# Thread: 2 norm of an orthogonal projector

1. ## 2 norm of an orthogonal projector

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

I know that for any projection P, $\displaystyle \left\|P \right\|_2 \geq 1,$ , and that if P is an orthogonal projection, then $\displaystyle \left\|P \right\|_2 = 1,$, 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 $\displaystyle \left\|P \right\|_2 = 1,$ then P is an orthogonal projection.

I know that for any projection P, $\displaystyle \left\|P \right\|_2 \geq 1,$ , and that if P is an orthogonal projection, then $\displaystyle \left\|P \right\|_2 = 1,$, but I am not sure how to prove the other way around.
If P is not orthogonal then $\displaystyle (\ker P)^\perp$, 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 $\displaystyle x\in(\ker P)^\perp$ such that $\displaystyle Px\ne x$. Then $\displaystyle Px = (Px-x) + x$. But $\displaystyle Px-x\in\ker(P)$ so it is orthogonal to x. It follows (Pythagoras' theorem) that $\displaystyle \|Px\|^2 = \|Px-x\|^2 + \|x\|^2$. Since $\displaystyle Px-x\ne0$, that tells you that $\displaystyle \|Px\|>\|x\|$ and hence $\displaystyle \|P\|_2>1$.

3. Originally Posted by Opalg
Also, those two spaces have the same dimension. So there must exist a vector $\displaystyle x\in(\ker P)^\perp$ such that $\displaystyle Px\ne x$.
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 $\displaystyle \text{ran}(P)$ and $\displaystyle (\ker(P))^\perp$ have the same dimension because they both have $\displaystyle \ker(P)$ as a complementary subspace. If $\displaystyle (\ker(P))^\perp \subseteq \text{ran}(P)$ 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 $\displaystyle x\in(\ker(P))^\perp$ such that $\displaystyle x\notin\text{ran}(P)$. Then $\displaystyle x\ne Px$.

5. Thanks a lot, now I got it

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# proof projection has norm 1

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