Eigenvalues of vectors with irrational entries I have been trying to find eigenvalues and eigenvectors of this matrix: $\begin{bmatrix}3 & -2\\1 & -1\end{bmatrix}$. So far I have got $\lambda_1=1+\sqrt2$ and $\lambda_2=1-\sqrt2$. I am stuck at finding eigenvectors at this point. Regular row-reduction method gives me hard to compute values. I am sure I am missing something obvious here. 
 A: You've gotten the eigenvalues correctly. Now, given a matrix $A$ with an eigenvalue $\lambda$, an eigenvector for $\lambda$ is just a non-zero element of the null space of $A-\lambda I$.
So, let's find the null space of $$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}-\begin{bmatrix}1+\sqrt{2} & 0\\0 & 1+\sqrt{2}\end{bmatrix}=\begin{bmatrix}2-\sqrt{2} & -2\\1 & -2-\sqrt{2}\end{bmatrix}$$
We row-reduce to simplify the matrix while keeping its null space the same:
$$\begin{bmatrix}2-\sqrt{2} & -2\\1 & -2-\sqrt{2}\end{bmatrix}\xrightarrow{\;\;\Large\mathsf{\text{row 2} \;-\; \left(\frac{1}{2-\sqrt{2}}\right)\,\text{row 1}}\;\;}\begin{bmatrix}2-\sqrt{2}\strut & -2\\0 & 0\end{bmatrix}$$
Obviously, for a non-zero vector $\Bigl[\begin{smallmatrix}x_1\\x_2\end{smallmatrix}\Bigr]$, we have that
$$\begin{align*}
\begin{bmatrix}2-\sqrt{2} & -2\\0 & 0\end{bmatrix}\begin{bmatrix}x_1\strut\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}&\iff (2-\sqrt{2})x_1+(-2)x_2=0\\\\\\
&\iff x_2=\frac{2-\sqrt{2}}{2}x_1
\end{align*}$$
and an easy choice of $x_1$ and $x_2$ that satisfy this is $x_1=2$ and $x_2=2-\sqrt{2}$.
Let's check that this worked:
$$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}\begin{bmatrix}2\strut\\2-\sqrt{2}\end{bmatrix}=\begin{bmatrix}3\cdot 2+(-2)\cdot(2-\sqrt{2})\strut\\1\cdot 2+(-1)\cdot(2-\sqrt{2})\end{bmatrix}=\begin{bmatrix}2+2\sqrt{2}\\2\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\begin{bmatrix}2\strut\\2-\sqrt{2}\end{bmatrix}$$
$$\checkmark$$
I leave it to you to do this process for the eigenvalue $1-\sqrt{2}$.
A: If you're curious how to get the row reduction method to work:
$\begin{bmatrix}3-\lambda & -2 \\1 & -1-\lambda\end{bmatrix} \sim \begin{bmatrix}1 & -1-\lambda\\3-\lambda & -2 \end{bmatrix} \sim \begin{bmatrix}1 & -1-\lambda\\0 & -2+(-1-\lambda)(-3+\lambda) \end{bmatrix}$.
This looks messy, but, as alex says, this matrix must be rank 1.  So $-2+(-1-\lambda)(-3+\lambda) = 2-(1-\lambda)^2=0$ when you write it all out.  Thus, the eigenvectors corresponding to the eigenvalue $\lambda$ are exactly the multiples of $\begin{bmatrix}1+\lambda\\1\end{bmatrix}$ (note that we switched the rows once, so we have to switch them back).
This actually tells you the eigenvalues as well, since they are just the two roots of the equation $2-(1-\lambda)^2=0$.
Of course alex's approach is much better than this.  I include it mostly to make the point that the row reduction method will always work, if done correctly.  This is worthwhile practice to ensure that your toolbox is functioning correctly.
