# Thread: 2 norm of an orthogonal projector

1. ## 2 norm of an orthogonal projector

Let P be a projector.
Show that if $\left\|P \right\|_2 = 1,$ then P is an orthogonal projection.

I know that for any projection P, $\left\|P \right\|_2 \geq 1,$ , and that if P is an orthogonal projection, then $\left\|P \right\|_2 = 1,$, but I am not sure how to prove the other way around.

2. Originally Posted by math8
Let P be a projector.
Show that if $\left\|P \right\|_2 = 1,$ then P is an orthogonal projection.

I know that for any projection P, $\left\|P \right\|_2 \geq 1,$ , and that if P is an orthogonal projection, then $\left\|P \right\|_2 = 1,$, but I am not sure how to prove the other way around.
If P is not orthogonal then $(\ker P)^\perp$, the orthogonal complement of the kernel of P, is not equal to the range of P. Also, those two spaces have the same dimension. So there must exist a vector $x\in(\ker P)^\perp$ such that $Px\ne x$. Then $Px = (Px-x) + x$. But $Px-x\in\ker(P)$ so it is orthogonal to x. It follows (Pythagoras' theorem) that $\|Px\|^2 = \|Px-x\|^2 + \|x\|^2$. Since $Px-x\ne0$, that tells you that $\|Px\|>\|x\|$ and hence $\|P\|_2>1$.

3. Originally Posted by Opalg
Also, those two spaces have the same dimension. So there must exist a vector $x\in(\ker P)^\perp$ such that $Px\ne x$.
I don't quite follow here, why is that true?

4. Originally Posted by math8
I don't quite follow here, why is that true?
The subspaces $\text{ran}(P)$ and $(\ker(P))^\perp$ have the same dimension because they both have $\ker(P)$ as a complementary subspace. If $(\ker(P))^\perp \subseteq \text{ran}(P)$ then (because they have the same dimension) these two subspaces would have to coincide, which would imply that P is orthogonal. Since P is not orthogonal, there must be an element $x\in(\ker(P))^\perp$ such that $x\notin\text{ran}(P)$. Then $x\ne Px$.

5. Thanks a lot, now I got it