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Eigenvalues and eigenvectors

  1. May 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A and B be similar matrices
    a)Prove that A and B have the same eigenvalues


    2. Relevant equations
    None


    3. The attempt at a solution
    Firstly, i dont see how this can even be possible unless the matrices are exactly the same :S
     
  2. jcsd
  3. May 28, 2009 #2

    matt grime

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    So you think that eigenvalues uniquely characterize a matrix? What about

    10
    01

    and

    11
    01

    for example?

    You've put 'none' for relevant equations. That isn't true - there's a definition of 'similar' and many for 'eigenvalue'. Try it. HINT: polynomials.
     
  4. May 28, 2009 #3
    EDIT: Changed "equation" to "polynomial"

    You have to show that A and B=P^-1AP (for some invertible matrix P) have the same characteristic polynomial.
     
    Last edited: May 28, 2009
  5. May 28, 2009 #4

    HallsofIvy

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    Oh, I think that way is much too complicated!

    Do it directly from the equation:
    If [itex]Av= \lambda v[/itex] then, for any invertible P, [itex]P^{-1}Av= \lambda P^{-1}v[/itex]. Now define [itex]u= P^{-1}v[/itex].
     
  6. May 28, 2009 #5
    Your way is too easy. :smile:
     
  7. May 28, 2009 #6
    or let [tex] Av = \lambda v [/tex]

    then [tex] AP^{-1}Pv = \lambda v [/tex] and go from there
     
  8. May 28, 2009 #7
    what do u mean by the same characteristic equation?
     
  9. May 28, 2009 #8

    EDIT: changed "equation" to "polynomial"


    The the characteristic polynomial of matrix A is [tex] det(A- \lambda I). [/tex]

    The characterisitc polynomial of matrix B is [tex] det(B - \lambda I) = det(PAP^{-1} - \lambda I) [/tex]

    so show that [tex] det(A- \lambda I) = det(PAP^{-1} - \lambda I) [/tex]


    But there are easier ways as HallsofIvy noted.


    Start with [tex] Av = \lambda v[/tex] where v is a nonzero vector

    then [tex] AP^{-1}Pv = \lambda v [/tex] since [tex] P^{-1}P = I [/tex]

    Do you know what to do next?
     
    Last edited: May 28, 2009
  10. May 28, 2009 #9

    jbunniii

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    A and B are similar if and only if they both represent the same linear map, with respect to two possibly different bases. Eigenvalues are defined independently of what basis, if any, you choose. QED.
     
  11. May 28, 2009 #10

    Mark44

    Staff: Mentor

    Make that "The characteristic equation of matrix A is [tex] det(A- \lambda I) = 0. [/tex]"
    For it to be an equation, it at least has to have an equals sign.
     
  12. May 28, 2009 #11
    Sorry. What I should have said is that they have the same characteristic POLYNOMIAL. :redface:
     
  13. May 28, 2009 #12
    Av = lambda v
    (AP^-1 P)v = lambda v
    Bv = lambda v

    i think?
     
  14. May 28, 2009 #13
    oh wait....B = P^-1 AP .....so what i said is wrong...

    how can i manipulate A P^-1 P to look like P^-1 AP?
     
  15. May 29, 2009 #14
    ohhh i see...

    since they can have the same eigenvalues, does this mean that the matrices can also have the same eigenvectors?
     
  16. May 29, 2009 #15
    They can, but it's not likely.
     
  17. May 29, 2009 #16
    I just tried several similar matrices but they all share the same eigenvector o_O
    Can i get an example where two similar matrices have different eigenvectors?
     
  18. May 29, 2009 #17

    HallsofIvy

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    I'm surprised you were able to find similar matrices that had the same eigenvectors!

    [tex]A= \begin{bmatrix}2 & 0 \\ 0 & 3\end{bmatrix}[/tex]
    has, obviously, 2 and 3 as eigenvalues with corresponding eigenvectors <1 0> and <0 1>.

    [tex]B= \begin{bmatrix}1 & -1 \\ 2 & 4\end{bmatrix}[/tex]
    has the same eigenvalues with corresponding eigenvectors <1, -1> and <1, -2>.

    All I did was start with the obvious diagonal matrix, A, choose a simple invertible P:
    [tex]P= \begin{bmatrix}2 & 1 \\ 1 & 1\end{bmatrix}[/tex]
    and calculate [itex]B= P^{-1}AP[/itex].
     
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