Alternative form of the Spectral norm of a symmetric, square matrix

I want to show that for a symmetric, square matrix $$A$$, $$|x^\top A y| \leq ||A||||x||_2||y||_2~,$$ where $$||A||$$ denotes the spectral norm of $$A$$. I have the following proof, so it would be really helpful if anyone can confirm whether this is correct:

Let $$A = P^\top DP$$ be the spectral decomposition of $$A$$, where $$P$$ is orthogonal and $$D$$ is diagonal. Make the change of variables: $$X = Px$$ and $$Y=Py$$. Now, we have: $$\begin{eqnarray} |x^\top A y| &=& |X^\top D Y|\\&=& |\sum_i D_{ii}X_iY_i|\\&\leq& \max_i |D_{ii}|\sum_i |X_iY_i| \\&\leq& ||A||||X||_2||Y||_2 = ||A|| ||x||_2||y||_2~. \end{eqnarray}$$ Can someone PLEASE verify whether what I have is correct? Thanks in advance.

• I think it is correct. Moreover, one can show that $\sup_{\Vert x\Vert\leq 1,\Vert y\Vert\leq 1}\vert x^T Ay\vert =\Vert A\Vert$. – TheWildCat Jul 19 at 6:31

Your proof is correct. More general, with a simpler proof:

If $$A$$ is a square matrix , we get by Cauchy-Schwarz:

$$(1) \quad|x^TAy| \le ||x||_2 \cdot ||Ay||_2.$$

Since $$||A||$$ is the spectral norm, we have

$$(2) \quad ||Ay||_2 \le ||A|| \cdot ||y||_2.$$

The result follows now from $$(1)$$ and $$(2)$$.

• Thanks a lot, @Fred! – Usermath Jul 19 at 7:51