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Det(1+AB) = det(1+BA)

  1. Jun 20, 2013 #1
    I have some reason to believe that

    [tex]
    \det(\textrm{id} + AB) = \det(\textrm{id} + BA)
    [/tex]

    is true even when AB and BA are not the same size. In other words, A and B are not neccessarily square matrices.

    For example, if

    [tex]
    A = \big(A_1,\; A_2\big),\quad\quad\quad
    B = \left(\begin{array}{c} B_1 \\ B_2 \\ \end{array}\right)
    [/tex]

    then

    [tex]
    \det(\textrm{id} + AB) = 1 + A_1B_1 + A_2B_2
    [/tex]

    and

    [tex]
    \det(\textrm{id} + BA) = \det\left(\begin{array}{cc}
    1 + B_1A_1 & B_1 A_2 \\
    B_2 A_1 & 1 + B_2 A_2 \\
    \end{array}\right)
    [/tex]
    [tex]
    = (1 + B_1A_1)(1 + B_2A_2) - A_1A_2B_1B_2 = 1 + B_1A_1 + B_2A_2
    [/tex]

    Anyone knowing how to prove the general case?
     
  2. jcsd
  3. Jun 20, 2013 #2

    micromass

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  4. Jun 20, 2013 #3
    So much names to be known.
     
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