$T^*T=TT^*$ and $T^2=T$. Prove $T$ is self adjoint: $T=T^*$

Question

$V$ is an inner product space of finite dimension over $\mathbb{R}$, and $T:V\to V$ a linear transformation which is normal, that is, $T^*T=TT^*$. In addition $T^2=T$. Prove $T$ is self adjoint, that is, $T=T^*$.

I tried to prove it algebraically by using the inner product but it didn't work for me. Then I tried to prove the statement that $T$ is self adjoint iff $\langle v,Tv\rangle$ is real for all $v$.

I know that $T$ is diagonalizable, therefore there is a basis of $V$ consisting of eigenvectors. In addition every eigenvector of $T$ is an eigenvector of $T^*$. Can I prove that $\langle v,Tv\rangle$ is real only for the vectors in the basis and then it means it is for all $v$?

Any help or further hints are very appreciated.

Answer 1

The fact that $T^2=T$, implies that the polynomial $p(x)=x^2-x$ annihilates $T$, which means that the eigenvalues of $T$ are all in the set $\{0,1\}$.

The fact taht $TT^*=T^*T$ means that $T$ is normal, and hence diagonalizable by a unitary matrix, i.e. $T=U^* D U$, where $U^*U=UU^*=I$ and $D$ diagonal. But all the elements of $D$ are in $\{0,1\}$, and hence $D^*=D$. Therefore $$ T^*=(U^*DU)^*=U^*D^*U=U^*DU=T. $$

Answer 2

$T$ is normal, meaning $T=UDU^*$, where $U$ is unitary (that is, $U^{-1}=U^*$) and $D$ is diagonal. Then

$$ T^2=UDU^*UDU^*=UD^2U^*=T=UDU^* $$

implies $D^2=D$. What can you argue now about $D$?