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.

  • $\begingroup$ 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

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 $(2)$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.