Let $R = M_2(\mathbb{R})$ the set of all square $2$x$2$ matrices with real coefficients.
$A,B \in R$, define $A$ ~ $B$, if and only if, exist $P \in R$ inversible such that $A=P^{-1}BP$.
Prove that ~ is a equivalence relation.
What I'm trying:
Proving the equivalence relation:
Reflexive: $A$~$A$, so we can write $P=I$, and get $A=A$
Symmetric: $A$~$B$, so $A=P^{-1}BP$, that implies $B=PAP^{-1}$, and here, we can just switch the notation and write $B=P^{-1}AP$. (I'm not sure about that at all)
Transitive: $A$~$B$ and $B$~$C$, so there's $P$, and $T$, such that, $A=P^{-1}BP$ and $B=T^{-1}CT$
Substituing $B$ into the first equation: $A=P^{-1}T^{-1}CTP$. Which helps nothing, because $P^{-1}T^{-1}$ isn't the inverse of $TP$. (This one I actually have no ideia how to show that.)
If there's something wrong in the first two please let me know.
Thanks in advance.