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
$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$
linear algebramatricesmatrix-normspositive-semidefinite
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
$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$
Best Answer
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.