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Thread: the algebra of Diagonalized matrices

  1. #1
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    the algebra of Diagonalized matrices

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

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

    Many Thanks

    Multiply both sides by $\displaystyle LD^{-1}L^{-1} $

    $\displaystyle C LD^{-1}L^{-1} =LD^{-1}L^{-1}LDL^{-1}$

    $\displaystyle C LD^{-1}L^{-1} =LL^{-1}DD^{-1}LL^{-1}$

    $\displaystyle C LD^{-1}L^{-1} = I$

    Now multiply both sides by $\displaystyle C^{-1} $

    $\displaystyle CC^{-1} LD^{-1}L^{-1} = IC^{-1}$

    $\displaystyle C^{-1} = LD^{-1}L^{-1} $

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

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

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

    So, D = I is not feasible?

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

    Could the Diagonal matrix be:

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

    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:

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

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

    Many Thanks
    Just a note, using the property that

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

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

    $\displaystyle D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\
    0 & \lambda_1 & \cdots & 0 \\
    0 & 0 & \ddots& 0 \\
    0 & \cdots & 0 & \lambda_n \end{bmatrix}$

    and L by

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

    Check with

    $\displaystyle C = \begin{bmatrix} 2 & 1 \\
    1 & 2 \end{bmatrix}$
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