Results 1 to 4 of 4

Math Help - Scalar multiplicaton of the identity operator

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    8

    Scalar multiplicaton of the identity operator

    The question I'm trying to answer says:

    Suppose the T is an element of L(V), is such that every non-zero vector in V is an eigenvector of T. Prove that T is scalar multiplication of the identity operator

    The way I look at it is that for T to be able to take on every vector of V as an eigenvector T must be the identity operator itself, or some multiple of it, otherwise it would not be able to take on these exact values of V. I'm wondering how I would go about proving this or if I am on the wrong track completely.

    any help is Greatly appreciated!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by falloutboy10 View Post
    The question I'm trying to answer says:

    Suppose the T is an element of L(V), is such that every non-zero vector in V is an eigenvector of T. Prove that T is scalar multiplication of the identity operator

    The way I look at it is that for T to be able to take on every vector of V as an eigenvector T must be the identity operator itself, or some multiple of it, otherwise it would not be able to take on these exact values of V. I'm wondering how I would go about proving this or if I am on the wrong track completely.

    any help is Greatly appreciated!!
    Let \{v_1,..,v_n\} be a basis of the linear space, so Tv_i=\lambda_iv_i.
    Now, first show that \lambda_i=\lambda_j\,\,\forall\,i\,,\,j - Show this first for v_1,v_2 and then a simple inductions does the work.-
    Next, after we know that Tv_i=\lambda v_i\,\,\forall i, check that T-\lambda I=0= the zero operator.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by falloutboy10 View Post
    The question I'm trying to answer says:

    Suppose the T is an element of L(V), is such that every non-zero vector in V is an eigenvector of T. Prove that T is scalar multiplication of the identity operator

    The way I look at it is that for T to be able to take on every vector of V as an eigenvector T must be the identity operator itself, or some multiple of it, otherwise it would not be able to take on these exact values of V. I'm wondering how I would go about proving this or if I am on the wrong track completely.

    any help is Greatly appreciated!!
    Suppose there were two distinct eigenvalues of T, \lambda_1 and \lambda_2 and that v_1 and v_2 are non-zero eigenvectors corresponding to \lambda_1 and \lambda_2, respectively. Then T(v_1+ v_2)= T(v_1)+ T(v_2)= \lambda_1v_1+ \lambda_2 v_2. But since every vector is an eigenvector of T, there must exist some \lambda_3 such that T(v_1+ v_2)= \lambda_3(v_1+ v_2)= \lambda_3v_1+ \lambda_3v_2. That gives \lambda_1v_2+ \lambda_2v_2 = \lambda_3v_1+ \lambda_3v_2. From that, (\lambda_1-\lambda_3)v_1= (\lambda_3- \lambda_2)v_2.

    Since v_1 and v_2 are non-zero eigenvectors corresponding to distinct eigenvalues, they cannot be multiples of one another. But that means we must have \lambda_1- \lambda_3= 0 and \lambda_3-\lambda_2= 0 leading to the conclusion that \lambda_1= \lambda_2, a contradiction. That is, if T has every vector as eigenvector, it must have only one eigenvalue.
    Last edited by HallsofIvy; November 10th 2009 at 05:16 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2009
    Posts
    8
    Thank you both! I completely get it now
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Identity operator
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 13th 2011, 02:03 AM
  2. Scalar Multiplication
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: January 22nd 2011, 05:23 AM
  3. scalar products
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 12th 2010, 07:43 PM
  4. identity linear operator
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 29th 2009, 11:01 PM
  5. what is a scalar?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: December 18th 2008, 09:19 AM

Search Tags


/mathhelpforum @mathhelpforum