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Complex Eigenvectors (2x2)

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Given A = [ (3,-7),(1,-2) ] and λa = [itex]\frac{1}{2}[/itex] + i [itex]\frac{\sqrt{3}}{2}[/itex] find a single eigenvector which spans the eigenspace.


    2. Relevant equations



    3. The attempt at a solution

    So I row reduced the matrix to get [(2, -5 + i[itex]\sqrt{3}[/itex]),(0,0 ] and from here we can write a solution as (x1,x2)=x2((1/2,1)) however that is not a complex eigenvector, it is just a real v etor. So Somewhere along the lines I am making a mistake in understanding how you solve for a complex eigenvector. I can do this in Rn very well, but this is throwing me off.
     
  2. jcsd
  3. Nov 17, 2013 #2

    Mark44

    Staff: Mentor

    I think you might be making a mistake in your row reduction. This is what I get:
    $$\begin{bmatrix} 5/2 - (\sqrt{3}/2)i & -7 \\ 0 & 0\end{bmatrix}$$
     
  4. Nov 17, 2013 #3
    May I ask how you did that because I keep getting the same answer even redoing the problem...
     
  5. Nov 17, 2013 #4
    I think I did the similar thing to what you did, but instead I multiplied the top row by the complex conjugate to row reduce.
     
  6. Nov 17, 2013 #5

    Mark44

    Staff: Mentor

    I started with this:
    $$ \begin{bmatrix}3 - (1/2)(1 + \sqrt{3}i) & -7 \\ 1 & -2 - (1/2)(1 + \sqrt{3}i) \end{bmatrix}$$

    That simplifies to
    $$\begin{bmatrix}5/2 -i\sqrt{3}/2 & -7 \\ 1 & -5/2 - i\sqrt{3}/2 \end{bmatrix}$$

    I added -1 times the top row to (5/2 - i√3/2) times the bottom row.

    I checked my eigenvector, and for the given matrix, Ax = λx.
     
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