If $P$ is positive definite, is$X^T P X$ also positive definite?

Question

Let $P\in\Re^{r\times r}$ be a symmetric positive definite matrix; i.e. $P=P^T\succ 0$. Also, let $X\in\Re^{r\times m}$ be a matrix such that $\mathrm{rank}(X)=r$ with $r\leq m$. My question is about the positive definiteness (or positive semi-definiteness) of the following product:

$X^TPX$

Can we say anything about the positive (semi-)definiteness of this product? As far as I tried with several numerical examples, we should be able to say that it is indeed semidefinite.

Wat is the point of considering two matrices $X,Y$ and two numbers $m,n$ instead of only $X$ and $m$? — Anne Bauval, Jun 01 '23 at 15:13

score 4 · Accepted Answer · answered Apr 21 '17 at 16:17

Yes, both are positive-semidefinite (definite if $r=m,n$).

This follows straight from the definition: for any vector $v$, $$v^TY^TPYv = (Yv)^TP(Yv) \geq 0$$ since $P$ is positive-definite. $Y^TPY$ is positive-definite whenever $Y$ has full column rank, since in that case $v\neq \mathbf{0} \Rightarrow Yv\neq \mathbf{0}\Rightarrow v^T(Y^TPY)v > 0$.

If $P$ is positive definite, is$X^T P X$ also positive definite?

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