Proposition 1

— Assume that E is finite-dimensional and H is nondegenerate.

a) There exists an endomorphism u of E characterized by $(x|y)_Q = (u(x)|y)_H$ for x, y in E.

b) $\operatorname{Tr}(Q/H) = \operatorname{Tr}(u)$.

c) $\operatorname{Tr}(Q/H) = \sum_{i=1}^m Q(e_i)$ for every basis $(e_1, \ldots, e_m)$ of E orthonormal for H.

a) follows from the fact that the bilinear form $(x, y) \mapsto (x|y)_H$ is non-degenerate. Every sequence in E orthonormal for H may be completed to a basis of E orthonormal for H. Consequently, $\operatorname{Tr}(Q/H)$ is the supremum of the set of numbers of the form $\sum_{i=1}^m Q(e_i)$ over all bases $(e_1, \ldots, e_m)$ of E orthonormal for H. To prove b) and c), it suffices to show that $\sum_{i=1}^m Q(e_i) = \operatorname{Tr}(u)$ for every basis of this kind. Set $a_{ij} = (u(e_i)|e_j)_H = (e_i|e_j)_Q$ for $1 \leq i, j \leq m$; then $u(e_i) = \sum_{j=1}^m a_{ij} e_j$ for $1 \leq i \leq m$, whence

Q.E.D.