Let $F$ be a field. Consider the a mapping on the algebra of all linear transformations on $F^n$ that are triangularizable as follows: $$(T,A)=trace(TA).$$ Then this is an inner product and non-singular.
Why?
( In fact, Trace function is $\sum a_{i}\lambda_{i}$, where $\lambda_i$'s are eigenvalues of $TA$ and $a_i$'s are algebriac multiplicities of $\lambda_i$'s.) Also, the eigenvalues are in $F$.
Note that trace is bi-linear, so it certainly gives a multilinear function $F^n \times F^n \to F$.
Perhaps the "best" thing about the trace function is that $tr(AB)=tr(BA)$ which can be proven from the explicit construction of trace, since the diagonals of such matrices do not change. This gives the second requirement for an inner product (symmetry.)
What is left to show: if $tr(A^2)=0$, then $A=0$, but only matrices of the form $XY-YX$ have trace zero.
Edit: using your definition of trace and the additional assumption, what are the eigenvalues for $A^2$? can their sum ever be zero?