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Geometric multiplicity > 1

  1. May 15, 2013 #1
    after finding out what geometric multiplicity was, I was surprised to notice that in every question I'd done it was always 1.
    So I'm trying to prove an example with g.m. > 1 to see why it works.
    I've found a matrix which definitely has an eigenvalue with g.m. = 2. I've checked everything with WolframAlpha, so the following is correct:

    Matrix A =
    [tex]
    \left( \begin{array}{ccc}
    5 & 4 & 2 \\
    4 & 5 & 2 \\
    2 & 2 & 2 \end{array} \right) [/tex]
    Determinant = 10

    Characteristic polynomial = [tex]-((x-10) (x-1)^2)[/tex]

    So eigenvalues =
    10
    1 < -- with a.m. = 2, and g.m. = 2

    So find the eigenvectors to find I'd start with:
    (A - 1 * I ) v = 0, the matrix being:
    [tex]
    \left( \begin{array}{ccc}
    4 & 4 & 2 \\
    4 & 4 & 2 \\
    2 & 2 & 1 \end{array} \right)[/tex]
     
    Last edited: May 15, 2013
  2. jcsd
  3. May 15, 2013 #2

    WannabeNewton

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    Science Advisor

    It is certainly not true that the geometric multiplicity is always 1 for each eigenvalue. All you know is that the geometric multiplicity is always less than or equal to the algebraic multiplicity; there is nothing that says, however, that the geometric multiplicity must be 1. Recall that if ##A## is a matrix and ##\lambda## is an eigenvalue of ##A## then the geometric multiplicity of ##\lambda## is ##\dim E_{\lambda}## where ##E_{\lambda}## is the associated eigenspace.

    As a simple counter example, consider ##A = I##. The only eigenvalue of ##I## is ##\lambda = 1 ## and every ##v\in \mathbb{R}^{n}## is an eigenvector of ##I## so ##E_{\lambda = 1} = \mathbb{R}^{n}## which has a geometric multiplicity of ##n## which is not 1 in general.
     
  4. May 15, 2013 #3
    I know, I was just looking for a normal example that showed otherwise.
    Thank you for your counter example

    How do I continue with my method?
     
  5. May 15, 2013 #4

    WannabeNewton

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    Science Advisor

    Well your example is also quite easy in fact. Taking ##B = \begin{pmatrix}
    4 & 4 &2 \\
    4& 4 &2 \\
    2 & 2 & 1
    \end{pmatrix}##
    we can very easily put this in reduced row echelon form as ##C = \begin{pmatrix}
    0 & 0 &0 \\
    0& 0 &0 \\
    2 & 2 & 1
    \end{pmatrix}##. Thus, the solutions to ##Cv = 0## are given by ##v_{1},v_{2} = \text{arbitrary}## and ##v_{3} = -2v_1 -2v_2## where ##v = (v_1,v_2,v_3)^{T}##. Therefore, setting ##v_1 = t,v_2 = s##, we can write any ##v\in \text{Null}C## as ##v = t(1,0,-2)^{T} + s(0,1,-2)^{T}## i.e. ##E_{\lambda = 1} = \text{Null}C = \text{Span}\{(1,0,-2)^{T},(0,1,-2)^{T}\}## which of course has a geometric multiplicity of 2.
     
    Last edited: May 15, 2013
  6. May 15, 2013 #5
    I see how it works now, thank you
     
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