We have the Euclidean norm (or $\displaystyle l_2-norm$) for a vector on the vector space $\displaystyle R^m$ ; $\displaystyle \parallel{v}\parallel_2 = \sqrt{\sum_{j=1}^m\mid{v_j}\mid^2}$.

Given a vector norm $\displaystyle ||\cdot||$ on $\displaystyle R^m$ the induced matrix norm for m x m matrices A is defined by

$\displaystyle ||A||\ =\ max_{v\neq{0}}\frac{||Av||}{||v||}$

That is , $\displaystyle ||A||$ is the smallest number $\displaystyle \alpha$ such that $\displaystyle ||Av||\leq\alpha||v||\ \forall\ v \in R^m$

So given the Euclidian-norm for a vector the induced matrix norm $\displaystyle ||A||_2=\sqrt{max.\ eigenvalue A^TA}$

Can anyone proof/explain why?

In my book it is given but I donīt see it directly, i think it has something to do with the fact that $\displaystyle A^TA$ is symmetric.