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Determining invertibility of a matrix

  1. May 31, 2012 #1
    Let A, B, C, D be matrices such that:

    AB + CD = 0

    and

    B is invertible. Moreover, consider the dimension restrictions:

    A(m x n), B(n x n), C(m x m), D(m x n)

    If C is a square matrix, is there a way to show that it is also invertible with only the above conditions?
     
    Last edited: May 31, 2012
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  3. May 31, 2012 #2

    AlephZero

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    If you take A = D = 0, then AB + CD = 0 for any matrix C, so you can't prove C is invertible.
     
  4. May 31, 2012 #3
    A and D are non-zero matrices, forget to say.
     
  5. May 31, 2012 #4

    micromass

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    Take

    [tex]A=\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right), B=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right), C=\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right), D=\left(\begin{array}{cc} -1 & 0\\ 0 & -1 \end{array}\right)[/tex]
     
  6. May 31, 2012 #5
    A and D are rectangular, not square.
     
  7. May 31, 2012 #6

    micromass

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    Take

    [tex]A=\left(\begin{array}{c} 1\\ 1\end{array}\right), B=\left(\begin{array}{c} 1 \end{array}\right), C=\left(\begin{array}{cc} 1 & 1\\ 1 & 1 \end{array}\right), D=\left(\begin{array}{c} -1\\ 0 \end{array}\right)[/tex]
     
  8. May 31, 2012 #7
    The example you gave yields incompatible dimensions: AB is 1x2 and CD is 2x1.
     
  9. May 31, 2012 #8

    micromass

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    I was editing. Check again.

    You can really find these things for yourself.
     
  10. May 31, 2012 #9
    That's what I am trying to do! B and C should have the same dimension.

    A(m x n), B(n x n), C(m x m), D(m x n)
     
  11. May 31, 2012 #10

    micromass

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    Here, m=2 and n=1.
     
  12. May 31, 2012 #11
    Yeah, you're right. Thanks a lot.

    It seems there is some more information in my physical problem to show that C should be invertible, but could not find it yet.
     
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