Suppose I have a finite (non-)commutative ring $R/k$ (over a field $k$ of char $0$) with a linear "trace" function $t: R \to k$. Can I always find an embedding $f: R \to M_r(k)$ compatible with the trace functions on both sides?
One restriction I can see for the trace function on $R$ is that it should be invariant under cyclic permutations : $t(a_1a_2\dots a_n) = t(a_2\dots a_na_1)$. Is this the only restriction?
You cannot always find such an embedding. Consider the ring $R=\mathbb{Q}\langle x,y\rangle$ subject only to the condition that any monomial in the letters $x$ and $y$ of degree $3$ is zero. This is a noncommutative ring, finite dimensional over $\mathbb{Q}$, and the natural factor map $t\colon R\to R/(x,y)\cong \mathbb{Q}$ is a ring homomorphism, and in particular it is linear. Let $f\colon R\to \mathbb{M}_r(k)$ be any (possibly non-unital) ring homomorphism respecting this trace. Then $f(1)$ must be a rank $1$ idempotent (since its trace is $1$). Thus, since $f(1)$ acts as the identity on $f(R)$, we see that $f(R)\subseteq f(1)\mathbb{M}_r(k)f(1) \cong k$. Hence $f$ is not an embedding, since $k$ has no zero-divisors.
