Let $Q_1,Q_2$ be a quadratic forms on an Euclidean space $X$ of finite dimension such that $Q_1|_V=Q_2|_V$ on a nonempty open subset $V\subset X$.

How to prove not using the differential calculus that $Q_1=Q_2$?



1 Answer 1


You want to use the fact that a quadratic form is fully determined by its values on the elements of a basis, that an open set contains an open ball, and that an open ball contains a basis.

In greater detail, consider the case that $V$ is an open ball, let $e_1, \dotsc, e_n \in V$ be a basis for $X$, and suppose $Q$ is a quadtratic form which is zero on every point in $V$. For any point $x \in X$, we can thus uniquely write $x$ as $x_1e_1 + \dotsi + x_ne_n$. We can then write $Q(x) = \sum_{i \leq j} a_{ij}x_ix_j$ for any $x \in X$. Therefore, it suffices to show that if $Q$ is zero on $V$, then each coefficient $a_{ij}$ is zero. This leaves two cases:

  • When $i = j$, we have $Q(e_i) = a_{ij} = 0$
  • When $i < j$, we have $Q(\frac{e_i + e_j}2) = \frac{a_{ii} + a_{ij} + a_{jj}}4$. Now, we already know that $a_{ii} = a_{jj} = 0$, and since $V$ is an open ball and thus is convex, $\frac{e_i + e_j}2$ is in $V$, and therefore $Q(\frac{e_i + e_j}2) = 0 = \frac{a_{ij}}4$. In particular, $a_{ij} = 0$.

Thus each coefficient is zero, so $Q$ must be zero on all of $X$.

Taking $Q = Q_1 - Q_2$ gives your desired result.


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