We are given $T:V \to V$ a normal linear transform (meaning $TT^*=T^*T$)
We are also given $T^2=T$.
Show that $T$ is self adjoint (meaning $T^*=T$).
What I did
I think I may have done something wrong
It is easy to show that the eigenvalues of $T$ are $1$ and $0$ (Assume $T$ is not the identity or the zero matrices).
So, for every eigenvector $v\in V$:
$<Tv,v>=\delta_{1,0}||v||^2$ where if $Tv=v$ then $<Tv,v>=||v||^2$ and if $Tv=0$ then $<Tv,v>=0$.
let's look at $<T^*v,v>$.
$<T^*v,v>=<v,Tv>=\bar \delta_{1,0}||v||^2$ where if $Tv=v$ then $<T^*v,v>=<v,Tv>=\bar 1 ||v||^2 = ||v||^2$ and if $Tv=0$ then $<T^*v,v>=<v,Tv>=\bar 0 ||v||^2=0$
So we see that $<T^*v,v>=<Tv,v>$ regardless of eigevector / value choice.
if $w$ is another eigenvector of different eigenvalue, then $<Tv,w>=<T^*v,w>=0$ because the basis of eigenvectors is orthonormal.
So really, we see that $<Tv,u>=<T^*v,u>$ for any combinations of eigenvector (even from different eigenvalues).
and since we have a basis of eigenvectors, every vector can be decomposed into combination of eigenvectors, and so for every vectors (not just eigenvectors) we get $<Tx,y>=<T^*x,y>$ which proves that $T=T^*$.
Is this correct?
Yes, it is correct. It might be nice (as you seem to assume it is trivial) to add in the end that since $<Tx,y>=<T^*x,y>$ for every possible $x$ and $y$, one can choose $x$ and $y$ from columns of the identity matrix and and hence that each entry of $T^*$ and $T$ depending on the chosen columns are the same.