Kenels, images, eigenvalues and eigenvectors

paddy543211 sent this p.m.:

quick question

hey hallsofivy, I am new to the forum and I'm trying to post a thread but I think I need activation? So im fairly urgently trying to post up this question and I thought seeing as your such a big helper could you please take a look at it or send it to somebody who you think might know? I don't have a clue, Thanks!

Let V be an F-vector space and let ϕ : V -> V be a linear map such that ϕ∘ϕ = ϕ

Show

(a) ker ϕ ∩ Imϕ = {0};[/quote]

Suppose v is in the image of ϕ. Then there exist some u so that ϕu= v. Suppose v is also in the kernel of v. Then ϕv= 0. But then ϕ∘ϕu= ϕv= 0 = ϕu= v. That is, v must be 0.

Quote:

(b) ker ϕ + Imϕ = V .

Let v be any vector in V. Let u= V- ϕv. ϕ(V-ϕv)= ϕv- ϕϕv= 0 so u is in the kernel of ϕ. Let w= v- u= v- (v- ϕv)= ϕv.

Quote:

Deduce that V = kerϕ ⊕ Imϕ and that ϕ is the projection on Imϕ along kerϕ .

(c) Show that the only possible eigenvalues of ϕ are 0 and 1.

(d) Determine the corresponding eigenspaces.

As before, any vector, v, can be written as u+w with u in kerϕ and w in imϕ. ϕv= ϕu+ ϕw= 0+ ϕw. Since w is in the image of ϕ, w= ϕx for some x and then ϕw= ϕϕx= ϕx= w. In particular, for any v in kerϕ, w= 0 so ϕv= 0. If v is in Imϕ, u= 0 so ϕv= ϕw= w= u.

**Moderator edit:** Then paddy543211 gets an Infraction for begging for help in a pm instead of posting him/herself. No doubt the question is urgent, they always are. Personally, it looks to me like a question from an assignment that counts towards a final grade.