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Distinct eigenvalues problem

  1. May 15, 2009 #1
    The real matrix A= [tex]

    \begin{pmatrix}\alpha & \beta \\ 1 & 0 \end{pmatrix}


    [/tex] has distinct eigenvalues [tex]\lambda1[/tex] and [tex]\lambda2[/tex].
    If P=[tex]

    \begin{pmatrix}\lambda1 & \lambda2 \\ 1 & 0 \end{pmatrix}


    [/tex]

    proove that P[tex]^{}^-^1[/tex]AP = D =diag{[tex]\lambda1[/tex] , [tex]\lambda2[/tex]}.

    deduce that, for every positive integer m, A[tex]^{}m[/tex] = PD[tex]^{}m[/tex]P[tex]^{}^-^1)[/tex]


    so i just tryed to multiply the whole lot out, (p^-1 is easy to find, just swap,change signs)
    and i got
    [tex]

    \begin{pmatrix}\lambda1(\alpha - \lambda2) + \beta & \lambda2(\alpha - \lambda2) + \beta \\ \lambda1(-\alpha + \lambda1) - \beta & \lambda2(-\alpha + \lambda1) - \beta \end{pmatrix}


    [/tex]

    am i going the right road with this or should i be approaching it differently?
     
  2. jcsd
  3. May 15, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Eigenvalues

    Just doing the calculation should show that, but I don't get that for the calculation.

    If
    [tex]P= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 0\end{bmatrix}[/tex]
    then
    [tex]P^{-1}= \begin{bmatrix}0 & 1 \\ \frac{1}{\lambda_1} & -\frac{\lambda_1}{\lambda_2}\end{bmatrix}[/tex]

    Is that what you got?

    You will also want to use the fact that the characteristic equation for A is [itex]x^2- \alpha x- \beta= 0[/itex] so [itex]\lambda_1^2- \alpha \lambda_1- \beta= 0[/itex] and [itex]\lambda_2^2- \alpha \lambda_2- \beta= 0[/itex].
     
  4. May 15, 2009 #3
    Re: Eigenvalues

    Nice reply!

    Perhaps, something like:

    [tex]\lambda_{1} = \frac{\alpha \pm \sqrt{\alpha^{2}+4\beta}}{2}[/tex]

    [tex]\lambda_{2} = \frac{\alpha \pm \sqrt{\alpha^{2}+4\beta}}{2}[/tex]

    not sure though how it will solve the problem.
     
  5. May 15, 2009 #4
    Re: Eigenvalues

    Couldn't you prove it by showing that the columns of P are the eigenvectors of A?
     
  6. May 15, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Eigenvalues

    Yes, if you already have that theorem. But the obvious way to do it is to just do the product [itex]P^{-1}AP[/itex].
     
  7. May 15, 2009 #6
    Re: Eigenvalues

    Thanks for all the replies. i had it wrong when getting p^-1 i only went and forgot 1/detA!!!

    so for P^1AP i get [tex]
    = \begin{bmatrix}\lambda_1 & \lambda_1 \\ \alpha-\lambda_1^2/\lambda_2+\alpha/\lambda_1 & \alpha - \lambda_1^2/\lambda_2\end{bmatrix}
    [/tex]

    I thought diag([tex]\lambda_1[/tex],[tex]\lambda_2[/tex]) = [tex]
    D= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}
    [/tex]

    i still think im doing something wrong
    confussed!
     
  8. May 18, 2009 #7
    Re: Eigenvalues

    After repeatedly trying and not getting anywhere im after realising i had the question wrong P should read [tex]


    \begin{pmatrix}\lambda1 & \lambda2 \\ 1 & 1 \end{pmatrix}



    [/tex]

    So im off the try this new version.

    But if i wanted to prove it like you said random variable how would i go about that?
    take
    [itex]
    \lambda_1^2- \alpha \lambda_1- \beta= 0 and
    \lambda_2^2- \alpha \lambda_2- \beta= 0
    [/itex] and let them = columns of p?
     
  9. May 19, 2009 #8
    Re: Eigenvalues

    right so now i got 1/([tex]\lambda_1-\lambda_2[/tex]) [tex]\begin{bmatrix}\lambda_1(\alpha - \lambda _2 ) + \beta & \lambda_2(\alpha - \lambda_2) + \beta \\ \lambda_1(-\alpha + \lambda_1) - \beta & \lambda_2(-\alpha + \lambda_1) - \beta\end{bmatrix}

    [/tex]

    not sure where to go from here...
     
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