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Decomposition of matrix

  1. Nov 30, 2013 #1

    Jano L.

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    I would like to learn bit more about matrices and their decomposition. Let ##\mathbf C## be symmetric real-valued square matrix. Let ##\mathbf R## be such that

    $$
    \mathbf R\mathbf R^T = \mathbf C.
    $$

    Is the matrix ##\mathbf R## necessarily lower triangular (I suspect not)?

    Cholesky decomposition leads to ##\mathbf R## that is lower triangular. Is there some other method of calculationg ##\mathbf R## ?
     
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  3. Nov 30, 2013 #2

    ShayanJ

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    Gold Member

    For two matrices A and B we have:

    [itex]
    (AB)_{ij}=\sum_k A_{ik}B_{kj}
    [/itex]

    Now we have [itex] B=A^T \Rightarrow B_{kj}=A_{jk} [/itex] so the above equation becomes:

    [itex]
    (AA^T)_{ij}=\sum_k A_{ik}A_{jk}
    [/itex]

    Now it is obvious that [itex] (AA^T)_{ij}=(AA^T)_{ji} [/itex]

    So the product of any square matrix with its transpose is a symmetric matrix.
     
  4. Nov 30, 2013 #3

    Jano L.

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    Gold Member

    OK, from your example now I see ##\mathbf R## is not necessarily lower triangular. Thanks.
     
  5. Nov 30, 2013 #4

    D H

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    Staff Emeritus
    Science Advisor

    Since your matrix C is a symmetric real-valued square matrix, it can be decomposed as C=UƩUT, where U is a real unitary matrix and Ʃ is a diagonal matrix. If C is positive semidefinite, then R=UƩ1/2UT is a symmetric real-valued matrix such that RR=C. Since it's symmetric, one also has RRT=C since RT=R.
     
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