Eigenvectors of this 3x3 matrix

**handsonstance** · May 18th 2010, 12:37 PM

The matrix is

alpha 0 1
0 Beta 0
-1 0 alpha

The eigenvalues of this is Beta, alpha±i which is fine.
I don't however understand how the eigenvectors end up being
For lambda = beta, (0,1,0)
For lambda = alpha ± i, (1,0,0),(0,0,1).

Any help would be appreciated as I have an exam tomorrow!

**dwsmith** · May 18th 2010, 01:24 PM

Originally Posted by handsonstance

The matrix is

alpha 0 1
0 Beta 0
-1 0 alpha

The eigenvalues of this is Beta, alpha±i which is fine.
I don't however understand how the eigenvectors end up being
For lambda = beta, (0,1,0)
For lambda = alpha ± i, (1,0,0),(0,0,1).

Any help would be appreciated as I have an exam tomorrow!

$\begin{vmatrix} \alpha-\lambda & 0 & 1\\ 0 & \beta-\lambda & 0\\ -1 & 0 & \alpha-\lambda \end{vmatrix}=(\alpha-\lambda)[(\beta-\lambda)(\alpha-\lambda)]+(\beta-\lambda)=0$

$\lambda_1=\alpha+\mathbf{i},$ $\lambda_2=\alpha-\mathbf{i}$ , and $\lambda_3=\beta$

$\lambda_1=\alpha+\mathbf{i}:\begin{vmatrix} \alpha-(\alpha+\mathbf{i}) & 0 & 1\\ 0 & \beta-(\alpha+\mathbf{i}) & 0\\ -1 & 0 & \alpha-(\alpha+\mathbf{i}) \end{vmatrix}=\begin{vmatrix} -\mathbf{i} & 0 & 1\\ 0 & \beta-\alpha-\mathbf{i} & 0\\ -1 & 0 & -\mathbf{i} \end{vmatrix}$

$rref=\begin{vmatrix} 1 & 0 & \mathbf{i}\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{vmatrix}\rightarrow x_3\begin{vmatrix} 0\\ 0\\ 1 \end{vmatrix}+\mathbf{i}\begin{vmatrix} -1\\ 0\\ 0 \end{vmatrix}$

$\lambda_2=\alpha-\mathbf{i}:\begin{vmatrix} \alpha-(\alpha-\mathbf{i}) & 0 & 1\\ 0 & \beta-(\alpha-\mathbf{i}) & 0\\ -1 & 0 & \alpha-(\alpha-\mathbf{i}) \end{vmatrix}=\begin{vmatrix} \mathbf{i} & 0 & 1\\ 0 & \beta-\alpha+\mathbf{i} & 0\\ -1 & 0 & +\mathbf{i} \end{vmatrix}$

$rref=\begin{vmatrix} 1 & 0 & -\mathbf{i}\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{vmatrix}\rightarrow x_3\begin{vmatrix} 0\\ 0\\ 1 \end{vmatrix}+\mathbf{i}\begin{vmatrix} 1\\ 0\\ 0 \end{vmatrix}$

$\lambda_3=\beta:\begin{vmatrix} \alpha-\beta & 0 & 1\\ 0 & \beta-\beta & 0\\ -1 & 0 & \alpha-\beta \end{vmatrix}=\begin{vmatrix} \alpha-\beta & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & \alpha-\beta \end{vmatrix}$

$rref=\begin{vmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{vmatrix}\rightarrow x_2\begin{vmatrix} 0\\ 1\\ 0 \end{vmatrix}$

**handsonstance** · May 18th 2010, 01:37 PM

how did you get to the last two lines of each of the calculations?

**dwsmith** · May 18th 2010, 01:37 PM

Originally Posted by handsonstance

how did you get to the last two lines of each of the calculations?

The rref lines?

**handsonstance** · May 18th 2010, 01:39 PM

yep the rref line and then the final eigenvector?

**dwsmith** · May 18th 2010, 01:44 PM

Originally Posted by dwsmith

$rref=\begin{vmatrix} 1 & 0 & \mathbf{i}\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{vmatrix}\rightarrow x_3\begin{vmatrix} \mathbf{i}\\ 0\\ 1 \end{vmatrix}$

rref is reduced row echelon form. To do so, you perform elementary row operations or plug your matrix in to a calculator select rref.

$rref=\begin{vmatrix} 1 & 0 & \mathbf{i}\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{vmatrix}$

From this point, the column vector tell us $x_1=-\mathbf{i}$ , $x_2=0$ , and $x_3$ is a free variable. This can be written as such:
$\begin{vmatrix} -\mathbf{i}\\ 0\\ x_3 \end{vmatrix}\rightarrow x_3\begin{vmatrix} 0\\ 0\\ 1 \end{vmatrix}+\mathbf{i}\begin{vmatrix} -1\\ 0\\ 0 \end{vmatrix}$

**handsonstance** · May 18th 2010, 02:20 PM

ah i get it. thanks very much for your help!

**dwsmith** · May 18th 2010, 02:21 PM

Originally Posted by handsonstance

ah i get it. thanks very much for your help!

Where did the original answers come from though for $\alpha\pm\mathbf{i}$ ?

**handsonstance** · May 18th 2010, 02:27 PM

the original answers came from my lecturer. the question was about plotting phase portraits of a 3D linear vector field and then arguing why they are topologically equivalent, with the results being alpha±i drawn in the X-Z plane, and beta drawn in the Y plane, so i am assuming what you got was what he meant by the X-Z plane. I had another go by hand and I got the same results as you did. I am not sure the of the relevance of the complex number however as it seems to be disregarded in order to get the same results as he did

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