Let T be a bounded operator on the Hilbert space H with the property that $T^*(T-I)= 0$. Show that T is an orthogonal projection.
Im not really sure how to show that an operator is an orthogonal projection? If $A:X\rightarrow U$ is an projection onto a closed subspace of X, then $<x- Ax, u_i > = 0, \;\; \forall u_i \in U?$
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 $Tx$, does T' and T has the same eigenvalues? There are a lot of question marks here. Thanks for the help!