# Thread: Eigenvalue of Real Vector Space Proof

1. ## Eigenvalue of Real Vector Space Proof

I'm trying to understand a proof from my textbook. They are showing that every operator on a finite dimensional real vector space with odd dimension has an eigenvalue. I already know that every operator on a finite dimensional real vector space has an invariant subspace of dimension two or one...

We have $\displaystyle V$, which is the finite dimensional vector space, and $\displaystyle T$, which is an operator on V. There is a subspace $\displaystyle U$ of $\displaystyle V$ that is invariant and has dimension 2 (otherwise we are done). There is some other subspace $\displaystyle W$ such that $\displaystyle U\bigoplus W=V$.

Anway, here is what I don't understand. Don't we know that either $\displaystyle W$ is invariant under $\displaystyle T$ or $\displaystyle T$ is non-injective (and has eigenvalue 0?). If $\displaystyle w\in W$ and $\displaystyle Tw=u\in U$, then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a $\displaystyle u'\in U$ such that $\displaystyle Tu'=u=Tw$, which shows that $\displaystyle T$ isn't injective?

The short version: if W is not invariant under T, then must T have eigenvalue 0?
Thanks.

2. Originally Posted by billa
I'm trying to understand a proof from my textbook. They are showing that every operator on a finite dimensional real vector space with odd dimension has an eigenvalue. I already know that every operator on a finite dimensional real vector space has an invariant subspace of dimension two or one...

We have $\displaystyle V$, which is the finite dimensional vector space, and $\displaystyle T$, which is an operator on V. There is a subspace $\displaystyle U$ of $\displaystyle V$ that is invariant and has dimension 2 (otherwise we are done). There is some other subspace $\displaystyle W$ such that $\displaystyle U\bigoplus W=V$.

Anway, here is what I don't understand. Don't we know that either $\displaystyle W$ is invariant under $\displaystyle T$ or $\displaystyle T$ is non-injective (and has eigenvalue 0?). If $\displaystyle w\in W$ and $\displaystyle Tw=u\in U$, then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a $\displaystyle u'\in U$ such that $\displaystyle Tu'=u=Tw$, which shows that $\displaystyle T$ isn't injective?

The short version: if W is not invariant under T, then must T have eigenvalue 0?
Thanks.
I think this is easier than that (if this isn't satisfactory, and I'm being a jerk for not reading your post and answering your actual question...feel free to let me know...and I'll read it more carefully :P)

See if you can see why the following lemma finishes the problem

Theorem: Let $\displaystyle p:\mathbb{R}\to\mathbb{R}$ be a real polynomial and $\displaystyle \deg p=2n+1$, then $\displaystyle p$ has a real zero.
Proof: Let

$\displaystyle \displaystyle p(x)=\sum_{j=0}^{2n+1}a_j x^j$

Since $\displaystyle p(x_0)=0$ if and only if $\displaystyle \frac{1}{a_{2n+1}}p(x_0)=0$ we may assume without loss of generality that $\displaystyle p$ is monic. Note then that $\displaystyle \displaystyle \lim_{x\to-\infty}p(x)=-\infty$ and $\displaystyle \displaystyle \lim_{x\to\infty}p(x)=\infty$. Thus, there exists some $\displaystyle T>0$ such that $\displaystyle x<-T\implies p(x)<0$ and $\displaystyle x>T\implies p(x)>0$. Thus, $\displaystyle p(-T-1)<0$ and $\displaystyle p(T+1)>0$. Thus, by the Intermediate Value Theorem there exists some $\displaystyle \xi\in(-T-1,T+1)$ such that $\displaystyle p(\xi)=0$. The conclusion follows. $\displaystyle \blacksquare$

3. I had typed a really long message here... but I deleted it. Here is the abridged version (I tend to type too much )

Because of the approach of my course and textbook, I have no definition of a characteristic polynomial or determinant of an operator on a real vector space. I tell you this because I suspect that I need these concepts to use your theorem... if not, let me know, and I'll keep trying.

Thanks again...