# Matrix inequality with square-root matrix

Let $P$ be an $n \times n$ positive semi-definite matrix, and let $x \in \mathbb{R}^n$ be such that $x^T P x > 0$.

Prove the following inequality: $${P^{1/2}} x \, x^T P^{1/2} \preceq (x^T P x) \, I.$$

What I have done: 1) numerically, the inequality always holds; 2) tried to apply the Schur complement.

If $x$ is zero, there is nothing to prove; if $x$ is nonzero, divide both sides by $\|P^{1/2}x\|^2$ to get $uu^T\preceq I$ for some unit vector $u$. This should be easy enough to prove. If you still have difficulty, transform the inequality again by picking an orthonormal basis that contains $u$ as a basis vector.
• Thank you. Does the strategy holds true also is $x$ is a complex vector, and $x^T$ is replaced by $x^*$? – user18453 Jan 19 '17 at 12:35