# Normalizing an eigenvector to have unit length

I worked out a problem where one of the eigenvectors is $(1+ \sqrt{2}, 1)$. How do I normalize this vector? Should I use approximate values?

## 1 Answer

There's no need to use approximate values. You have all the exact values you need to normalize the vector.

Here's the general formula for normalizing a vector: If $v$ is the non-zero vector $(a,b)$, then the normalized vector $v$ is

$$\frac{1}{\sqrt{a^2+b^2}}(a,b).$$

What do you get?

• Yes, I know that. However, with this vector, the "a" value is 1+ sq rt 2, so I would be squaring the binomial and then adding 1, and the taking the square root of the entire expression. Very clumsy. – user230049 Apr 9 '15 at 17:58
• Yeah, but that's probably the answer they're looking after. – Mankind Apr 9 '15 at 18:07
• I think you're right. Maybe I'll write both answers. – user230049 Apr 9 '15 at 18:44