# 2 norm of an orthogonal projector

#### 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.

#### Opalg

MHF Hall of Honor
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 $$\displaystyle \|P\|_2>1$$.

• math8

#### math8

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?

#### Opalg

MHF Hall of Honor
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 $$\displaystyle x\ne Px$$.

#### math8

Thanks a lot, now I got it 