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Math Help - Make a Matrix Defective

  1. #1
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    Make a Matrix Defective

    Find all possible values for \alpha that make the matrix defective or show that no such value exist.

    \begin{bmatrix}<br />
1 & 1 & 0\\ <br />
1 & 1 & 0\\ <br />
0 & 0 & \alpha<br />
\end{bmatrix}\Rightarrow \begin{vmatrix}<br />
1-\lambda & 1 & 0\\ <br />
1 & 1-\lambda & 0\\ <br />
0 & 0 & \alpha-\lambda<br />
\end{vmatrix}=\lambda(\alpha-\lambda)(\lambda-2)=0

    Therefore, \lambda_1=0, \lambda_2=\alpha, and \lambda_3=2.

    \begin{bmatrix}<br />
1-\alpha & 1 & 0\\ <br />
1 & 1-\alpha & 0\\ <br />
0 & 0 & 0<br />
\end{bmatrix} what I am thinking is that this can only occur if \alpha=0 or 2. Then if that doesn't work, there is no such value; however, I am not sure if that thought is entirely true. Would it be possible for other values of alpha to make the matrix defective?
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Find all possible values for \alpha that make the matrix defective or show that no such value exist.

    \begin{bmatrix}<br />
1 & 1 & 0\\ <br />
1 & 1 & 0\\ <br />
0 & 0 & \alpha<br />
\end{bmatrix}\Rightarrow \begin{vmatrix}<br />
1-\lambda & 1 & 0\\ <br />
1 & 1-\lambda & 0\\ <br />
0 & 0 & \alpha-\lambda<br />
\end{vmatrix}=\lambda(\alpha-\lambda)(\lambda-2)=0

    Therefore, \lambda_1=0, \lambda_2=\alpha, and \lambda_3=2.

    \begin{bmatrix}<br />
1-\alpha & 1 & 0\\ <br />
1 & 1-\alpha & 0\\ <br />
0 & 0 & 0<br />
\end{bmatrix} what I am thinking is that this can only occur if \alpha=0 or 2. Then if that doesn't work, there is no such value; however, I am not sure if that thought is entirely true. Would it be possible for other values of alpha to make the matrix defective?
    A matrix is defective if there are any repeated eigenvalues.

    So you have

    \left|\begin{matrix}<br />
1-\lambda & 1 & 0\\ <br />
1 & 1-\lambda & 0\\ <br />
0 & 0 & \alpha-\lambda<br />
\end{matrix}\right|

     = (\alpha - \lambda)\left|\begin{matrix}<br />
1-\lambda & 1\\ <br />
1 & 1-\lambda\end{matrix}\right|

     = (\alpha - \lambda)[(1 - \lambda)^2 - 1^2]

     = (\alpha - \lambda)(1 - \lambda + 1)(1 - \lambda - 1)

     = -\lambda(\alpha - \lambda)(2 - \lambda).


    So to have repeated eigenvalues, then \alpha = 2 or \alpha = 0.
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    Solved
    Last edited by dwsmith; April 24th 2010 at 08:14 PM.
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