Question

`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?

Answer 1

`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.