# Thread: 2 norm of a projector

1. ## 2 norm of a projector

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 $P^{2}=P$

But I am not sure how to proceed.

2. Originally Posted by 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 $P^{2}=P$

But I am not sure how to proceed.
Geometrically, there is a good way to see this. Translating it into algebra is not as easy.

For any vector $x\in\mathbb{C}^m$, you can write $x = Px + (x-Px)$. This is the unique expression for x as the sum of a vector in the range of P (namely Px) and a vector in the kernel of P ( namely x–Px). The projection P is orthogonal if and only if its range and kernel are orthogonal subspaces of $\mathbb{C}^m$.

If P is orthogonal then you can apply Pythagoras' theorem to deduce that $\|x\|^2 = \|Px\|^2 + \|x-Px\|^2 \geqslant \|Px\|^2$. So $\|Px\|\leqslant\|x\|$ (for all x), which tells you that $\|P\|=1$.

If P is not orthogonal then you want to find a vector x such that $\|Px\|>\|x\|$. The way to do that is to take x to be a unit vector in the orthogonal complement of the kernel of P. That's the hard part of the problem, and I don't have time now to explain how the argument goes.

3. For the first part, can we say that since P is a projector (assuming $P \neq 0$ ), there is a vector $v \neq 0$ such that $Pv=v$.

Now take $x = \frac{v}{\left\|v \right \|}$, so $\left\|Px \right \|_{2} = 1$ and using the definition of $\left\|P \right \|_{2}$, we get the inequality ?