2 norm of an orthogonal projector

Feb 2009
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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

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Aug 2007
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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 \(\displaystyle \|P\|_2>1\).
 
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Feb 2009
98
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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
Aug 2007
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2,789
Leeds, UK
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 \(\displaystyle x\ne Px\).