# Thread: vector space, eigenvector

1. ## vector space, eigenvector

Hey, I can't seem to get anywhere with this question so any help you can provide would be much appreciated.

$\displaystyle T$ is a linear operator on a finite dimensional vector space $\displaystyle V$, whereby $\displaystyle T^2$ equals the identity operator.

1. Prove that for any given vector u in $\displaystyle V$, $\displaystyle -(Tu -u)$ is the $\displaystyle 0$ vector or an eigenvector with eigenvalue $\displaystyle -1$

2. Prove that V is indeed the direct sum of the eigenspaces $\displaystyle V(-1)$ and $\displaystyle V(1)$ where $\displaystyle V(\lambda)$ is considered the set of eigenvectors with eigenvalue $\displaystyle \lambda$, together with $\displaystyle 0$.

I am really stuck here, so thank you for any help.

2. ## Re: vector space, eigenvector

Originally Posted by Juneu436
2. Prove that V is indeed the direct sum of the eigenspaces $\displaystyle V(-1)$ and $\displaystyle V(1)$ where $\displaystyle V(\lambda)$ is considered the set of eigenvectors with eigenvalue $\displaystyle \lambda$, together with $\displaystyle 0$
Hint For any given vector $\displaystyle u\in V$ we have $\displaystyle T(u+Tu)=Tu+T^2u=Tu+u$ so, $\displaystyle u+Tu$ is the zero vector or an eigenvector associated to the eigenvalue $\displaystyle \lambda=1$ .

3. ## Re: vector space, eigenvector

Originally Posted by FernandoRevilla
Hint For any given vector $\displaystyle u\in V$ we have $\displaystyle T(u+Tu)=Tu+T^2u=Tu+u$ so, $\displaystyle u+Tu$ is the zero vector or an eigenvector associated to the eigenvalue $\displaystyle \lambda=1$ .
Thanks!

So for Q1:
$\displaystyle -T(Tu-u)=Tu-T^2u=Tu-u$, so $\displaystyle -u+Tu$ is the zero vector or an eigenvector with eigenvalue $\displaystyle -1$.

And for Q2:
From your hint, and my working to Q1, bring them both together and that is the answer?

4. ## Re: vector space, eigenvector

i think you need a bit more than that. you should, however, be able to write down the minimal polynomial for T, which will then tell you T cannot have any other eigenvalues.

i think it only fair to point out that for this result to be strictly true, we must require V to be over a field of char(F) ≠ 2 (in that case V(1) = V).

5. ## Re: vector space, eigenvector

Originally Posted by Deveno
i think you need a bit more than that. you should, however, be able to write down the minimal polynomial for T, which will then tell you T cannot have any other eigenvalues.
How do I do that? Sorry for a stupid question deveno, sometimes I don't think great.

EDIT: Is it p(x)=x for the minimal polynomial for T?

6. ## Re: vector space, eigenvector

the minimal polynomial for a linear transformation T:V-->V, is the monic polynomial m(x) of least degree such that m(T) = 0 (the 0-transformation).

in this case, we have T^2 = I, so T^2 - I = 0, that is, T satisfies x^2 - 1 = 0.

now it could well be that T = I (there is nothing in your question that rules this out), or T = -I, in which case m(x) = x - 1, or m(x) = x + 1.

otherwise x^2 - 1 is the minimal polynomial for T, so in any case the only possible eigenvalues for T are in the set {-1,1}.

this shows V = V(1) + V(-1), but it does not prove that we have a direct sum, unless V(1)∩V(-1) = {0}. prove the intersection only has the 0-vector in it

(this is true, incidentally, for any two eigenspaces, can you see why?).