If matrix $P$ is positive semidefinite, is $C^TPC$?

Question

Or if it's a conditionally true statement, what must be true about the matrix $C$ to make $C^TPC$ PSD?

edit - all matrices are made of real numbers.

edit - here is the issue I'm running into that is made me want to confirm this (MATLAB programming):

eigs(H)

ans =

0.1128
0.0001
0.0001
0.0001

$P = C^THC$

% Where H is symmetric, 4x4, and the multiplying matrix is size 4x330, mostly zeros

chol(P) Error using chol Matrix must be positive definite.

So it seems multiplication did not preserve being positive definite.

Answer 1

Based on the fact that you used transpose and not adjoint, I'll assume that you are working with real matrices over $\Bbb R^n$. (If you want this to work for complex inner product spaces, you'll need to use the adjoint in place of transpose in general.) Let's consider what happens with $C^TPC$ in an inner product:

$$\langle C^TPC x,x\rangle = \langle PCx,(C^T)^T x\rangle = \langle PCx,Cx\rangle.$$

Do you see where to go from here?

Answer 2

Write $P= P^{1/2} P^{1/2}$ where $P^{1/2}$ is a symmetric square root of $P$. Since $P$ is positive definite, it has such a symmetric square root (take its eigen-decomposition and replace the eigenvalues with their square roots, for example).

Clearly, $C^T P C$ is symmetric.

Then, $x^T C^T P C x= x^T C^T P^{1/2} P^{1/2} C x = (P^{1/2} C x)^T (P^{1/2} C x) = ||P^{1/2} C x||^2 \geq 0$.

So, $C^T P C$ is positive semi-definite as well.

Note that theres a difference between positive definite and positive semi-definite (the latter allowing zero eigenvalues, which chol in matlab does not like).