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A question in prooving invertion

  1. Mar 8, 2008 #1
    http://img364.imageshack.us/my.php?image=aaauo9.jpg

    i need to proove that this operator is invertable
    i solved it by putting a standart basis and findind that each line in the matrix
    is independant.

    is it ok??

    is there any easier way to slve it??

    when i tried to find the invert of this operator i got the same matrix as before
    very weird result
    is that ok??
     
  2. jcsd
  3. Mar 8, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    [tex]P\left(\left[\begin{array}{cc}a & b \\c & d\end{array}\right]\right)= \left[\begin{array}{cc}d & c \\ b & a\end{array}\right][/tex]
    Yes, that's okay. The multiplicative inverse of 1 is 1 and the multiplicative inverse of -1 is -1. The inverse of the identity operator is obviously the indentity operator. It's quite possible for for a linear operator to be its own inverse.

    And it isn't necessary to write this in terms of a basis. You can "reason" it out:

    We can think of P as doing two things to a matrix: 1) Swap the columns so that
    [tex]\left[\begin{array}{cc}a & b \\c & d\end{array}\right][/tex]
    becomes
    [tex]\left[\begin{array}{cc}b & a \\d & c\end{array}\right][/tex]
    Then Swap rows so that becomes
    [tex]\left[\begin{array}{cc}d & c \\b & a\end{array}\right][/tex]

    The inverse must do just the opposite, going back from
    [tex]\left[\begin{array}{cc}d & c \\b & a\end{array}\right][/tex]
    to
    [tex]\left[\begin{array}{cc}a & b \\c & d\end{array}\right][/tex]

    But the inverse (opposite) of "swap the rows" is "swap the rows" and the inverse of "swap the columns" is "swap the columns"!
    What is
    [tex]P\left(P\left(\left[\begin{array}{cc}a & b \\c & d\end{array}\right]\right)\right)[/tex]
     
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