Let $B$ be a $N \times K$ matrix and $\Sigma$ be a $K \times K$ (symmetric) covariance matrix that is positive definite. How can I show that the matrix of the form $B \Sigma B^T$ is also a covariance matrix (and hence at least positive semi-definite)?
I have tried plugging in specific examples for $B$ and $\Sigma$ and I can see that it always returns a covariance matrix. But, how can I prove this?
Since $\Sigma$ is assumed positive semi-definite you know that for every non-zero $K$-vector $\mathbf{z}$ you have $$ \mathbf{z}^T \Sigma \mathbf{z} > 0, $$ then let $\mathbf{y}$ be an arbitrary non-zero $n$-vector and $\mathbf{z} = \mathbf{B}^T \mathbf{y}$, then \begin{align*} \mathbf{y}^T \mathbf{B} \Sigma \mathbf{B}^T\mathbf{y} &=\left(\mathbf{B}^T \mathbf{y} \right)^T \Sigma \left( \mathbf{B}^T \mathbf{y}\right) \\ &=\mathbf{z}^T \Sigma \mathbf{z} \geq 0, \end{align*} allowing for the degenerate case where there are non-zero vectors $\mathbf{y}$ such that $\mathbf{B}^T \mathbf{y} = \mathbf{0}$. In particular you have that $\operatorname{rank}(\mathbf{B}\Sigma\mathbf{B}^T) = \operatorname{rank}(\mathbf{B})$.