# 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?

Apr 3, 2017

${f}^{- 1} \left(X\right) = {A}^{- 1} X A$

#### Explanation:

Note that if we define:

$g \left(X\right) = {A}^{- 1} X A$

then:

$g \left(f \left(X\right)\right) = {A}^{- 1} \left(f \left(X\right)\right) A = {A}^{- 1} A X {A}^{- 1} A = I X I = X$

$f \left(g \left(X\right)\right) = A \left(g \left(X\right)\right) {A}^{- 1} = A {A}^{- 1} X A {A}^{- 1} = I X I = X$

So $g \left(X\right)$ is inverse to $f \left(X\right)$.

The domains of $f \left(X\right)$ and $g \left(X\right)$ are both the whole of ${M}_{2} \left(\mathbb{R}\right)$, so the ranges of both are the whole of ${M}_{2} \left(\mathbb{R}\right)$ and they are bijections.