Prove: For full rank matrix $A\in\mathbb{C}^{m\times n}$ with $m>n$ and diagonal matrix $B\in\mathbb{C}^{n\times n}$ we have $rank(ABA^H)=n$

Question

For full rank matrix $A\in\mathbb{C}^{m\times n}$ with $m>n$ and diagonal matrix $B\in\mathbb{C}^{n\times n}$ with $det(B)\ne0$ how can we prove that $rank(ABA^H)=n$?

I have already proven that $rank(AB)=n$ and that $(A^HA)^{-1}$ and $B^{-1}$ are defined. These properties are easy enough to show. It is also clear that the column rank of $AB$ and the row rank of $A$ are equal. I also have a hunch that $ABA^H$ is Hermitian, but I haven't proven this.

I cannot see how to put any of these facts together to prove that $rank(ABA^H)=n$. Any help would be greatly appreciated!

Answer 1

This is a bit of an indirect solution but one might go with the following:

Hint first: Try multiplying matrices from the right and left to $ABA^H$ and then look at the resulting rank.

Solution:

Consider the Matrix $A^H A B A^H A$. Since $A^H A$ is invertible, $A^H A B A^H A = (A^H A) B (A^H A)$ is invertible and as such of $rank$ $n$. Now suppose that $rank(A B A^H) < n$. In that case $dim(image(A B A^H)) < n$ and as such $dim(image( (A)(A B A^H)(B))) < n$. This would be a contradiction to $A^H A B A^H A$ being invertible. As such the proof is concluded

Answer 2

If $x\in\mathbb{C}^m$ and $ABA^H x = 0$, then $x^H ABA^H x = 0$, so $(B^{\frac12}A^Hx)^2 = 0$, where we used the fact that in complex domain, the square root of any diagonal matrices exist. Starting from there, try to show the left null space of $ABA^H$ is a subspace of left null space of $AB^{\frac12}$, so $rank(ABA^H)\ge rank(AB^{\frac12}) = n$. The rest is easy.

Answer 3

We can actually prove a stronger statement directly, namely that $\text{Col}(A)=\text{Col}(ABA^H)$.

Clearly we have $\text{Col}(ABA^H)\subseteq \text{Col}(A)$ so it really only suffices to show that $\text{Col}(ABA^H)\supseteq \text{Col}(A)$.

Choose $y\in \text{Col}(A)$ arbitrarily, then find $x\in \mathbb{C}^n $ such that $y=Ax$.

Since $\text{rank}(A)=\text{rank}(A^H)=n$ we know the columns of $A^H\in \mathbb{C}^{n\times m}$ span all of $\mathbb{C}^n$, so we may find $v$ such that $A^Hv=B^{-1}x$. Evidently, $BA^Hv=x$ and so $ABA^Hv=Ax=y$ which shows $y\in \text{Col}(ABA^H)$ as required.