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Thread: Make a Matrix Defective

  1. #1
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    Make a Matrix Defective

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

    $\displaystyle \begin{bmatrix}
    1 & 1 & 0\\
    1 & 1 & 0\\
    0 & 0 & \alpha
    \end{bmatrix}\Rightarrow \begin{vmatrix}
    1-\lambda & 1 & 0\\
    1 & 1-\lambda & 0\\
    0 & 0 & \alpha-\lambda
    \end{vmatrix}=\lambda(\alpha-\lambda)(\lambda-2)=0$

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

    $\displaystyle \begin{bmatrix}
    1-\alpha & 1 & 0\\
    1 & 1-\alpha & 0\\
    0 & 0 & 0
    \end{bmatrix}$ what I am thinking is that this can only occur if $\displaystyle \alpha=0$ or $\displaystyle 2$. 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 $\displaystyle \alpha$ that make the matrix defective or show that no such value exist.

    $\displaystyle \begin{bmatrix}
    1 & 1 & 0\\
    1 & 1 & 0\\
    0 & 0 & \alpha
    \end{bmatrix}\Rightarrow \begin{vmatrix}
    1-\lambda & 1 & 0\\
    1 & 1-\lambda & 0\\
    0 & 0 & \alpha-\lambda
    \end{vmatrix}=\lambda(\alpha-\lambda)(\lambda-2)=0$

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

    $\displaystyle \begin{bmatrix}
    1-\alpha & 1 & 0\\
    1 & 1-\alpha & 0\\
    0 & 0 & 0
    \end{bmatrix}$ what I am thinking is that this can only occur if $\displaystyle \alpha=0$ or $\displaystyle 2$. 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

    $\displaystyle \left|\begin{matrix}
    1-\lambda & 1 & 0\\
    1 & 1-\lambda & 0\\
    0 & 0 & \alpha-\lambda
    \end{matrix}\right|$

    $\displaystyle = (\alpha - \lambda)\left|\begin{matrix}
    1-\lambda & 1\\
    1 & 1-\lambda\end{matrix}\right|$

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

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

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


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