Trace of AB = Trace of BA We can define trace if $A =\sum_{i} \langle e_i, Ae_i\rangle$ where $e_i$'s are standard column vectors, and $\langle x, y\rangle =x^t y$ for suitable column vectors $x, y$. With this set up, I want to prove trace of AB and BA are same, so it's enough to prove that $$\sum_{i} \langle e_i, ABe_i\rangle =\sum_{i} \langle e_i, BAe_i\rangle$$ but how to conclude that?

Though this isn't the approach you might want to see as how you stated the question, I thought I would throw this in as well.
$\mathrm{Tr}(AB) = \sum_{i=1}^n (AB)_{ii} = \sum_{j=1}^n\sum_{k=1}^n A_{jk}B_{kj}$
Now you can manipulate that expression to show that it equals $\mathrm{Tr}(BA)$.
