Proposition 2

— Assume that E is finite-dimensional. There exist a basis $(e_1, \ldots, e_n)$ of E and an integer m with $0 \leq m \leq n$ such that

for $t_1, \ldots, t_n$ real. If, moreover, the relation $\operatorname{H}(x) = 0$ implies $Q(x) = 0$ for $x \in E$, then $\operatorname{Tr}(Q/H) = \sum_{i=1}^m Q(e_i)$.

There exists a basis $(e'_1, \ldots, e'_n)$ of E orthogonal for H. One can assume the basis to be indexed in such a way that $\operatorname{H}(e'_i) > 0$ for $1 \leq i \leq m$ and $\operatorname{H}(e'_i) = 0$ for $m < i \leq n$. Then set $e_i = e'_i / \operatorname{H}(e'_i)^{1/2}$ for $1 \leq i \leq m$ and $e_i = e'_i$ for $m < i \leq n$; the relation (1) is satisfied.

Let F be the subspace of E generated by $e'_{m+1}, \ldots, e'_n$; it is the set of $x \in E$ such that $\operatorname{H}(x) = 0$. Denote by $\pi$ the canonical mapping of E onto $E_1 = E/F$. Since Q and H are zero on F, there exist two positive quadratic forms $Q_1$ and $H_1$ on $E_1$ such that $Q = Q_1 \circ \pi$ and $H = H_1 \circ \pi$. Moreover, $(\pi(e_1), \ldots, \pi(e_m))$ is a basis of $E_1$ orthonormal for $H_1$, and therefore $H_1$ is nondegenerate.

By Prop. 1 and Remark 2,

Q.E.D.