Operator norm of symmetric matrix

Question

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 $x$.

Answer 1

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||$

Answer 2

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).

Answer 3

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}