Let $A$ be a symmetric positive semidefinite matrix and $u$ be a unit vector.

Then $u^TAu \leq \|A\|_2$ ?

Here $\Vert \cdot \Vert_2$ is the induced/operator 2-norm defined as

$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$


Yes, we have $u^TAu=\langle Au,\overline{u}\rangle\le\|Au\|_2\|\overline{u}\|_2\ \text{(Cauchy-Schwarz)}=\|Au\|_2\le\|A\|_2$. This is true for every complex square matrix $A$. We don't need $A$ to be positive semidefinite.

  • $\begingroup$ So A doesn’t have to be symmetric either? $\endgroup$ – Pew Dec 15 '20 at 9:40
  • $\begingroup$ @Pew No, it doesn't. Any complex square matrix will be fine. $\endgroup$ – user1551 Dec 15 '20 at 9:49

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