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Thread: Prove that a matrix in NOT invertible if an eignvalue is 0

  1. #1
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    Prove that a matrix in NOT invertible if an eignvalue is 0

    ^^

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  2. #2
    Moo
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    Quote Originally Posted by ishan View Post
    ^^

    Thank you very much!
    In the basis that diagonalizes the matrix, the matrix can be written :

    $\displaystyle \begin{pmatrix}
    \lambda_1 & 0 & 0 &\ldots & 0 \\
    0 & \lambda_2 & 0 & \ldots & 0 \\
    0 & 0 & \lambda_3 & \ldots & 0 \\
    \vdots & \vdots & \vdots & \ddots & 0 \\
    0 & 0 & 0 & \ldots & \lambda_n \end{pmatrix}$

    Where $\displaystyle \lambda_1, \dots, \lambda_n$ are the eigenvalues of the matrix.
    The determinant in this base is obviously the product of the $\displaystyle \lambda_i$
    So if one of the eigenvalues is 0, the determinant is 0.
    And since the determinant doesn't change with the basis, we have proved that det A = 0, and hence it is not invertible.




    Edit : waaaah 39th post
    Last edited by Moo; Dec 7th 2008 at 03:25 AM.
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  3. #3
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    The determinant of a matrix is equal to the product of the eigenvalues. Hence the determinant is 0 and the matrix is not invertible.
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  4. #4
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    If $\displaystyle A$ has $\displaystyle 0$ as an eigenvalue then it means $\displaystyle A\bold{x} = 0\bold{x}$ for $\displaystyle \bold{x} \not = \bold{0}$.
    Therefore, $\displaystyle A\bold{x} = \bold{0}$ for $\displaystyle \bold{x}\not = \bold{0}$.
    This means $\displaystyle A$ is not invertible.
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