# Complex Eigenvectors (2x2)

1. Nov 17, 2013

### Dgray101

1. The problem statement, all variables and given/known data

Given A = [ (3,-7),(1,-2) ] and λa = $\frac{1}{2}$ + i $\frac{\sqrt{3}}{2}$ 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$\sqrt{3}$),(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. Nov 17, 2013

### Staff: Mentor

I think you might be making a mistake in your row reduction. This is what I get:
$$\begin{bmatrix} 5/2 - (\sqrt{3}/2)i & -7 \\ 0 & 0\end{bmatrix}$$

3. Nov 17, 2013

### Dgray101

May I ask how you did that because I keep getting the same answer even redoing the problem...

4. Nov 17, 2013

### Dgray101

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. Nov 17, 2013

### 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.