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Math Help - more matrix algebra

  1. #1
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    more matrix algebra

    Ok, so, given that, C^{-1} = LD^{-1}L^{-1}, and C is a real valued non-singular n x n matrix, D is the diagonal matrix of C's n distinct real non-zero eigenvalues, and L have as columns a linearly independent set of eigenvectors in the order given by eigenvalues down D's diagonal, how do I find eigenvalues and the corresponding eigenspaces for C^{-1} ?

    Is it just the normal method  [A - \lambda I_n]? (I can't see a result out of this method...)

    Your help is much appreciated
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  2. #2
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    Quote Originally Posted by tsal15 View Post
    Ok, so, given that, C^{-1} = LD^{-1}L^{-1}, and C is a real valued non-singular n x n matrix, D is the diagonal matrix of C's n distinct real non-zero eigenvalues, and L have as columns a linearly independent set of eigenvectors in the order given by eigenvalues down D's diagonal, how do I find eigenvalues and the corresponding eigenspaces for C^{-1} ?

    Is it just the normal method  [A - \lambda I_n]? (I can't see a result out of this method...)

    Your help is much appreciated
    Do you realise that C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}?

    Now read this matrix equation as C^{-1} acting on each column of L.... Voila! You have C^{-1}'s eigen vectors
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  3. #3
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    Quote Originally Posted by Isomorphism View Post
    Do you realise that C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}?

    Now read this matrix equation as C^{-1} acting on each column of L.... Voila! You have C^{-1}'s eigen vectors

    So the eigenvalues =  LD^{-1}<br />
?
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    Do you realise that C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}?

    Now read this matrix equation as C^{-1} acting on each column of L.... Voila! You have C^{-1}'s eigen vectors
    also is the eigenspace = x_n = {t(v_n)}?
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Do you realise that C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}?

    Now read this matrix equation as C^{-1} acting on each column of L.... Voila! You have C^{-1}'s eigen vectors
    I realised what you said, but I still need more explanation on how to do this question. Sorry Isomorphism. But how does one find eigenvalues and eigenspaces with no values....this is the complex part for me....

    Thanks for all your help
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by tsal15 View Post
    I realised what you said, but I still need more explanation on how to do this question. Sorry Isomorphism. But how does one find eigenvalues and eigenspaces with no values....this is the complex part for me....

    Thanks for all your help
    Let l_1, l_2, ... l_n be columns of L, then observe that C^{-1} L = L D^{-1} = C^{-1} [l_1 \, l_2 \, ... \, l_n] = [l_1 \, l_2 \, ... \, l_n] D^{-1} Let d_1, d_2,.. d_n be the eigen values of C, then C^{-1} [l_1 \, l_2 \, ... \, l_n] = [l_1 \, l_2 \, ... \, l_n] \text{diag}(\frac1{d_1},\frac1{d_2},..,\frac1{d_n}  ). This clearly means C^{-1} l_i = \frac1{d_i} l_i...

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