Operator norm of symmetric matrix The operator norm of a $n \times n$ matrix $A$ can be written as $\max_{x, y \in \mathbb{S}^{n-1}} y^T A x$.  If $A$  is symmetric, can one show that this definition reduces to $\max_{x \in \mathbb{S}^{n-1}} x^T A x$.  I suspect that due to the symmetry, we need not consider a $y$ that is different from $x$. 
 A: I suppose you are considering a real matrix (with coefficients in the real field). Spectral theorem says there exists an ortogonal matrix $Q$ that diagonalizes $A$, so $QAQ^T=D$ is diagonal. Since $Q$ is ortogonal, it sends vectors of norm 1 to vectors of norm 1. Hence $A$ has the same norm than $D$, and we are reduced to this case. But in this case it is easy (and in fact the norm is the maximum eigenvalue of $A$, with your definition: usually it is taken the max of the absolute values of what you takes).  
A: The norm of A is 
$||A||^2=\sup_{x\neq 0} \frac{||Ax||^2}{||x||^2}= \sup_{x\neq 0} \frac{\langle Ax,Ax \rangle}{||x||^2}=$
$ \sup_{x\neq 0} \langle A\frac{x}{||x||} ,A \frac{x}{||x||}\rangle=\sup_{x\in \mathbb{S}^{n-1}}\langle Ax, Ax\rangle=$
$ \sup_{x\in \mathbb{S}^{n-1}}(Ax)^tAx= \sup_{x\in \mathbb{S}^{n-1}}x^t (A^tA) x =||A^tA||$
A: Since $A$ is symmetric, By SVD we have $A=Q D Q^T$, where $Q$ is orthonormal and $D$ is diagonal. Let $S^{n-1}$ be such that $x^{T}x=1$ if $x\in S^{n-1}$. Also call $y=Q^Tx$. observe that if $x^Tx=1$ then $y^Ty=1$. Now,
\begin{eqnarray}
\sup_{x \in S^{n-1}}|x^T A x| = \sup_{y \in S^{n-1}} |y^T D y| =|\lambda|_{\max}\\ 
\sup_{x \in S^{n-1}}||Ax||^2_2 = \sup_{x \in S^{n-1}}|x^T A^TA x| = \sup_{y \in S^{n-1}} |y^T D^2 y| = \lambda^2_{\max}\\
\implies \sup_{x\in S^{n-1}}||Ax||_2 = \sup_{x \in S^{n-1}}|x^T A x| = \sup_{x \in S^{n-1}}<Ax,x>
\end{eqnarray}
