prove that $T^2=T$ diagonalizeable without using Jordan

Question

Question: Given that $T:V \to V$ and $T^2=T$ prove that $T$ is diagonalizable.

What I know:

$T^2-T=0=T(T-I)$.

$\operatorname{Im}(T-I) \subseteq\operatorname{Ker}(T)$ therefore $\dim V=\dim\operatorname{Ker}(T-I)+\dim\operatorname{Im}(T-I)\leqslant \dim\operatorname{Ker}(T-I)+\dim\operatorname{Ker}T$

Can I just say now that since the the sum of the dimensions of these two kernels is $\dim V\leqslant$ so it's equal to $\dim V$ and that the map is diagonalizable? I feel like something is missing with this explanation.

Thanks for any enlightening comments.

An operator is diagonaliable over a field if and only if its minimal polynomial is separable and split over that field. The minimal poly of $T$ must divide $x(x-1)$, and so trivially splits into distinct linear factors. — Alex Youcis, Oct 28 '13 at 20:21
@AlexYoucis, perhaps it is a matter of language, but I think the theorem says a matrix is diagonalizable over some field iff its minimal polynomial is the product of different linear factors...What has "separable" to do here? — DonAntonio, Oct 28 '13 at 20:26
@Norbert, I see that answer. How's that simpler than what Alex's proposes? — DonAntonio, Oct 28 '13 at 20:26
@DonAntonio Presumably the OP is just learning about diagonalization and doesn't know about minimal polynomials. At least that is the order I learned things in. I have no opinion otherwise. — Bruno Joyal, Oct 28 '13 at 20:27
That's a possibility, @Marie ...but as for simplicity the minimal polynomial test for diagonability has no, imo, adversaries...not even close. — DonAntonio, Oct 28 '13 at 20:31
A simpler alternative is to note that is probably a duplicate. — Git Gud, Oct 28 '13 at 20:35
@DonAntonio I agree. But from the point of view of an experienced mathematician like yourself, all things are susceptible to becoming trivialities. Good for the advancement of mathematics, but not so adapted to pedagogy. ;) Anyhow, in this case I think that my solution is ideal. This problem is important because it gives the OP a chance to familiarize themselves with idempotent operators, which have a very important geometric interpretation (orthogonal projections etc). To understand a new animal, one must catch it with a net and dissect it carefully. Not blow it up with a bazooka! :-) — Bruno Joyal, Oct 28 '13 at 20:38
(That being said, I had nothing to do with what happened to my neighbor's cat.) — Bruno Joyal, Oct 28 '13 at 20:39
I agree with the part when you say your answer gives a rather nice way, @Marie, specially on the light of what the OP added to his question under "what I know" (and for this I'm upvoting you). I just think the minimal polynomial, which btw is not that advanced a notion within this subject, simplifies greatly stuff here. This being said, I may have stared at your neighbor's cat, it freaked out and jumped into the street, the bus came on and... — DonAntonio, Oct 28 '13 at 20:43

score 5 · Accepted Answer · answered Oct 28 '13 at 20:23

5

Hint: show instead that every element of V can be uniquely written as $u+v$ where $u$ is in the kernel and $v$ is in the image. How does $T$ act on such a decomposition?

answered Oct 28 '13 at 20:23

Bruno Joyal

54,711

Hey thanx for the answer! So I said: Let $v \in V$, and so $v=v-T(v)+T(v)$ but $v-T(v) \in KerT$ and $T(v) \in ImT$. So that means that the direct sum of KerT and ImT is V. I know that a theorem says a map is diagonalizable iff the direct sum of it's eigenspaces is the entire vector space. I just need a hint on how to formalize the fact that KerT and ImT are indeed eigenspaces. – jreing Oct 29 '13 at 04:40
and also - I know what the characteristic polynomial is - I can't seem to understand how to connect it to this question (I usually use it more with matrices, or matrices representing maps) – jreing Oct 29 '13 at 04:43
@user1685224 You are welcome! Your reasoning is correct, but you didn't show uniqueness of the representation, which is required for a direct sum. But you have the right idea. To see that the kernel and image are eigenspaces, remark that $T$ acts on the kernel as the zero map (so all vectors in it are eigenvectors with eigenvalue $0$), and it acts on the image as the identity (so all vectors in it are eigenvectors with eigenvalue $1$). – Bruno Joyal Oct 29 '13 at 04:46
@user1685224 That being said, I notice that you have accepted practically none of the 60 or so questions you have asked on this site. When you are satisfied with an answer, you should click on the checkmark next to it. I suggest that you go back through your old questions to accept the answers you are happy with. All the best, – Bruno Joyal Oct 29 '13 at 04:47
Remark taken, although many times (esp. in calculus questions) I don't get an answer that really makes it clear (and no one responds or sees the question afterwards..) – jreing Oct 29 '13 at 04:53
@user1685224 No worries. You can always edit a question to "bump" it back to the front page (but don't do it too much). ;) – Bruno Joyal Oct 29 '13 at 04:54

prove that $T^2=T$ diagonalizeable without using Jordan

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