Complex Eigenvectors (2x2)

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Homework Statement



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.


Homework Equations





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.
 

Answers and Replies

  • #2
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Homework Statement



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.


Homework Equations





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.

I think you might be making a mistake in your row reduction. This is what I get:
$$\begin{bmatrix} 5/2 - (\sqrt{3}/2)i & -7 \\ 0 & 0\end{bmatrix}$$
 
  • #3
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May I ask how you did that because I keep getting the same answer even redoing the problem...
 
  • #4
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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.
 
  • #5
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May I ask how you did that because I keep getting the same answer even redoing the problem...

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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