A question about a HPD matrix and the rank

Question

Let $H$ be an $n\times n$ Hermitian positive definite matrix and $A$ an $n\times m$ matrix. How to prove that $\mathrm{rank}(A)=\mathrm{rank}(A^*HA)$?

Answer 1

The fact that $H$ is Hermitian (i.e. $H^* = H$) implies that $H$ is normal (i.e. $H^*H = HH^*)$. It is "well known" that normal matrices can be diagonalized with a unitary transformation, i.e. there are a matrix $U$ with $U^*U = UU^* = I$ (here, $I$ denotes the identity matrix) and a diagonal matrix $D$ such that $D = U^*HU$. Again, since $H$ is Hermitian, we have $D^* = (U^*HU)^* = U^*H^*U = U^*HU = D$. So the diagonal entries of $D$ are real. Since $H$ is positive definite, we have that $D$ also is positive definite (proof: $x^*Dx = x^*U^*HUx = (Ux)^*H(Ux) > 0$ for any $x \neq 0$ since $U$ is invertible). This implies that the diagonal entries of $D$ are real and positive. So there is a diagonal matrix $F$ with real and positive diagonal entries such that $D = F^2 = F^*F$. Then we have $F^*F = U^*HU$ or equivalently $H = UF^*FU^*$.

With all this, we put $$B:= FU^*A.$$ Note that $rank\ B = rank\ A$, since $FU^*$ is invertible. Now $$A^*HA = A^*UF^*FU^*A = B^*B,$$ so $rank\ A^*HA = rank\ B^*B$. Next, I claim that $rank\ B = rank\ B^*B$. This can be seen as follows. $B$ is an $n\times m$ matrix, so $B^*B$ is an $m \times m$ matrix. Now, $rank\ B = m - \dim ker\ B$ and $rank\ B^*B = m - \dim ker\ B^*B$. But $ker\ B = ker\ B^*B$ since $$ x \in ker\ B \Rightarrow Bx = 0 \Rightarrow B^*Bx = 0 \Rightarrow x \in ker\ B^*B$$ and conversely $$ x \in ker\ B^*B \Rightarrow B^*Bx = 0 \Rightarrow x^*B^*Bx = (Bx)^*(Bx) = 0 \Rightarrow Bx = 0 \Rightarrow x \in ker\ B.$$

All in all we have shown $$ rank^\ A = rank\ B = rank\ B^*B = rank\ A^*HA,$$ as desired.