Solving Complex Eigenvector for (-1 + i \sqrt{11})

[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.

 

[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.

 

[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 $(12, \sqrt{x^2 + 1})$).

 

[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.

 

[CompuChip]

That's easily checked. What was your answer?

 

[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 $$A \vec{V} = \lambda \vec{V}$$.

 

[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.

 

[hotcommodity]

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.

I see, I appreciate the help. Thank you.