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Math Help - Linear Algebra: Determinant question

  1. #1
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    Linear Algebra: Determinant question

    Hello,

    Problem:

    If A,B,C,D are matrices such that C and D commute and D is invertible, then


    \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}(AD - BC)

    There is a hint that says " multiply on the right by \begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix}".

    My effort:

    So I observed that:

    \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \text{det}\begin{pmatrix} A & B \\ C & D \end{pmatrix}\cdot\text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix}

    If we choose X = -D^{-1}C, then \text{det}\begin{pmatrix} A+BX & B \\ C+DX & D \end{pmatrix} = \text{det}\begin{pmatrix} A- BD^{-1}C & B \\ 0 & D \end{pmatrix}

    Now since C and D commute, D^{-1}C = CD^{-1}. Thus

    \text{det}\begin{pmatrix} A-BD^{-1}C & B \\ 0 & D \end{pmatrix} = \text{det}\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} = \text{det}(A-BCD^{-1})\text{det}(D) = \text{det}(AD - BC)

    It looked like a convincing proof until I realised that I have used the hypothesis in the penultimate step. I have also used it while setting \text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1.

    So how can I fix those steps? Or do we need a different direction?

    Thank you,
    Srikanth
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  2. #2
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    Quote Originally Posted by Isomorphism View Post
    I have also used it while setting \text{det}\begin{pmatrix} 1 & 0 \\ X & 1 \end{pmatrix} = 1.

    So how can I fix those steps? Or do we need a different direction?

    Thank you,
    Srikanth
    The matrix (assume its square):

    \begin{pmatrix} I_n & 0_{m,n} \\ X & I_m \end{pmatrix}

    is lower triangular, and so its determinant is the product of the elements on the diagonal, but these are all 1 so the determinant is 1 without having to assume what was to be proven.
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  3. #3
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    Assume A,\ B,\ C,\ D are all square and multiplicativly conformant (I believe that is implied by the conditions in the problem)

    \det\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} =\det \left[\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix}\begin{pmatrix} I & 0 \\ 0 & D^{-1} \end{pmatrix}\right] \det(D)  <br />
=\text{det}\begin{pmatrix} A-BCD^{-1} & BD^{-1} \\ 0 & I \end{pmatrix} \det(D)<br />

    expanding the first determinant in the last expression, from the bottom right gives:

    \text{det}\begin{pmatrix} A-BCD^{-1} & B \\ 0 & D \end{pmatrix} =\det(A-BCD^{-1})\det(D)

    etc...
    Last edited by Constatine11; December 25th 2008 at 03:31 AM.
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