Let ${\bf A} \in \mathbb{C}^{m \times n}, {\bf D} \in \mathbb{C}^{m \times m}$ where ${\bf D}$ is a full-rank diagonal matrix and $m > n$. Consider the following product: $$ {\bf M} = {\bf A}^H {\bf D} {\bf A} $$ Where the superscript $H$ denotes conjugate transposition. I have the following hypotheses:
- If ${\rm rank}\left({\bf A}\right) = n$, then ${\rm rank}\left({\bf M}\right) = n$
- ${\rm rank}\left({\bf M}\right) = {\rm rank}\left({\bf A}\right)$
Is it known whether my any of my hypotheses are true or wrong? If so, where can I find proofs or counterexamples?
Thanks in advance.
For a simple counterexample, let $m=2,n=1$, $A=\begin{pmatrix} 1 \\ -1\end{pmatrix}$, and $D=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$. Then $A^HDA=0$ has lower rank than $A$.
To motivate this example, note that for any $v,w\in\mathbb{C}^n$, $$\langle A^HDAv,w\rangle=\langle DAv,Aw\rangle,$$ so the kernel of $A^HDA$ is just the vectors $v$ such that $DAv$ is orthogonal to the range of $A$. So, to make $A^HDA$ have smaller rank, you want to pick $D$ so that it makes something in the image of $A$ become orthogonal to the image of $A$. In particular, in the case $n=1$, if you have two orthogonal vectors that have the same set of coordinates that are zero (say $(1,-1)$ and $(1,1)$), you can turn one into the other by a diagonal matrix with nonzero entries, as in the example above.