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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 ring homomorphism respecting this trace. Then $f(1)$ must be a rank $1$ idempotent. 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.