Let $k$ be a field and $A$ a $k$algebra with unit. The trace module is $$ T(A)=A/[A,A], $$ where $[A,A]$ is the left $A$module generated by all elements of the form $abba$ for $a,b\in A$. The natural trace map is the projection $T:A\to T(A)$. For an $A$module $P$ one wants to construct a trace map $$ Tr_P: End_A(P) \to T(A), $$ which is $k$linear and satisfies $Tr_P(uv)=Tr_P(vu)$. If $P=A^n$ is finite free, one has $End_A(A^n)\cong M_n(A^{op})$ and a natural trace map is $$ Tr_{A^n}((a_{i,j}))=\sum_jT(a_{j,j}). $$ Now if $P$ if finite projective, there exists $Q$ such that $P\oplus Q\cong A^n$ and so one can define a trace for $P$ by $$ Tr_P(u)=Tr_{A^n}(u\oplus 0). $$ My question is, whether this trace map does depend on the choice of $Q$?

There's a standard way to define the trace (look up "HattoriStallings trace") that agrees with yours, but is clearly independent of choices. For any (left) $A$module $P$, there's a natural map $$\operatorname{Hom}_A(P,A)\otimes_AP\to\operatorname{End}_A(P),$$ sending $\varphi\otimes y$ to the endomorphism $x\mapsto\varphi(x)y$, which is an isomorphism when $P$ is finitely generated projective. Composing its inverse with the map $$\operatorname{Hom}_A(P,A)\otimes_AP\to A/[A,A]$$ induced by the evaluation map $$\operatorname{Hom}_A(P,A)\otimes_kP\to A,$$ gives the trace map $$\operatorname{Tr}_P:\operatorname{End}_A(P)\to A/[A,A].$$ 

