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Math Help - the algebra of Diagonalized matrices

  1. #1
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    the algebra of Diagonalized matrices

    if C = LDL^{-1} , where D is the Diagonal Matrix, how do I rearrange the expression to have C^{-1} in terms of L, L^{-1} and D. (also, what specific diagonal matrix can I give? is the I_n ok?)

    Many Thanks
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    Quote Originally Posted by tsal15 View Post
    if C = LDL^{-1} , where D is the Diagonal Matrix, how do I rearrange the expression to have C^{-1} in terms of L, L^{-1} and D. (also, what specific diagonal matrix can I give? is the I_n ok?)

    Many Thanks

    Multiply both sides by  LD^{-1}L^{-1}

     C LD^{-1}L^{-1} =LD^{-1}L^{-1}LDL^{-1}

     C LD^{-1}L^{-1} =LL^{-1}DD^{-1}LL^{-1}

     C LD^{-1}L^{-1} = I

    Now multiply both sides by  C^{-1}

     CC^{-1} LD^{-1}L^{-1} = IC^{-1}

     C^{-1} = LD^{-1}L^{-1}

    If  D is the identity matrix then  D^{-1} = D , and hence:

     C^{-1} = LDL^{-1}
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    Quote Originally Posted by Mush View Post
    If  D is the identity matrix then  D^{-1} = D , and hence:

     C^{-1} = LDL^{-1}
    Having said that, theoretically, C^{-1} = C, right?

    So, D = I is not feasible?

    Thanks for your helpful reply
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  4. #4
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    This question goes on to say that I need to specify the diagonal matrix.

    Could the Diagonal matrix be:

    \begin{bmatrix} \lambda_1 & 0 & n \\0 & \lambda_2 & n \\ n & n & \lambda_n \end{bmatrix}

    I mentioned this question in an earlier post. I'm just wondering if I can get some further assistance, if possible

    Thanks to all assistance received
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  5. #5
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    Quote Originally Posted by tsal15 View Post
    This question goes on to say that I need to specify the diagonal matrix.

    Could the Diagonal matrix be:

    \begin{bmatrix} \lambda_1 & 0 & n \\0 & \lambda_2 & n \\ n & n & \lambda_n \end{bmatrix}

    I mentioned this question in an earlier post. I'm just wondering if I can get some further assistance, if possible

    Thanks to all assistance received
    A matrix of eigenvalues normally works, yes.
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  6. #6
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    Quote Originally Posted by tsal15 View Post
    if C = LDL^{-1} , where D is the Diagonal Matrix, how do I rearrange the expression to have C^{-1} in terms of L, L^{-1} and D. (also, what specific diagonal matrix can I give? is the I_n ok?)

    Many Thanks
    Just a note, using the property that

    \left( AB \right)^{-1} = B^{-1}A^{-1}

    gives your result pretty quickly. To go on further, if C has n distict eigenvalues, say \lambda_n, then to each we find an associatede eigenvector, say \bold{e}_n, then the diagonal matrix is given by

    D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\<br />
0 & \lambda_1 & \cdots & 0 \\<br />
0 & 0 & \ddots& 0 \\<br />
0 & \cdots & 0 & \lambda_n \end{bmatrix}

    and L by

    L = \begin{bmatrix} \bold{e}_1 & \bold{e}_1 & \cdots & \bold{e}_n\end{bmatrix}.

    Check with

    C = \begin{bmatrix} 2 & 1 \\<br />
1 & 2 \end{bmatrix}
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