Complex Eigenvectors

**polarbear73** · December 24th 2010, 04:04 AM

I am solving for the eigenvector of the following array:
$\left(\begin{array}{cc}1-8i&-13\\5&-1-8i\end{array}\right)\left(\begin{array}{cc}n_1\\n_ 2\end{array}\right)=\left(\begin{array}{cc}0\\0\en d{array}\right)$
The rows are multiples of each other, and if I understand this either row can be solved for the eigenvector. If I use the bottom row I get
$\left(\begin{array}{cc}1+8i\\5\end{array}\right)$
If I use the top row I get
$\left(\begin{array}{cc}13\\1-8i\end{array}\right)$
Is this correct? Are those two vectors equal? I do not have this problem when the eigenvalues are all real numbers.

**Ackbeet** · December 24th 2010, 04:29 AM

Two questions/comments:

1. What are the eigenvalues?
2. The non-zero scalar multiple of any eigenvector is also an eigenvector with the same eigenvalue. So if your two answers are multiples of each other, then you could use either as the eigenvector.

**FernandoRevilla** · December 24th 2010, 04:31 AM

Originally Posted by polarbear73

Is this correct? Are those two vectors equal? I do not have this problem when the eigenvalues are all real numbers.

The corresponding determinant is 0. So, those vectors are proportionals. Choose one of them and you'll obtain a basis for the eigenspace.

Fernando Revilla

Edited: I didn't see the previous post.

**polarbear73** · December 24th 2010, 12:14 PM

The original problem was
$x'=\left(\begin{array}{cc}3&-13\\5&1\end{array}\right)x$
$x(0)=\left(\begin{array}{cc}3\\-10\end{array}\right)$
If I did it correctly, the determinant of the array was $=\lambda^2-4\lambda+68$
Thus $\lambda_{1,2}=2\pm8i$
I was working with $2+8i$ The book says that a 2x2 array will have 2 eigenvalues, or 1 doubled one, and that each value will return a single eigenvector for that specific array. What puzzles me is that if I use real number eigenvalues and solve either row I get the same eigenvector. I failed to notice that the two results in my first post form a singular matrix, so I guess they're multiples of each other. I'm wondering if it would be possible to choose the right values of $n_1,n_2$ to make those two equal.

**Ackbeet** · December 24th 2010, 04:57 PM

You have two distinct eigenvalues, which you've correctly computed, which guarantees that the eigenvectors will be linearly independent. You have correctly found the first eigenvector. You could use either of the eigenvectors you found in the OP. They are both correct. If you multiply the first by the scalar $(1-8i)/5,$ you get the second eigenvector. Now you should find the eigenvector corresponding to the other eigenvalue $2-8i.$

The solution to the DE is

$\mathbf{x}=e^{At}\mathbf{x}_{0},$

so you're in the process of computing the matrix exponential. Make sense?

**FernandoRevilla** · December 24th 2010, 11:12 PM

Originally Posted by polarbear73

Thus $\lambda_{1,2}=2\pm8i$

You can find inmediately the eigenvector associated to $\lambda_2=\bar{\lambda}_1$ conjugating the coordinates of the eigenvector associated to $\lambda_1$ .

In fact, suppose

$z=\begin{bmatrix}{z_1}\\{z_2}\end{bmatrix}$ is an eigenvector of $\lambda_1$

then, $Az=\lambda_1 z$

Taking conjugates and using that $A$ is a real matrix:

$\bar{A}\bar{z}=\bar{\lambda}_1\bar{z}\Rightarrow A\bar{z}= \lambda_2\bar{z}$

( of course, this result is valid for all $A\in\mathbb{R}^{n\times n}$ )

In our problem, and taking into account that $(1+8i,5)^t$ is an eigenvector for $2+8i$ , then $(1-8i,5)^t$ is an eigenvector for $2-8i$ .

As a consequence the transition matrix $P$ such that $P^{-1}AP=\textrm{diag}\;(2+8i,2-8i)$ , is:

$P=\begin{bmatrix}{1+8i}&{1-8i}\\{5}&{5}\end{bmatrix}$

Fernando Revilla

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