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Eigenvector of complex Eigenvalues

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    ##A=\begin{bmatrix} 16 &{-6}\\39 &{-14} \end{bmatrix}##



    2. Relevant equations



    3. The attempt at a solution

    I did ##A=\begin{bmatrix} 16-\lambda &{-6}\\39 &{-14-\lambda} \end{bmatrix}##

    and got that ##\lambda_1=1+3i## and ##\lambda_2=1-3i##

    The solution is partially given for both the vectors:

    ##x_1=\begin{bmatrix} 1+i \\ ?+?i\end{bmatrix}##
    I should fill in the "?". I tried placing 1+3i in to replace lambda:

    ##\begin{bmatrix} 16-(1+3i) &{-6}\\39 &{-14-(1+3i) } \end{bmatrix}##
    Which I then get:
    ##\begin{bmatrix} 15-3i &{-6}\\39 &{-15-3i } \end{bmatrix}##


    I don't know what to do from here. I've tried all sorts of ways to get this figured out but I just keep getting the wrong answers.
     
  2. jcsd
  3. Nov 14, 2013 #2

    Mark44

    Staff: Mentor

    Add -39 times the first row to 15 - 3i times the second row. That should result in the second row being all zeros. This seems to be a pretty messy problem to do by hand.
     
  4. Nov 14, 2013 #3
    For whatever reason all of the problems for this section are like this. I just hope our test tomorrow is simpler...

    Got ahead of myself on posting. 1 sec while I do what you said.

    They did cancel so I'm left with

    ##(-585+117i)x_1=-234x_2##

    ##x_1=1+i=\frac{-234x_2}{-585+117i}##

    which is ##x_2(\frac{135}{338}+\frac{27i}{338})=1+i##

    so ##x_2=\frac{26}{9}+\frac{52i}{27}##

    I got it wrong somewhere.
     
    Last edited: Nov 14, 2013
  5. Nov 14, 2013 #4
    I figure it out:

    ##(-585+117i)x_1+234x_2=0## where ##x_1=1+i##

    ##\frac{(-585+117i)(1+i)}{-234}=x_2##

    ##x_2=3+2i##

    This is really exactly the same so I'm not sure what happened up there. Probably just calculations. I used my calculator once I had it in the second equation form.
     
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