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Find the characteristic polynomial

  1. Jan 20, 2008 #1
    1. The problem statement, all variables and given/known data
    If I have a n x n matrix B, and I must show that a vector a is an eigenvector for the matrix B and I have to find the corresponding eigenvalue, what is the easiest way of doing this?

    3. The attempt at a solution
    I know I can find the characteristic polynomial, but I thought that there is perhaps an easier way?
     
  2. jcsd
  3. Jan 20, 2008 #2

    morphism

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    Look at Ba?
     
  4. Jan 20, 2008 #3
    Ahh yes, thanks :-)
     
  5. Jan 20, 2008 #4

    HallsofIvy

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    Write out the defining equation for an eigenvalue, Bx= [itex]\lambda[/itex]x, of course. You know x= a and you know B. Go ahead and do the multiplication on the left and compare it to the right side!

    For example if you know that
    [tex]A= \left[\begin{array}{cc} -3 & 15 \\ -2 & 8 \end{array}\right][/tex]
    and
    [tex]a= \left[\begin{array}{c} 5 \\ 2\end{array}\right][/tex]
    is an eigenvector of B, and want to find the corresponding eigenvalue, [itex]\lambda[/itex]

    [tex]\left[\begin{array}{cc} -3 & 15 \\ -2 & 8 \end{array}\right]\left[\begin{array}{c} 5 \\ 2\end{array}\right]= \lambda\left[\begin{array}{c} 5 \\ 2\end{array}\right][/tex]

    [tex]\left[\begin{array}{c} 15 \\ 6\end{array}\right]= \left[\begin{array}{c} \lambda 5 \\ \lambda 2\end{array}\right][/tex]
    So we must have [itex]15= 5\lambda[/itex] and [itex]6= 2\lambda[/itex]. Obviously, [itex]\lambda= 3[/itex] satisfies both. (If the same [itex]\lambda[/itex] had not satisfied both, then the given vector was not an eigenvector.)

    (Edit: Or I could have just said "look at Ba"!)
     
  6. Jan 20, 2008 #5
    Nono, because in the example with "Ba" the eigenvalue was 0, so I thought the method was different from the one you showed in your example.

    Thanks to both of you.
     
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