# A question in prooving invertion

1. Mar 8, 2008

### transgalactic

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. Mar 8, 2008

### HallsofIvy

Staff Emeritus
$$P\left(\left[\begin{array}{cc}a & b \\c & d\end{array}\right]\right)= \left[\begin{array}{cc}d & c \\ b & a\end{array}\right]$$
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
$$\left[\begin{array}{cc}a & b \\c & d\end{array}\right]$$
becomes
$$\left[\begin{array}{cc}b & a \\d & c\end{array}\right]$$
Then Swap rows so that becomes
$$\left[\begin{array}{cc}d & c \\b & a\end{array}\right]$$

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

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
$$P\left(P\left(\left[\begin{array}{cc}a & b \\c & d\end{array}\right]\right)\right)$$