# Orthogonal projection

1. Jul 8, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Let P$$\in$$L(V). If P^2=P, and llPvll<=llvll, prove that P is an orthogonal projection.

2. Relevant equations

3. The attempt at a solution
I think that regarding llPvll<=llvll is redundant. For example, consider P^2=P
and let v be a vector in V. Doesn't P^2=P kind of give it away by itself?
I mean v=a1v1+...+amvm+...+anvn. Consider the subspace that P projects to whose
dimension is less than V's. So for P^2 to =P, and P^2v=/=0,
PPv=a1v1+...+amvm=Pv=a1v1+...+amvm. Notice how the scalars from 1 to m
do not change to make P^2=P possible. Isn't it obvious enough from P^2=P
that the length of the vector a1v1+..+amvm is < a1v1+...+amvm+...+anvn?
I know that that's not were trying to prove, but why is llPvll<=llvll important
when we know P^2=P?

Last edited: Jul 8, 2009
2. Jul 8, 2009

### cipher42

The condition is not redundant. The condition that $$P^2=P$$ is precisely the definition of a projection, but you most definitely need the second piece, that $$||Pv||\leq||v||$$ to show that it is orthogonal. As a counterexample, consider the oblique (as opposed to orthogonal) projection in $$\mathbb{R}^2$$

$$P=$\left( \begin{array}{ccc} 0 & 0 \\ c & 1 \end{array} \right)$$$

You can quickly check that it is indeed a projection by computing $$P^2=P$$ but if we look at its action on a vector we find:

$$Pv=$\left( \begin{array}{ccc} 0 \\ cv_1+v_2 \end{array} \right)$$$

So for c>1, we would have $$||Pv||>||v||$$. This goes to show that an projection need not always reduce the magnitude of a vector upon which it acts and hence the second condition is not derivable from the first. In fact, we have to hope that the added condition is precisely what you need to show that the projection is orthogonal, or else that's going to be one tricky homework problem

3. Jul 8, 2009

### evilpostingmong

Thank you! There had to be some motive for the author to put llPvll<=llvll, I just
couldn't find it.