Exercise 7.B.7

Proof. Let $e_1,\dots ,e_n$ be an orthonormal basis of $V$ consisting of eigenvalues of $T$ and let ${\lambda }_1,\dots ,{\lambda }_n$ denote their corresponding eigenvalues. We have

for each $j=1,\dots ,n$. This implies that ${\lambda }_j=0$ or ${\lambda }_j=1$. Hence, every eigenvalue of $T$ is a real number and by the previous exercise $T$ is self-adjoint. Moreover, ${\lambda }_j^2={\lambda }_j$, regardless if ${\lambda }_j$ is $0$ or $1$. Therefore $T^2=T$. □