Proving a hermitian operator is an orthogonal projection

Question

Let $V$ be a finite dimensional vector space over $\Bbb C$ and $T:V \to V$ a linear transformation, such that $T^2=T$ and T is hermitian ($T=T^*)$. Prove that $T$ is an orthogonal projection over some subspace $U \subseteq V$.

Answer 1

By the spectral theorem, $T$ is diagonalizable with respect to a basis $(e_1,e_2,\dots,e_n)$ of eigenvectors. For each $k\in\{1,2,\ldots,n\}$, let $\lambda_k\in\mathbb C$ be such that $T(e_k)=\lambda_ke_k$. Then$$\lambda_ke_k=T(e_k)=T\bigl(T(e_k)\bigr)=T(\lambda_ke_k)={\lambda_k}^2e_k$$and so $\lambda_k=0$ or $\lambda_k=1$. We can assum without loss of generality than, for some $m\in\{1,\ldots,n\}$, $\lambda_k=1$ if $k\leqslant m$ and that $\lambda_k=0$ otherwise. Let $U=\langle e_1,e_2,\ldots,e_m\rangle$. Then $T$ is the orthogonal projection over $U$.