Results 1 to 3 of 3

Math Help - Eigenvalue of Real Vector Space Proof

  1. #1
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100

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

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

    The short version: if W is not invariant under T, then must T have eigenvalue 0?
    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by billa View Post
    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 V, which is the finite dimensional vector space, and T, which is an operator on V. There is a subspace U of V that is invariant and has dimension 2 (otherwise we are done). There is some other subspace W such that U\bigoplus W=V.

    Anway, here is what I don't understand. Don't we know that either W is invariant under T or T is non-injective (and has eigenvalue 0?). If w\in W and Tw=u\in U, then since U is invariant under T (and T is injective or else 0 is an eigenvalue) there is a u'\in U such that Tu'=u=Tw, which shows that 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 p:\mathbb{R}\to\mathbb{R} be a real polynomial and \deg p=2n+1, then p has a real zero.
    Proof: Let


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


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

  3. #3
    Member billa's Avatar
    Joined
    Oct 2008
    Posts
    100
    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...
    Last edited by billa; December 8th 2010 at 05:57 PM. Reason: I found it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complexification of a real vector space
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 5th 2011, 07:23 PM
  2. Vector Space proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 12th 2011, 06:26 PM
  3. Replies: 3
    Last Post: November 25th 2010, 11:38 PM
  4. Replies: 4
    Last Post: September 19th 2010, 03:02 PM
  5. Vector space proof
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 13th 2006, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum