Is this a Projection operator on Hilbert space?

Question

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.

Answer 1

Your equation is $T^*T=T^*$. The key observation here is that $(T^*T)^*=T^*T$. Then $$ T=(T^*)^*=(T^*T)^*=T^*T=T^*. $$ so $T$ is selfadjoint, and moreover now your original equation reads $T^2=T$, so $T$ is an orthogonal projection.

Answer 2

Here is one way to see this. We have

$T^*T - T^* = T^*(T-I) = 0 = (T^*(T-I))^* = (T^*-I)T = T^* T - T$

hence cancelling the $T^*T$ you get that $T$ is self-adjoint, in particular normal, and so you may appeal to the spectral theorem, which in this case tells you that the fuction $\bar{x}(x-1) = 0$ on the spectrum of $T$.

We want to show that $\sigma(T) \subset \{0,1\}$ for $T$ to be a projection.

Now we have $\bar{x}(x-1) = |x|^2 - \bar{x} = 0$ i.e., since $|x|^2$ is real that for all $x \in \sigma(T)$ you have $x = |x|^2$.

This can happen only if $x = 0$ or $x = 1$ as we wanted.

EDIT: Marc van Leeuwen pointed out nicely that after the first line we actually already have $T^* = T$ and using this $T^2 = T$ as well, hence the claim.