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Math Help - Matrix in partitioned form.

  1. #1
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    Matrix in partitioned form.

    Let B be an m * m matrix with inverse B^-1, c an m-vector and 0 the m dimensional null vector. A is given in the following partitioned form...

    A = \left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right]

    Determine A^{-1} symbolically in a partition form.

    For which I get the following...

    A^{-1} = \left[ \begin{array}{cccc} d & e \\ f & g \end{array} \right]

    \left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right] \left[ \begin{array}{cccc} d & e \\ f & g \end{array} \right] =  \left[ \begin{array}{cccc} 1 & 0 \\ 0 & 1 \end{array} \right]

    Bd = 1

    d = B^{-1}

    Be = 0

    Since B is non-singular, B does not equal 0...

    e = 0

    c^Td + f = 0

    c^TB^{-1} + f = 0

    f = -c^TB^{-1}

    c^Te + g = 1

    g = 1

    A^{-1} = \left[ \begin{array}{cccc} B^{-1} & 0 \\ -c^TB^{-1} & 1 \end{array} \right]

    Is this correct so far?

    The question then asks the dimension of A and the dimensions of the submatrices in A^{-1}.

    Am I right in thinking the dimension of A is m+1, the dimension of B^{-1} is m and the dimension of -c^TB^{-1} is 1?

    Thanks for the help.
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  2. #2
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    Quote Originally Posted by sean.1986 View Post
    Let B be an m * m matrix with inverse B^-1, c an m-vector and 0 the m dimensional null vector. A is given in the following partitioned form...

    A = \left[ \begin{array}{cccc} B & 0 \\ c^T & 1 \end{array} \right]

    Determine A^{-1} symbolically in a partition form.

    For which I get the following...

    ... [SNIP] ...

    A^{-1} = \left[ \begin{array}{cccc} B^{-1} & 0 \\ -c^TB^{-1} & 1 \end{array} \right]

    Is this correct so far? Yes.

    The question then asks the dimension of A and the dimensions of the submatrices in A^{-1}.

    Am I right in thinking the dimension of A is m+1, the dimension of B^{-1} is m and the dimension of -c^TB^{-1} is 1?
    A matrix is a two-dimensional structure. So the dimensions of A are that it is an (m+1)\mathord\times(m+1) matrix; B^{-1} is mm; and -c^TB^{-1} is 1m (or the transpose of an m-vector). Also, the 0 is an m1 matrix (or an m-vector), and the 1 is a 1×1 matrix.
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  3. #3
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    Ah that makes sense, thank you.
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