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Math Help - Eigenvectors of this 3x3 matrix

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by handsonstance View Post
    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}<br />
\alpha-\lambda  & 0 & 1\\ <br />
0 & \beta-\lambda  & 0\\ <br />
-1 & 0 & \alpha-\lambda <br />
\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}<br />
\alpha-(\alpha+\mathbf{i})  & 0 & 1\\ <br />
0 & \beta-(\alpha+\mathbf{i})  & 0\\ <br />
-1 & 0 & \alpha-(\alpha+\mathbf{i}) <br />
\end{vmatrix}=\begin{vmatrix}<br />
-\mathbf{i}  & 0 & 1\\ <br />
0 & \beta-\alpha-\mathbf{i}  & 0\\ <br />
-1 & 0 & -\mathbf{i} <br />
\end{vmatrix}

    rref=\begin{vmatrix}<br />
1 & 0 & \mathbf{i}\\ <br />
0 & 1  & 0\\ <br />
0 & 0 & 0<br />
\end{vmatrix}\rightarrow x_3\begin{vmatrix}<br />
0\\ <br />
0\\ <br />
1 <br />
\end{vmatrix}+\mathbf{i}\begin{vmatrix}<br />
-1\\ <br />
0\\ <br />
0 <br />
\end{vmatrix}

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

    rref=\begin{vmatrix}<br />
1 & 0 & -\mathbf{i}\\ <br />
0 & 1  & 0\\ <br />
0 & 0 & 0<br />
\end{vmatrix}\rightarrow x_3\begin{vmatrix}<br />
0\\ <br />
0\\ <br />
1 <br />
\end{vmatrix}+\mathbf{i}\begin{vmatrix}<br />
1\\ <br />
0\\ <br />
0 <br />
\end{vmatrix}

    \lambda_3=\beta:\begin{vmatrix}<br />
\alpha-\beta  & 0 & 1\\ <br />
0 & \beta-\beta & 0\\ <br />
-1 & 0 & \alpha-\beta <br />
\end{vmatrix}=\begin{vmatrix}<br />
\alpha-\beta  & 0 & 1\\ <br />
0 & 0 & 0\\ <br />
-1 & 0 & \alpha-\beta<br />
\end{vmatrix}

    rref=\begin{vmatrix}<br />
1 & 0 & 0\\ <br />
0 & 0 & 1\\ <br />
0 & 0 & 0<br />
\end{vmatrix}\rightarrow x_2\begin{vmatrix}<br />
0\\ <br />
1\\ <br />
0 <br />
\end{vmatrix}
    Last edited by dwsmith; May 21st 2010 at 06:14 PM. Reason: Not paying attention
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  3. #3
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    how did you get to the last two lines of each of the calculations?
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  4. #4
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    Quote Originally Posted by handsonstance View Post
    how did you get to the last two lines of each of the calculations?
    The rref lines?
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  5. #5
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    yep the rref line and then the final eigenvector?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    rref=\begin{vmatrix}<br />
1 & 0 & \mathbf{i}\\ <br />
0 & 1  & 0\\ <br />
0 & 0 & 0<br />
\end{vmatrix}\rightarrow x_3\begin{vmatrix}<br />
\mathbf{i}\\ <br />
0\\ <br />
1 <br />
\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}<br />
1 & 0 & \mathbf{i}\\ <br />
0 & 1  & 0\\ <br />
0 & 0 & 0<br />
\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}<br />
  -\mathbf{i}\\ <br />
  0\\ <br />
  x_3 <br />
  \end{vmatrix}\rightarrow x_3\begin{vmatrix}<br />
 0\\ <br />
 0\\ <br />
 1 <br />
 \end{vmatrix}+\mathbf{i}\begin{vmatrix}<br />
-1\\ <br />
0\\ <br />
0 <br />
\end{vmatrix}
    Last edited by dwsmith; May 21st 2010 at 06:14 PM.
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  7. #7
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    ah i get it. thanks very much for your help!
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  8. #8
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    Quote Originally Posted by handsonstance View Post
    ah i get it. thanks very much for your help!
    Where did the original answers come from though for \alpha\pm\mathbf{i}?
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  9. #9
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    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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