Normalising eigenvector to length 1

Question

For this matrix:

$\begin{bmatrix}2 & -4\\-4 & 8\end{bmatrix}$

The eigenvalues are 0 and 10.

The first eigenvector is then:

0 $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$

The normalised eigenvector is

$\begin{bmatrix} \frac{2}{\sqrt5} \\ \frac{1}{\sqrt5} \end{bmatrix}$

Can someone tell me where the $\sqrt5$ comes from? How do you work it out? Thanks.

From Q1 here: http://edshare.soton.ac.uk/2161/1/Ex8_qu1.pdf

Answer 1

The norm (Euclidean distance) of the eigenvector is $\sqrt{2^2+1^2} = \sqrt{5}$. We need that $k \sqrt{5} = 1$ so $k = 1/\sqrt{5}$, so if $v$ is the vector, $v_{1} = \frac{2}{\sqrt5}$ and $v_{2} = \frac{1}{\sqrt5}$.