<s>[INST] 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. 
 [/INST] 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 $1-\sqrt{2}$.
</s>