Show that the range of $\mathbf{F}=\mathbf{H}^*\mathbf{T}\mathbf{H}$ and $\mathbf{H}^*$ are the same I am reading an applied mathematics paper and I do not understand a seemingly trivial statement. Sorry if this is very obvious, but I am not a mathematician.
Let $\mathbf{H}\in \mathbb{C}^{MxN}$ have maximal rank $M\leq N$, assume $\mathbf{T}\in \mathbb{C}^{MxM}$ is a diagonal matrix, show that the range of $\mathbf{H}^*$ coincides with the range of $\mathbf{F}=\mathbf{H}^*\mathbf{T}\mathbf{H}$. And more generally show that it works for any invertible $\mathbf{T}\in \mathbb{C}^{MxM}$.
 A: I am not sure what the authors had in mind, but here is a fairly quick proof of the fact. First of all, we note that $F = H^*(TH)$ necessarily has a range contained within that of $H^*$. Thus, it suffices to show that the dimension of the range (i.e. the rank of $F$) is equal to that of $H^*$. We already know that $\operatorname{rank}(F) \leq \operatorname{rank}(H)$.
The fact that $H$ has full row rank implies that $HH^*$ is invertible. With that, we have
$$
\operatorname{rank}(F) \geq \operatorname{rank}(FH^*) = \operatorname{rank}(H^*THH^*).
$$
Now, $T(HH^*)$ is a product of invertible matrices and is thus invertible. It follows that $FH^* = H^*THH^*$ has rank equal to that of $H^*$. Thus, we have
$$
\operatorname{rank}(F) \geq \operatorname{rank}(H^*THH^*) =\operatorname{rank}(H^*).
$$
We have $\operatorname{rank}(F) \leq \operatorname{rank}(H^*)$ and $\operatorname{rank}(F) \geq \operatorname{rank}(H^*)$, so that $\operatorname{rank}(F) = \operatorname{rank}(H^*)$, which was what we wanted.

Another proof, which perhaps is a bit more in the typical style for numerical linear algebra, uses the properties of the Moore-Penrose pseudoinverse.
Because $H$ has full row-rank, it follows that $HH^+ = I$ (where $H^+$ denotes the MP-pseudoinverse of $H$). It is clear that the image of $H^*$ contains that of $F$ because $F$ is of the form $F = H^*A$ for some matrix $A$ (namely, $A = TH$). In order to show that the image of $F$ contains that of $H^*$, it suffices to find a matrix $M$ such that $FM = H^*$. To that end, we take $M = H^+T^{-1}$ and note that
$$
FM = (H^*TH)(H^+T^{-1})\\
= H^*T(HH^+)T^{-1}\\
= H^*T I T^{-1}\\
= H^*(T T^{-1})\\
= H^*I = H^*.
$$
So, we have $FM = H^*$, which means that the range of $F$ must contain that of $H^*$.
