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Matrix with repeated eigenvalues

  1. Nov 10, 2008 #1
    The question is asking for what values of x will the matrix have at least one repeated eigenvalue (algebraic multiplicity of 2 or greater). The matrix is
    | 3 0 0 |
    | 0 x 2 | So naturally a normal attempt to find the eigenvalue in a question with only intergers
    | 0 2 x | I would continue with:

    | [tex]\lambda[/tex]-3 0 0 |
    | 0 [tex]\lambda[/tex]-x -2|
    | 0 -2 [tex]\lambda[/tex]-x|

    And then finding the determinant would continue with ([tex]\lambda[/tex]-3)([tex]\lambda[/tex]-x)([tex]\lambda[/tex]-x) - (-2)(-2)([tex]\lambda[/tex]-3) etc...

    Except with 2 unknown variable it inevitably becomes a problem that I run into with more theoretical questions vs questions dealing only with numbers. As far as I know an eigenvalue with an algebraic multiplicity of >1 doesn't even have a geometric significance. Clearly I am approaching this entire question from the wrong angle and could use a push in the right direction if anyone is so able. Muchly appreciated.
     
  2. jcsd
  3. Nov 11, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: Eigenvalues

    I don't see why you need to be that theoretical!

    The characteristic equation for the eigenvalues is obviously [itex](\lambda- 3)((\lambda- x)^2- 4)= 0. 3 is an obvious solution to that so in order to have a double root, either 3 is a root of [itex](\lambda- x)^2- 4= 0[/itex] or that equation has a double root. If 3 is a root then we must have [itex](3- x)^2- 4= x^2- 6x+ 5= (x- 3)(x- 2)= 0[/itex] so that x= 2 or 3. If [itex](\lambda- x)^2- 4= \lambda^2- 2x\lambda+ x^2- 4= 0[/itex] has a double root, then its discriminant must be 0: [itex]4x^2- 4(x^2- 4)= 4[/itex] is never 0 so that can't happen.
     
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