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Homework Help: Similar Matrices & Geometric Multiplicity

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that if two matrices are similar then they have the same eigenvalues with the same algebraic and geometric multiplicity.

    2. Relevant equations
    Matrices A,B are similar if A = C[tex]\breve{}[/tex]BC for some invertible C (and C inverse is denoted C[tex]\breve{}[/tex] because I tried for a long time to figure out how to get an inverse sign in latex but couldn't figure it out...).

    3. The attempt at a solution
    To show that two similar matrices have the same eigenvalues with the same geometric multiplicities, I need to show that their characteristic polynomials are the same.
    Let A,B be similar matrices. Then,
    A = C(inverse)BC
    A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex]BC - [tex]\lambda[/tex]I
    A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex]BC - [tex]\lambda[/tex]C[tex]\breve{}[/tex]C since C[tex]\breve{}[/tex]C=I
    A-[tex]\lambda[/tex]I = C[tex]\breve{}[/tex][B-[tex]\lambda[/tex]]C
    det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex][B-[tex]\lambda[/tex]I]C)
    det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex]det(B-[tex]\lambda[/tex]I)det(C)
    det(A-[tex]\lambda[/tex]I) = det(C[tex]\breve{}[/tex])det(C)det(B-[tex]\lambda[/tex]I)
    Therefore, det(A-[tex]\lambda[/tex]I) = det(B-[tex]\lambda[/tex]I)
    So, A, B have the same characteristic polynomials. This implies that they have the same eigenvalues with the same algebraic multiplicity. However, I do not think that this implies that they have the same geometric multiplicity because it doesn't same anything about the dimension of the eigenspace. Does it? Any suggestions on how I should try to show that the geometric multiplicity is the same?

    So does anyone have any advice? I spent a while to get the first part, I just could use a nudge in the right direction for the next part... Thanks in advance.
     
  2. jcsd
  3. Feb 27, 2010 #2

    Dick

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    No, algebraic multiplicity doesn't tell you anything about geometric multiplicity. The geometric multiplicity is the number of linearly independent eigenvectors corresponding to a given eigenvalue lambda. Hint: can you show if A=C^(-1)BC and x is an eigenvector of A, then Cx is an eigenvector of B with the same eigenvalue?
     
  4. Feb 27, 2010 #3

    So now that I've proven that A, B have same eigenvalues [tex]\lambda[/tex]1-[tex]\lambda[/tex]n and that they appear the same amount of times, consider the eigenvectors for those eigenvalues. As you said, let x be an eigenvector of A. Need to show that Cx is an eigenvector of B with the same eigenvalue...

    If x is an eigenvector of A, then:
    Ax = [tex]\lambda[/tex]x where [tex]\lambda[/tex] is the corresponding eigenvalue.

    Since A=C^(-1)BC, Ax=C^(-1)BCx=[tex]\lambda[/tex]x
    Then, CAx=BCx=C[tex]\lambda[/tex]x

    So, BCx = Cx[tex]\lambda[/tex]

    Shows that Cx is an eigenvector of B with the same eigenvalue [tex]\lambda[/tex].

    So since if x is an eigenvector of A with eigenvalue [tex]\lambda[/tex] then Cx is an eigenvector of A with the same eigenvalue [tex]\lambda[/tex] then the geometric multiplicities are the same? The same eigenvalue will correspond to a similar eigenvector and eigenspace so the geometric multiplicity will be the same?
     
  5. Feb 27, 2010 #4

    Dick

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    Basically. The geometric multiplicity is the dimension of the eigenspace of lambda. You've shown C maps the eigenspace of A to the eigenspace of B. C^(-1) does the reverse. C is a nonsingular matrix. That shows the dimensions are the same, right?
     
  6. Feb 27, 2010 #5
    Yes, I understand now. Thank you for your help. I figured out the first part and just needed that little hint to help me with the rest!
     
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