Solns to complex eigenvector eq's

[cue928]

I have the following complex eigenvector:
-1+2i -5 0
1 1+2i 0

What is the best way to go about solving these problems? I've done a few by inspection/trial&error, but I believe there has to be a more formal way to do it.

 

[Mark44]

I have the following complex eigenvector:
-1+2i -5 0
1 1+2i 0

What is this? It doesn't look like a vector. It might be helpful for us to see the whole problem.

What is the best way to go about solving these problems? I've done a few by inspection/trial&error, but I believe there has to be a more formal way to do it.

 

[cue928]

Sorry in advance for not knowing how to do the formatting on here...
x1' = x1 - 5x2
x2' = x1 + 3x2
(1's, 2's are subscripts)
Matrix form:
1 -5
1 3

Then this matrix:
1-lambda -5
1 3-lambda
I solved this for lambda 2+- 2i
using 2-2i, I came up with the following matrix:
-1+2i 5
1 1+2i

 

[Mark44]

OK, I see what you're doing now - you are trying to row-reduce two matrices like this (this is the one for the 2 + 2i eigenvalue):
$$\begin{bmatrix} -1 - 2i & -5\\1 & 1 - 2i\end{bmatrix}$$

Note - you can click the matrix to see how I formatted the LaTeX.

If you add the first row to (1 + 2i) times the second row, you get this matrix:
$$\begin{bmatrix} -1 - 2i & -5\\0 & 0\end{bmatrix}$$

Since you're trying to find a solution for Ax = 0, where x = <x₁, x₂>, the matrix above indicates that (-1 - 2i)x₁ -5x₂ = 0, or x₁ = -5/(1 + 2i) *x₂.


If my arithmetic is all right, this says that x₁ = [-1 + 2i]x₂
EDIT: It wasn't correct, but it's fixed now.

Set x₂ = 1 to get a value for x₁, and you'll have one of your eigenvectors.

Do the same for the matrix, using the other eigenvalue.

 

[cue928]

The answer the book provides is:
x1(t) = e^2t(-5c1 Cos 2t - 5C2 sin 2t)
x2(t) = e^2t[(c1+2c2) cos 2t + (-2c1 + c2) sin 2t]

Deconstructing that, I thought the eigenvector would be: [-5 1+2i]
Multiplying that by -1+2i -5 = 0

 

[cue928]

For example, here's another one on which I was working::
$$\begin{bmatrix} 2-4i & -5\\4 & -2-4i\end{bmatrix}$$

By trial/error, I got V=[5 2-4i]

But, I am worried that there is a different way I should be doing it.

 

[Mark44]

My arithmetic was in error, but I've fixed it now.

An eigenvector for the eigenvalue 2 + 2i is
$$\begin{bmatrix} -1 + 2i \\ 1\end{bmatrix}$$

I have checked this by comparing Ax and λx (they were equal), where
$$A = \begin{bmatrix} 1 & -5 \\ 1&3\end{bmatrix}$$
λ = 2 + 2i, and x as above.

 

[Mark44]

For example, here's another one on which I was working::
$$\begin{bmatrix} 2-4i & -5\\4 & -2-4i\end{bmatrix}$$

By trial/error, I got V=[5 2-4i]

But, I am worried that there is a different way I should be doing it.

Yes, you should not be doing these by trial and error. Assuming the matrix above is A - λI, row reduce the matrix to get a row of zeroes, just as I did in my earlier post.

 

[cue928]

I definitely follow the row reduction you used on the one earlier but let me make sure I have the next step right b/c I don't see how you got from there to the eigenvector:
You had (-1-2i)x1 - 5x2 =
You are using a value of 1 for x2 so (-1-2i)x1 = 5?

 

[cue928]

EDIT: I think I see this now, will be trying a few more to see if I have this method down. Just out of curiosity, is this considered material that would be covered in linear algebra? Or did I miss it somewhere else?

 

[Mark44]

I definitely follow the row reduction you used on the one earlier but let me make sure I have the next step right b/c I don't see how you got from there to the eigenvector:
You had (-1-2i)x1 - 5x2 =
You are using a value of 1 for x2 so (-1-2i)x1 = 5?

Yes, so x₁ = 5/(-1 - 2i) and an eigenvector would be <5/(-1 - 2i), 1>. This should be cleaned up a bit, though, by multiplying the first coordinate by (-1 + 2i)/(-1 + 2i).

EDIT: I think I see this now, will be trying a few more to see if I have this method down. Just out of curiosity, is this considered material that would be covered in linear algebra? Or did I miss it somewhere else?

Yes, linear algebra, but some courses might not do much with complex numbers.

 

[HallsofIvy]

I have never particularly liked "row-reduction" for finding eigenvectors- perhaps just because that was not the method I first learned.

Instead, I use the basic definition of "eigenvalue". If 2+ 2i is an eigenvalue, then there exist a vector such that
$$\begin{bmatrix}1 & -5 \\ 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}(2+ 2i)x \\ (2+ 2i)y\end{bmatrix}$$

That gives the two equations: x- 5y= (2+ 2i)x and x+ 3y= (2+ 2i)y. The first equation becomes -5y= (1+ 2i)x or y= (-1/5- 2i/5)x. If we take x= 5, y= -1- 2i and then, just as a check, the second equation is 5- 3- 6i= 2- 6i and (2+2i)(-1- 2i)= -2-4i- 2i+ 4= 2- 6i, which is correct. An eigenvector corresponding to eigenvalue 2+ 2i is
$$\begin{bmatrix}5 \\ -1- 2i\end{bmatrix}$$