1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra- determining if a eigenbasis exists

  1. Dec 4, 2011 #1
    I have the matrix:
    1 1 0
    0 1 0
    0 0 0

    First question: Is it correct that this is a 3 x 3 matrix (as opposed to a 2 x 2 matrix, since the last row and column are 0s)?

    I have found the eigenvalues to be λ1 = 1, λ2 = 1, λ3 = 0. Then I have found the corresponding eigenspaces to be E1 = E2 = span (1 0 0)^T and E3 = span (0 0 1)^T.

    Second question: Are λ1 and λ2 considered different eigenvalues? If so, is each considered to have its own space, even if they are just equal?

    I have the following theorem (part b. on p324 LA w/ Otto Bretscher):
    There exists an eigenbasis for an n x n matrix A if (and only if) the geometric multiplicities of the eigenvalues add up to n.

    Final question: Is there an eigenbasis for the above matrix (which I have already found the eigenvalues and eigenvectors for)? Here are my two trains of thought and I do not know which is correct:

    Train 1: the multiplicity of λ1 is 1, the multiplicity of λ2 is 1, and the multiplicity of λ3 is 1. Therefore, 1+1+1 = 3 = n. Therefore, an eigenbasis exists.

    Train 2: the repeated eigenvalue (λ1 = λ2 = 1) is treated as 1 eigenvalue. Then we have the multiplicity of the repeated eigenvalue is 1 and the multiplicity of the other eigenvalue is 1. 1+1 = 2 < n = 3. Therefore, an eigenbasis does not exist.

    Which is correct?

    Thanks.
     
  2. jcsd
  3. Dec 5, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Can you find two independent vectors so as Av=1*v and
    independent from that for λ=0?

    ehild
     
  4. Dec 5, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No. This matrix has two eigenvalues, 1 and 0. 1 has algebraic multiplicity (multplicity as a root of the characteristic polynomial) 2 and 0 has algebraic multiplicity 1.

    That is the "algebraic" multiplicity, not the "geometric" multiplicity.

    You are confusing "algebraic" multiplicity, the multiplicity as a root of the characteristic equation, with the "geometric" multiplicity, the dimension of the eigenspace for that eigenvalue. The geometric multiplicity of an eigenvalue may be less than the algebraic multiplicity. Through "train 1" the geometric multiplicity would be the same as the geometric multiplicity for every matrix, there would always exist an eigenbasis, and every matrix would be diagonalizable!

    The fact that every eigenvector corresponding to eigenvalue 1 is a multiple of [1, 0, 0] tells you the eigenspace has dimension 1 so the geometric multiplicity of eigenvalue 1 is 1, not 2.

    "Train 2" is correct if you specify geometric multiplicity.
     
    Last edited: Dec 5, 2011
  5. Dec 5, 2011 #4

    Deveno

    User Avatar
    Science Advisor

    you have just 2 eigenspaces, E1 and E0.

    now if we call our matrix A, E0 is the _____ of A.

    by the ___-____ theorem, we know that the dimension of _____ is__?

    so the real question becomes: what is dim(E1)?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Algebra- determining if a eigenbasis exists
Loading...