How to proof equivalence relation?

Question

I need help with this problem:

Let $S=\left\{\left[\begin{matrix} a & b \\ c & d \end{matrix}\right] : a,b,c,d \in \mathbb{C}\right \}$ and $M=\left\{\left[\begin{matrix} a & b \\ -\overline{b} & \overline{a} \end{matrix}\right] : a,b \in \mathbb{C}, |a|+|b|\neq 0 \right\}$.

For every $X,Y \in S$ is $$X\rho Y \Leftrightarrow (\exists A \in M)AXA^{-1}=Y$$ Prove that $\rho$ is equivalence relation in $S$.

I know that I need to check if it is reflexive, symmetric and transitive, but I don't know how.

Thanks for replies.

Answer 1

Since the $2 \times 2$ identity matrix $I \in M$ is so that $IXI^{-1}=X$, we have that $X \rho X$. If $A,X$, and $Y$ are so $X \rho Y$ (that is, $AXA^{-1}=Y$), then $A^{-1} \in M$ and $A^{-1}YA=X \implies Y \rho X$. Finally, if $X\rho Y$ and $Y\rho Z$ then there are $A,B \in M$ so that $AXA^{-1}=Y$ and $BYB^{-1}=Z$. Hence $BAXA^{-1}B^{-1}=(BA)X(BA)^{-1}=Z$. Provided you can show $BA \in M$, you are done.

Answer 2

Hints

$\rho$ is reflexive if $\rho(x,x)$ holds for all $x$, and symmetric if $\rho(x,y) = \rho(y,x)$ for all $x,y$. Do you know what transitive means?

Can you prove that each of these 3 actually hold?