Find a subgroup of $M_2(R)$, the group of $2 \times 2$ matrices with real entries under addition, called $H$, if for all $A,B$ in $M_2(R)$, $A-B$ is an element of subgroup $H$ implies that $\operatorname{trace}(A)=\operatorname{trace}(B)$, and vice versa.
I know this means $A$ and $B$ must be similar and I'm sure the similar matrix has something to do with $A-B$ and this is how I'd find the subgroup and prove that it's a subgroup with the given bijection, but I'm not sure how to proceed.
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$\begingroup$The answer follows from the fact that the trace splits over matrix addition, so $\mathrm{trace}(A) = \mathrm{trace}(B)$ if and only if $\mathrm{trace}(A - B) = \mathrm{trace}(A) - \mathrm{trace}(B) = 0$. So you want $H$ to be the subgroup of trace zero matrices.
Using the fact that the trace splits over addition it should be very easy for you to prove that the set of trace zero matrices is indeed a subgroup and then what I've written above shows that this subgroup has the property you desire.
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