# 2-norm of a projector is greater than 1

1. May 6, 2010

### math8

Let $$P \in \textbf{C}^{m\times m}$$ be a projector. We prove that $$\left\| P \right\| _{2}\geq 1$$ , with equality if and only if P is an orthogonal projector.

I suppose we could use the formula $$\left\| P \right\| _{2}= max_{\left\| x \right\| _ {2} =1} \left\| Px \right\| _{2}$$ and use the fact that $$P^{2}=P$$ and $$P=P^{*}$$ (P* is the transpose conjugate of P).

But I am not sure how to use these.

2. May 6, 2010

### jbunniii

Technically this isn't true since $P = 0$ is a projection matrix with $||P||_2 = 0$. But assuming $P \neq 0$, there exists some $v \neq 0$ such that

$$Pv = v$$

If you use this fact along with the definition of the 2-norm of a matrix, your inequality follows pretty much immediately.