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Math Help - 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
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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 :

    \begin{pmatrix}<br />
\lambda_1 & 0 & 0 &\ldots & 0 \\<br />
0 & \lambda_2 & 0 & \ldots & 0 \\<br />
0 & 0 & \lambda_3 & \ldots & 0 \\<br />
\vdots & \vdots & \vdots & \ddots & 0 \\<br />
0 & 0 & 0 & \ldots & \lambda_n \end{pmatrix}

    Where \lambda_1, \dots, \lambda_n are the eigenvalues of the matrix.
    The determinant in this base is obviously the product of the \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; December 7th 2008 at 04:25 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 A has 0 as an eigenvalue then it means A\bold{x} = 0\bold{x} for \bold{x} \not = \bold{0}.
    Therefore, A\bold{x} = \bold{0} for \bold{x}\not = \bold{0}.
    This means A is not invertible.
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