Remark 4

Suppose $E$ is finite dimensional. When the quadratic form $H$ is invertible, prop. 15 applies. Let $(e_1, \ldots, e_n)$ be a basis of $E$. Put $q_{ij} = \langle e_i|e_j\rangle_Q$ and $h_{ij} = \langle e_i|e_j\rangle_H$ and introduce the matrices $q = (q_{ij})$ and $h = (h_{ij})$. Let $u$ be an endomorphism of $E$ such that $Q(x) = \langle x|u(x)\rangle_H$ for all $x \in E$. We have

and hence the matrix of $u$ with respect to the basis $(e_1, ..., e_n)$ of E is equal to $h^{-1} q$. By prop. 15, we have

If the basis $(e_1, ..., e_n)$ is orthonormal for H, then $h$ is the unit matrix of order $n$, and we get

so that we get formula (44) in this case.

Now suppose that the quadratic form H is not invertible. Let N be the kernel of H, and let $\pi$ be the canonical mapping from E onto E/N. There exists an invertible quadratic form $H_1$ on E/N such that $H = H_1 \circ \pi$. Let $(e_1, ..., e_n)$ be a sequence of elements of E such that the sequence $(\pi(e_1), ..., \pi(e_m))$ is a basis of E/N, which is orthonormal for $H_1$. Let $(e_1, ..., e_n)$ be a basis of N. Then $(e_1, ..., e_n)$ is a basis of E and we have

for all real numbers $\xi_1, ..., \xi_n$.

Suppose that for all $x \in E$, the relation $H(x) = 0$ implies $Q(x) = 0$; in other words, suppose that there exists a quadratic form $Q_1$ on E/N such that $Q = Q_1 \circ \pi$. By remark 2 and prop. 15, we have,

Exercises