MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $A$ symmetric and semidefinite positive, for each $x$ $$ x'Ax \geq \frac{1}{\Vert A\Vert} \Vert Ax \Vert^2 $$

This inequality appears at page 24 of "Introduction to Optimization" from Boris T. Polyak. I haven't been able to prove it. Any idea?

Thanks, Giovanni

share|cite|improve this question
up vote 3 down vote accepted

The claim is that $|A\| x' A x \ge \|A x\|^2 = x' A^2 x$, i.e. that $\|A\| A \ge A^2$. This comes from $\|A\| I \ge A$ by multiplying on the left and right by the positive semidefinite square root of $A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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