TITLE: Norm of a symmetric matrix equals spectral radius
QUESTION [36 upvotes]: How do I prove that the norm of a matrix equals the absolutely largest eigenvalue of the matrix? This is the precise question:

Let $A$ be a symmetric $n \times n$ matrix. Consider $A$ as an operator in $\mathbb{R}^n$ given by $x \mapsto Ax$. Prove that $\|A\| = \max_j |\lambda_j|$, where $\lambda_j$ are the eigenvalues of $A$.

I've read the relevant sections in my literature over and over but can't find any clue on how to begin. A solution is suggested here but the notion of diagonal operator is not in my literature so it doesn't tell me very much. So, any other hints on how to solve the question? Thanks.

REPLY [2 votes]: I don't have enough reputation to comment. But I'd like to fix a mistake in @yoknapatawpha's answer. The overall idea is correct.
Take the singular value decomposition of $A$ as $$A=VDW^T$$ and then we are looking for $$\max_{||u||=1}||Au||=\max_{||u||=1}||VDW^Tu||.$$
Here V and W are both orthogonal matrices.
Because V is orthogonal, we have
$$||VDW^Tu||=||DW^Tu||.$$
Now denote $$x=W^Tu,$$ because W is orthogonal, we have $$||x||=||W^Tu||=||u||=1$$ and also there is an one-to-one mapping between $x$ and $u$. Thus the problem is equivalent to
$$\max_{||x||=1} ||Dx||$$
and the rest of the answer follows @yoknapatawpha's.
However, generally it's not right to say $\|Av\| = \|D v\|$ is true for any $v$. For example when
$$V=\left[\begin{matrix}1& 0\\0 & 1\end{matrix}\right], D=\left[\begin{matrix}2& 0\\0 & 1\end{matrix}\right], W=\left[\begin{matrix}0& 1\\1 & 0\end{matrix}\right], v=\left[\begin{matrix}1\\0\end{matrix}\right]$$
in this case you can check that $||Av||=1$ but $||Dv||=2$.