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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.

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  • $\begingroup$ 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 $. $\endgroup$ – TheWildCat Jul 19 at 6:31
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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 $(2)$.

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

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