Let $Q \in \mathbb{C}^{m \times m}$ be a given positive (semi)-definite matrix such that $Q = CC^H,$ for some known matrix $C \in \mathbb{C}^{m \times r}$ having full column rank and orthogonal columns.
Suppose we decompose $Q$ as $Q = BB^H$, where the matrix $B \in \mathbb{C}^{m\times r}$ is full column rank. Then is it necessary that $B = CR,$ for some invertible $R$?
What I have tried:
By showing that null space of $Q$ is same as null space of $C^H,$ $Q = CC^H \implies rank(Q) = rank(C).$ Also column space of $Q$ is contained in column space of $C$. Therefore column space of $Q$ is same as column space of $C$. Repeating above, for $Q= BB^H$, we see that column space of $Q$ is same as column space of $B$. Hence every solution $B$ is of the form $B = CR$ for some matrix $R$. Furthermore, as it is a change of basis, $R$ can be shown to be invertible.
Is it safe to conclude that $B$ has to be of the form of $B = CR$?