If matrix $A\in\mathbb{C}^{m\times n}$ with $m>n$ is full rank and $B\in\mathbb{C}^{n\times n}$ is diagonal with $det(B)\ne0$, and $(B)_{ii}>0$ how can we prove that $ABA^H$ is positive semi-definite?
It is easy to show that diagonal matrix $B$ is positive definite. I also have a hunch that $ABA^H$ is Hermitian, but I have not proven this. I don't know how to apply either of these facts to shown that $ABA^H$ is positive semi-definite?
Any help would be greatly appreciated!
$ A : \mathbb{C}^n \rightarrow \mathbb{C}^m $, $ A^H : \mathbb{C}^m \rightarrow \mathbb{C}^n $
We need to show that $ <ABA^H x, x > \; \geq \; 0 $ for any $ x \in \mathbb{C}^m $, where $<,>$ denotes the standard Hermitian dot product on $\mathbb{C}^n$. We have $ <ABA^H x, x > \; = \; <BA^H x, A^H x > \; = \sum_{i=1}^n B_{ii}y_i \bar{y}_i $ where $y_i$are coordinates of $A^Hx$. This sum is non-negative, because all $ B_{ii} > 0 $