Lemma 1

Let V be a vector space of finite dimension over K and B a bilinear form on V x V. Let $(v_1, \ldots, v_n)$ be a basis of V over K and $\Delta = \det B(v_i, v_j)$. Then $A \neq 0$ if and only if, for each $x \neq 0$ in V there exists y in V such that $B(x, y) \neq 0$.

We have $A \neq 0$ if and only if the system of linear equations

has only the solution $A, = \cdots = A, = 0$ in K. If we put $x = \sum_{i=1}^n \lambda_i v_i$, the above system is equivalent to $B(x, v_j) = 0$ for $1 \leq j \leq n$, or also, since $(v_1, \ldots, v_n)$ is a basis of V over K, to $B(x, y) = 0$ for all $y \in V$, whence the lemma.

Let us show that condition c) implies that A is reduced. Let x be a nilpotent element of A : for any $y \in A$ the element xy is nilpotent, and so the endomorphism $L_{xy}$ of the vector space A is nilpotent. Now the following lemma implies that $\mathrm{Tr}(xy) = 0$ for all $y \in A$, whence $x = 0$ under hypothesis c).