The trace pairing on a finite endomorphism space is nondegenerate

If $\dim V=0$, [F1] gives $V=\{0\}$, so its only endomorphism is $D=0$. Assume henceforth $\dim V=m>0$ and choose an ordered basis.

If every entry of the matrix of $D$ were zero, [F2] would make $D$ vanish on every basis vector and hence on every vector. Thus if $D\mathrm{\ne }0$ there are indices $i,j<m$ with matrix entry $D_{ij}\mathrm{\ne }0$. Define $A({\sum }_{k<m}{a}_k{b}_k):=a_i{b}_j$ using the ordered basis chosen in step 1.1. Unique coordinates [F7] make this well-defined; for scalars $\alpha ,\beta$ and vectors with coordinate lists $(a_k)$ and $(c_k)$, the coordinate list of their linear combination is $(\alpha a_k+\beta c_k)$, so the formula gives $A(\alpha x+\beta y)=\alpha A(x)+\beta A(y)$. Thus $A$ is an endomorphism. Its matrix sends $b_i$ to $b_j$ and every other basis vector to $0$, so that matrix is $E_{ji}$.

By [F4], the matrix of $D\circ A$ is $[D][A]=[D]E_{ji}$. Row-by-column multiplication [F5] gives $([D]E_{ji}{)}_{kk}=D_{kj}{\delta }_{ik}$, so every diagonal entry is $0$ except the $(i,i)$ entry, which is $D_{ij}$. By [F6], $\mathrm{tr}(D\circ A)=D_{ij}\mathrm{\ne }0$. This contradicts the hypothesis for every $A$. Therefore $D=0$, including the zero-dimensional case from step 1.1.