Is $Q=V^TWV$ positive definite?

Question

I have the real symmetric $k \times k$ matrix $Q = V^T W V$, where I know $V$ is a $n \times k$ orthogonal matrix (its columns are orthogonal) and $W$ is a $n \times n$ diagonal matrix with all its elements positive (all $Wii>0$). $k \ll n$.

I am applying Cholesky decomposition on $Q$ and it works, so it must be positive definite, but I don't know how to prove it (or show that it was just luck with my data).

A matrix is positive definite if all it eigenvalues are positive 1, but I think I can't apply this here. $V$ is orthogonal, so $V^T=V^{-1}$ (pseudo-inverse), so $Q = V^{-1} W V$. Then I can say that every row of $V$ is an eigenvector of $Q$, and every element of $W$ is an eigenvalue of $Q$. But I know that $Q$ must have $k$ linearly independent eigenvectors, which I don't know if there are in the rows of $V$ (only the columns of $V$ are known to be orthogonal).

Is $Q$ positive definite or I just had lucky data?

Answer 1

$\forall x$, $x^TQx=(Vx)^TW(Vx)=\sum_{i}w_{ii}y_i^2>0$ since $y=0$ iff $x=0$ since $y=Vx$ and $V$ is orhtogonal. so yes, $Q$ is positive definite.

Answer 2

Write $W=D^2$ with $D$ diagonal; then $V^TWV=V^TD^TDV=(DV)^T(DV)$ is positive definite.