I'm having some difficulty proving the following:
Let $P \in M_n(\mathbb C)$ be idempotent. Prove that all nonzero singular values of $P$ satisfy $\sigma_i \ge 1$.
By definition I know that $P$ being idempotent means $P^2 = P$. Likely I have to invoke the Singular Value Decomposition Theorem to prove the problem. So, let the singular value decomposition of $P$ be given by $$P = U\Sigma V^* = P^2 = U\Sigma V^*U\Sigma V^*.$$ By definition I know that the singular values of $P$ are the square roots of the eigenvalues of $P^*P.$ And unfortunately I am not sure where to go from here.
I was inclined to say that $$\sigma_1 = \|P\|_2 = \|P^2\|_2 \le \|P\|_2\|P\|_2 = \sigma_1^2 \implies 1 \le \sigma_1,$$ for nonzero $\sigma_1$, but this doesn't tell me enough. What about $\sigma_2$? and further?
Can anyone provide a hint?