# Homework Help: Linear Algebra

1. Nov 1, 2005

### Stephan Hoyer

Ignore this post (it was irrecoverable due to bad LaTeX markup). See the one below.

Last edited: Nov 1, 2005
2. Nov 1, 2005

### Stephan Hoyer

Here's the question I'm stuck in on my latest problem set:

Suppose P is an operator on V and $P^2=P$. Prove that $V = null P \oplus range P$.

I'm pretty sure that P is simply the projection operator (consisting of the identity matrix replacing some 1's with 0's), in which case the conclusion follows easilly. But I've been looking at this one for a while and I can't see how I can prove that P must be the projection operator.

Thanks for your help.

3. Nov 2, 2005

### Hurkyl

Staff Emeritus
If I recall correctly, P² = P is the definition of a projection operator!

Not all projections have the form you gave: for example, consider in R² the orthogonal projection onto the line y = x given by [x, y] -> (1/2)[x+y, x+y]