Let $A$ be a tracial von Neumann algebra with trace $\tau$. If I'm not mistaken, this trace is unique. For $X=[x_{ij}]\in M_n(A)$, define
$$\tau_n(X):=\frac{1}{n}\sum_{i=1}^n\tau(x_{ii}).$$
I find this a very natural construction.
My questions are then:
Does this make $M_n(A)$ a tracial von Neumann algebra? If so, is it the unique one?
Best Answer
You don't say what "trace" is, so it is a bit hard to discuss uniqueness.
With the mostly usualy meaning "state with the tracial property" (and this leaves out whether you want your trace to be normal), traces are not unique. For starters, in any commutative von Neumann algebra, any state is a trace. Even in non-commutative von Neumann algebras it is easy to find multiple traces. In fact, every time that the centre is not trivial, if there is one trace there are infinitely many others. For instance in $M_2(\mathbb C)\oplus \mathbb C$ given any $t\in[0,1]$ you have that $$ \tau_t(A\oplus\lambda)=\tfrac12\Big[t\,\operatorname{Tr}(A)+(1-t)\lambda\Big] $$ is a tracial state for any $t$.
Tracial states are unique on factors, because in such case any two projections are comparable and the factors is the norm-closure of the linear span of the projections.
Your $\tau_n$ is a tracial state if $\tau$ is. It's uniqueness depends on the uniqueness of $\tau$.
As a final comment, your construction is as natural as it can get, because what you are doing is define $\tau_n=\operatorname{tr}\otimes \tau$ on the algebra $M_n(\mathbb C)\otimes A$.