Let $A$ be a finite dimensional algebra over some field $k$. Then any element $a \in A$ defines two $k$-linear maps $a_l, a_r$ on $A$ by left and right multiplication, respectively. So there are two ways of assigning a trace to $a$: $\DeclareMathOperator{\tr}{Tr} \tr_l(a) := \tr(a_l)$, and $\tr_r(a) := \tr(a_r)$. In all examples I am familiar with these two definitions of the trace map coincide (if I am not mistaken, they coincide at least for all separable $k$-algebras). Is this a general fact or are there examples of finite dimensional $k$-algebras, where we get two distinct trace maps in this way?
Of course one could ask the same question for the determinant or the characteristic polynomial of $a$ instead of the trace map, and I am equally interested in an answer to this question.
Thank you in advance!
Consider $A := \begin{pmatrix} \ast & \ast \\ 0 & \ast\end{pmatrix}\subset\text{Mat}_{2\times 2}({\mathbb k})$ and put $e_1 := \tiny\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $e_2 := \tiny\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$ and $\alpha := \tiny\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$. Then, with respect to the ordered basis $\{e_1,e_2,\alpha\}$ the right-multiplication by $e_1$ is given by $\ \scriptsize\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}^t$, while the left-multiplication by $e_1$ is given by $\scriptsize\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}^t$, so $\text{tr}_r(e_1) = 1\neq 2=\text{tr}_l(e_1)$.
More generally, if you take $A = {\mathbb k}Q$ the path algebra of a finite quiver without cycles and relations, and $e_v\in A$ is the idempotent associated to a vertex $v\in Q$, then the trace of $e_v\cdot -$ is the number of paths in $Q$ ending in $v$, while the trace of $-\cdot e_v$ is the number of paths starting in $v$. The above algebra $A$ arises in case $Q = \bullet\to\bullet$, the $A_2$-quiver. Such path algebras are never separable unless the quiver has no edges at all, so it does not contradict your previous observations.