Let $T$ be a bounded operator on a Hilbert space with the property that $T^*(T-I)= 0$. I'd like to show that $T$ is an orthogonal projection.
I'm not really sure how to show that an operator is an orthogonal projection. If $A:X\rightarrow U$ is a projection onto a closed subspace of X, then $\langle x- Ax, u_i\rangle = 0 \;\;\forall u_i \in U$ ?
Expanding we get $T'T = T' \Longleftrightarrow TT' = T\ (T'' = T$ in Hilbert spaces?)
$$Tx = T'(Tx) \Longleftrightarrow y = T'y$$
Hence $T'$ has $\lambda = 1$ after $Tx$, do $T'$ and $T$ have the same eigenvalues?
There are a lot of question marks here.