Let $k$ be a field and $A$ a finite dimensional $k$-algebra. Let $J$ be the Jacobson ideal. Let $t: A \to k$ be a morphism of $k$-vector spaces such that $t(ab) = t(ba), \forall a, b \in A$. Assume $ker(t)$ does not contain a nonzero left ideal. Let $M = \{a \in A \text{ such that } \forall x \in J, t(xa) = 0\}$. Now we need to show $M$ is the largest semisimple left $A$-submodule of $A$.
So to show $M$ is a left ideal, for $a \in M, b \in A, x \in J$ then $t(bax) = t(axb)$. Since $J$ is a two sided ideal, $xb \in J \Rightarrow t(axb) = 0$. $M$ being additively closed is also easy. But I don't know how to show $M$ is the largest semisimplicity. For semisimpleness I think it's equivalent to $M \cap J = 0$. I was thinking about some dimension argument involving $ker(t)$ and $M$. In case $t =0, J = 0$. Thus $M$ is automatically closed. But I don't know how to do the other case
Semisimplicity of $M$ is not equivalent to $M\cap J=0$, but to $JM=0$ (since for any left module $N$, $N/JN$ is the largest semisimple quotient of $N$).
By the definition of $M$, we have $JM\leq\ker(t)$, and since $JM$ is a left ideal and $\ker(t)$ contains no nonzero left ideals, we must have $JM=0$. So $M$ is semisimple.
Conversely, if $N$ is any semisimple left ideal of $A$, then $JN=0$, and so $N\leq M$. So $M$ is the largest semisimple left ideal.
Actually, the fact that $t(ab)=t(ba)$ is not needed. To prove that $M$ is a left ideal, let $m\in M$ and $r\in A$; for any $x\in J$, $t(xrm)=0$ since $xr\in J$ (as $J$ is a right ideal), and so $rm\in M$.