First let us recall some terminologies. Let $A$ be a finite dimensional algebra over a field $k$. A linear map $t: A\to k$ is called symmetric (resp. non-degenerate) if the bilinear map $Q_t: A\times A\to k$ given by $(x,y)\mapsto t(xy)$ is symmetric (resp. non-degenerate).
The trace form of $A$ is the linear map $T_A: A\to k$ given by $a\mapsto tr(\lambda_a)$, where $\lambda_a: A\to A$ is the left multiplication of $a$ on $A$ and $tr$ is the trace of linear maps. Note that $T_A$ is always symmetric. However, it is not necessary non-degenerate even in the case that $A$ is separable.
Now suppose $A$ is separable. One may choose an element $\sum u_i\otimes v_i \in A\otimes A$ such that $\sum au_i\otimes v_i = u_i\otimes v_ia$ for all $a\in A$ and $\sum u_iv_i=1$. It follows that $\sum u_ixv_i$ lies in the center $C$ of $A$ for all $x\in A$. Note that $T_C:C\to k$ is symmetric and non-degenerate. My question is that is the map $t: A\to k$ given by $x\mapsto T_C(\sum u_ixv_i)$ symmetric and non-degenerate?
Any comments are welcome. Thanks a lot.
