#f:M_2(RR)->M_2(RR);f(X)=AXA^(-1);A inM_2(RR);A-#inversable;How to demonstrate that #f# is an bijective function?

1 Answer
Apr 3, 2017

#f^(-1)(X) = A^(-1)XA#

Explanation:

Note that if we define:

#g(X) = A^(-1)XA#

then:

#g(f(X)) = A^(-1)(f(X))A = A^(-1)AXA^(-1)A = IXI = X#

#f(g(X)) = A(g(X))A^(-1) = A A^(-1)XA A^(-1) = IXI = X#

So #g(X)# is inverse to #f(X)#.

The domains of #f(X)# and #g(X)# are both the whole of #M_2(RR)#, so the ranges of both are the whole of #M_2(RR)# and they are bijections.