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Geometric multiplicity of an eigenvalue

  1. Nov 9, 2011 #1
    Say we have an eigenvalue [itex]\lambda[/itex] and corresponding eigenvectors of the form [itex](x,x,2x)^T[/itex].

    What is the geometric multiplicity?
     
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  3. Nov 9, 2011 #2

    Dick

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    Well, what's the definition of geometric multiplicity?
     
  4. Nov 9, 2011 #3
    If we have a matrix [itex]A[/itex] and eigenvalue [itex]\lambda[/itex] then by definition the geometric multiplicity of [itex]\lambda[/itex] is the dimension of [itex]\text{Ker}(A-\lambda I)[/itex] which is just the dimension of the eigenspace.

    So if we have found an eigenvector, say [itex]\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}[/itex] then what is the geometric multiplicity? i.e. what is the dimension of the eigenspace? Is it the number of non-zero elements of an eigenvector or the number of different non-zero elements?
     
  5. Nov 9, 2011 #4

    Dick

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    It's neither of those. It's the number of elements in a basis for your eigenspace. You've figured out that all of the eigenvectors are multiples of a single vector. So what's the dimension?
     
  6. Nov 9, 2011 #5
    How do I write a basis for the eigenspace? Won't the eigenspace for an eigenvalue always be multiples of an eigenvector?
     
  7. Nov 9, 2011 #6

    Dick

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    You just wrote a basis for the eigenspace. It's [1,1,2]. It spans the eigenspace. You might want to review the definition of basis and dimension. If the eigenspace had been given by [x,y,2x]^T, it would be two dimensional. What would be a basis for that?
     
  8. Nov 9, 2011 #7

    Dick

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    Ack. I misread your post. I thought you'd written [1,1,2]. All of the vectors in [x,x,2x]^T are multiples of the single vector [1,1,2]^T. Yes?
     
  9. Nov 9, 2011 #8
    A question I've just done has an eigenvector [tex]\begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}[/tex] for an eigenvalue (from the equation [itex]x=\frac{1}{4}y=\frac{1}{2}z[/itex]).

    How do I know whether this is the only linearly independent eigenvector in a basis for the eigenspace? (I know the algebraic multiplicity is 2 and that the geometric multiplicity must be less than or equal to this).
     
  10. Nov 9, 2011 #9

    Dick

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    It's one dimensional again. All the eigenvectors have the form [x,4x,2x]^T. They are all multiples of a single vector [1,4,2]^T. So that single vector is a basis.
     
  11. Nov 9, 2011 #10
    In another question all the eigenvector equations are multiples of [itex]2x-y-2z=0[/itex]

    In this case the geometric multiplicity is 2 - how do you know there are 2 linearly independent eigenvectors?

    [tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}[/tex] are 2 linearly independent eigenvectors but isn't[tex]\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}[/tex] a 3rd?
     
    Last edited: Nov 9, 2011
  12. Nov 9, 2011 #11

    Dick

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    Solve for z in terms of x and y. So z=x-y/2. That means if you know x and y then z is determined. So it's a two parameter solution, it's a plane. It's two dimensional. Now you tell me a basis for it. Two linearly independent solutions that span the plane.
     
  13. Nov 9, 2011 #12
    [tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] and [tex]\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}[/tex] are 2 linearly independent eigenvectors but isn't[tex]\begin{bmatrix}0 \\ 2 \\ -1 \end{bmatrix}[/tex] a 3rd?
     
  14. Nov 9, 2011 #13

    Dick

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    [0,2,-1]=(1)*[1,2,0]+(-1)*[1,0,1]. No, the third vector isn't linearly independent of the first two.
     
  15. Nov 9, 2011 #14
    So it is!

    If you have a zero in an eigenvector, say[tex]\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}[/tex] then how does this affect things? Does it reduce the dimension by 1?
     
  16. Nov 9, 2011 #15

    Dick

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    I really think you should review vector spaces and linear independence. Why do you think having a zero component in a vector has anything to do with it????
     
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