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 $ab-ba$ 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 $Q$?



Trace of finitely generated projective module
There's a standard way to define the trace (look up "Hattori-Stallings 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 
$$\operatorname{Tr}\_P:\operatorname{End}\_A(P)\to A/[A,A].$$


