Compare quadratic term and norm $x^TPx$ and $\|P\|\|x\|^2$ with $P\succ 0$

Question

How to compare the following: "$x^TPx$ and $\|P\|\|x\|^2$" with $P\succ 0$?

I mean $$x^TPx \geq \|P\|\|x\|^2 \text{ or } x^TPx \leq \|P\|\|x\|^2, \forall x\ne 0, P\succ 0$$

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I know that $$x^TPx = \langle P,X\rangle = \operatorname{tr}(PX)=\operatorname{tr}(Pxx^T)$$

But I still cannot figure out how to go to the step to compare both of them.

Answer 1

When $P$ is positive it is in particular symmetric, whence diagonalizable in an orthonormal basis and having non-negative eigenvalues $\lambda_1\geq \lambda_2 \geq ... 0 $: $$ P = \sum_i \lambda_i e_i e_i^T $$ So you have (using 2-norm) $$x^T Px = \sum_i \lambda_i (x^T e_i)^2 \leq \lambda_1 \sum_i (x^T e_i)^2 = \lambda_1 |x|^2 \leq \|P \| \; |x|^2$$