\frac{\sqrt{2}}{2}\right)\end{array}\right) &\text{(subtract }\sqrt{2}\text{ times the first row from row 2)}\\
&=\left(\begin{array}{cc}
1 & \frac{\sqrt{2}}{2}\\
0 & 0
\end{array}\right).
\end{align*}$$
The system has infinitely many solutions (as it should, since there have to be eigenvectors associated to the eigenvalue). They are all vectors $(x_1,x_2)^t$ such that $x_1+\frac{\sqrt{2}}{2}x_2 = 0$. That means that we must have $x_1 = -\frac{\sqrt{2}}{2}x_2$. If we set $x_2=\sqrt{2}$, then $x_1=-1$. So one eigenvector is $(-1,\sqrt{2})^t$. 
Now you need to do the same thing with $A-4I$, that is, with the matrix
$$\left(\begin{array}{rr}
-1 & \sqrt{2}\\
\sqrt{2} & -2
\end{array}\right).$$
(No, $(1,1)$ is not an eigenvectors; note that $A(1,1)^t = (3+\sqrt{2},2+\sqrt{2})^t$, which is not a scalar multiple of $(1,1)$)
A: Indeed, if you solve $(A-I)x = 0$, i.e.
$$ \begin{pmatrix} 2 & \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} x = 0$$
you find that $x = (1, -\sqrt{2})^{T}$ is an eigenvector.
For $(A - 4I)x = 0$, i.e.
$$ \begin{pmatrix} -1 & \sqrt{2} \\ \sqrt{2} & -2 \end{pmatrix} x = 0$$
you find that  $x = (\sqrt{2}, 1)^{T}$ is another eigenvector.
A: Just form the matrix $\lambda I -A$ and look for any element in the null space.
For example, with $\lambda = 1$, you have $\begin{pmatrix}
-2 &  -\sqrt{2} \\
-\sqrt{2}  &  -1 \\
\end{pmatrix}$.
You might notice that the first row is $\sqrt{2}$ times the second, so there is an eigenvector $(1,-\sqrt{2})^T$.
Repeat the process for the other eigenvalue. Since the eigenvalues are different, the eigenvectors are linearly independent and so span $\mathbb{R}^2$.
