partial isometry

Question

If $A\in M_{r\times m}(\mathbb{C}),B\in M_{m\times r}(\mathbb{C})$,$AB=Id_r$,then $A,B$ is a partial isometry.

My question:how to show $A^*A,B^*B$ is a projection?

Answer 1

This is not true. If $m=r=1$, you can take $A=2$, $B=1/2$, and none of them are partial isometries. Even if you require $\|A\|\leq1$, $\|B\|\leq1$, then statements fails: take $$ A=\begin{bmatrix} \tfrac12&\tfrac12\end{bmatrix} B=\begin{bmatrix} 1\\ 1\end{bmatrix}. $$ What you can tell from $AB=I$ is that $BA$ is an idempotent: $$ (BA)^2=B(AB)A=BA. $$ It is not necessary that $BA$ is selfadjoint, though. For instance take $$ A=\begin{bmatrix} \tfrac12&1\end{bmatrix} B=\begin{bmatrix} 1\\ \tfrac12\end{bmatrix}. $$ then $AB=1$, and $BA=\begin{bmatrix} 1/2 & 1\\ 1/4& 1/2\end{bmatrix}$.

You do have that $B^*A^*AB=I_r$, so $AB$ is a partial isometry, and $ABB^*A^*$ is a projection, since $$ (ABB^*A^*)^2=ABB^*A^*ABB^*A^*=ABI_rB^*A^*=ABB^*A^*. $$