# Complex Eigenvector

1. Mar 18, 2008

### hotcommodity

[SOLVED] Complex Eigenvector

I need to solve for an eigenvector using the complex eigenvalue $$-1 + i \sqrt{11}$$. I have a matrix:

$$A = \left(\begin{array}{cc}-3 & -5 \\3 & 1\end{array}\right)$$

From the equation $$A \vec{V} = \lambda \vec{V}$$, where $$\vec{V} = (x, y)$$ I get :

$$-3x - 5y = -1x + i \sqrt{11}x$$

$$3x + y = -1y + i \sqrt{11}y$$

Which gives:

$$-2x - i \sqrt{11}x - 5y = 0$$

$$3x + 2y - i \sqrt{11}y = 0$$

When I solve this system for x and y, I get a solution of (0, 0). The book agrees with the eigenvalue that I found, but has an eigenvector solution of $$(-2 + i \sqrt{11}, 3)$$. Can anyone spot what I'm doing wrong?

Any help is appreciated.

2. Mar 19, 2008

### foxjwill

It's much easier to just plug in the eigenvalue to the matrix

$$\begin{pmatrix} -3 - \lambda & -5 \\ 3 & 1 - \lambda\end{pmatrix}$$

and then solve it that way.

Also, switch the first and second rows, again to make it easier. You're equations are correct, and they shouldn't come up with 0,0.

Last edited: Mar 19, 2008
3. Mar 19, 2008

### CompuChip

If I solve the first equation for y, I get
$$y = -\frac{1}{5} (2 + \mathrm{i}\sqrt{11}) x$$
Then plugging this into the second equation gives an equation just for x. Don't forget to simplify the prefactor as much as possible, then solve for x. You'll see that though (x, y) = (0, 0) is a possibility (which you don't want, because you want it to be an eigenvector), but there are also others for which x is non-zero.

Also note, that once you found one eigenvector, you can take any multiple and it will be an eigenvector again. So you can multiply the whole thing by a factor to make the vector look nicer (e.g. if you'd get $(12/\sqrt{1 + x}, \sqrt{1 - x})$ I'd multiply by $\sqrt{1 + x}$ and write it as $(12, \sqrt{x^2 + 1})$).

4. Mar 19, 2008

### hotcommodity

Thanks for the replies. I realize what I was doing wrong (to some degree). I made an algebraic mistake along the way. My answer wasn't the same answer that the book had, but it worked such that $$A \vec{V} = \lambda \vec{V}$$, which makes me assume that my answer was a multiple of the books answer.

5. Mar 20, 2008

### CompuChip

That's easily checked. What was your answer?

6. Mar 20, 2008

### hotcommodity

I used the eigenvalue $$\lambda = 2 + i \sqrt{11}$$ to obtain an eigenvector of:

$$\vec{V_0} = \begin{pmatrix} 5\\ -2 - i \sqrt{11}\end{pmatrix}$$

As was noted, this works such that $$A \vec{V} = \lambda \vec{V}$$.

7. Mar 21, 2008

### CompuChip

And as I said, they are multiples of each other, namely:
$$\begin{pmatrix} -2 + i \sqrt{11} \\ 3 \end{pmatrix} = \frac{i \sqrt{11} - 2}{5} \cdot \begin{pmatrix} 5 \\ - 2 - i \sqrt{11} \end{pmatrix}$$

So your answer is equivalent to that in the book, only they chose a multiple to make the second component look simple, while you make the first component look nice.

8. Mar 21, 2008

### hotcommodity

I see, I appreciate the help. Thank you.

Last edited: Mar 21, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook