- #1

- 106

- 0

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

Last edited:

- Thread starter Stephan Hoyer
- Start date

- #1

- 106

- 0

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

Last edited:

- #2

- 106

- 0

Suppose P is an operator on V and [itex]P^2=P[/itex]. Prove that [itex]V = null P \oplus range P[/itex].

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

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

If I recall correctly, P² = P is theI'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.

Not all projections have the form you gave: for example, consider in

- Last Post

- Replies
- 9

- Views
- 732

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 15

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 978

- Last Post

- Replies
- 8

- Views
- 805

- Last Post

- Replies
- 0

- Views
- 720

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 774

- Last Post

- Replies
- 1

- Views
- 908