# Thread: Eigenvectors of this 3x3 matrix

1. ## Eigenvectors of this 3x3 matrix

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!

2. 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 $\lambda_3=\beta$

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

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

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

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

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

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

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

4. Originally Posted by handsonstance
how did you get to the last two lines of each of the calculations?
The rref lines?

5. yep the rref line and then the final eigenvector?

6. 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:
$\begin{vmatrix}
-\mathbf{i}\\
0\\
x_3
\end{vmatrix}\rightarrow x_3\begin{vmatrix}
0\\
0\\
1
\end{vmatrix}+\mathbf{i}\begin{vmatrix}
-1\\
0\\
0
\end{vmatrix}$

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

8. Originally Posted by handsonstance
ah i get it. thanks very much for your help!
Where did the original answers come from though for $\alpha\pm\mathbf{i}$?

9. 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