A question in prooving invertion

  • #1
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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??
 

Answers and Replies

  • #2
HallsofIvy
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[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]
when i tried to find the invert of this operator i got the same matrix as before
very weird result
is that ok??
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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