1

I'm working on proving the following:

If $P:V\to V$ is a linear operator and $P^2=P$ and $\|Pv\|\le\|v\|$ for all $v \in V$ then $P$ is an orthogonal projection.

My thoughts on this are that a linear map with the following properties is an orhtogonal projection: $null(P) $ is the orthogonal complement of its range, $P^2 = P$ and $\|Pv\|\le\|v\|$.

Therefore it should be enough to show that the range of $P$ is orthogonal to its null space. How to do this? Let $n \in null(P)$ and $u \in range(P) = U$. Want to show: $\langle u,n \rangle = 0$.

Am I doing this right? And how can I show $\langle u,n \rangle = 0$?

newb
  • 1,651
  • First $P$ is a projection because of $P^2=P$. Then you can see here. – Wei Zhan Aug 20 '13 at 07:27
  • Right, what you link to is an alternative proof of what I am trying to prove. But I don't see how it answers either of my two questions? – newb Aug 20 '13 at 07:30

1 Answers1

1

Suppose $u\in range(P)$ and $n\in null(P)$. If $u$ and $n$ are not orthogonal, then we have some scalar $c$ such that $\|u+cn\|<\|u\|$. But $P(u+cn)=Pu=u$ and so $\|P(u+cn)\|>\|u+cn\|$, contradicting your assumption.

Alex Becker
  • 61,883
  • How do you get a scalar $c$ such that $|u + cn|< |u|$? I can only get $|u + cn| \le |u|$ – newb Aug 26 '13 at 09:58
  • @newb We have $$|u+cn|^2=\langle u+cn,u+cn\rangle = \langle u,u\rangle + 2c\langle u,n\rangle + c^2\langle n,n\rangle = |u|^2+2c\langle u,n\rangle + c^2|n|^2$$ which has derivative with respect to $c$ at $c=0$ equal to $2\langle u,n\rangle\ne 0$. If this is positive then for small negative $c$ we have $|u+cn|<|u|$, while if this is negative then for small positive $c$ we have $|u+cn|<|u|$. – Alex Becker Aug 26 '13 at 17:48