Partial isometry and projection

Question

The following is a Theorem of Murphy's C*-algebras and operator theory:

Let $H_1, H_2$ be Hilbert spaces and $u\in B(H_1,H_2)$. If $u^*u$ is a projection, then $uu^*u=u$.

To show it, for $\xi\in H_1$, we have $$\|u\xi\|^2=\langle u^*u\xi,\xi\rangle = \|u^*u\xi\|^2$$ Then the author concludes $u(1-u^*u)=0$, but I do not know how he concludes it. Please help me. Thanks for your help.

Answer 1

Notice that $\|u(1-u^*u)\xi'\|^2=\|u^*u(1-u^*u)\xi'\|^2=\|(u^*u-u^*u)\xi'\|^2=0$, for every $\xi'$. Thus, $u(1-u^*u)=0$.

Answer 2

Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$