A Symmetric Matrix is Determined by its Quadratic Form, and a Second-Order Expansion by its Coefficients

We work in the setting of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background, whose notation is fixed for every dimension and is used here with a natural number $n$ satisfying $1\le n$: the real numbers, and the Euclidean norm $\lVert\,\cdot\,\rVert$, dot product and notion of openness, are as fixed there. Write $\mathcal{S}(n)$ for the set of symmetric real $n\times n$ matrices and $Mh$ for the matrix-vector product. Then the following hold.

1. (A symmetric matrix is determined by its quadratic form) (i) Let $B,C\in\mathcal{S}(n)$ satisfy

$$h\cdot(Bh)=h\cdot(Ch)\qquad\text{for every }h\in\mathbb{R}^{n}.$$

Then $B=C$.

2. (Uniqueness of a second-order expansion) (ii) Let $U\subseteq\mathbb{R}^{n}$ be open, let $y\in U$, let $f:U\to\mathbb{R}$, let $p,p'\in\mathbb{R}^{n}$ and let $B,B'\in\mathcal{S}(n)$. Suppose that for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\delta\in\mathbb{R}$ with $0<\delta$ such that every $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\delta$ satisfies $y+h\in U$ and both

$$\Bigl|f(y+h)-f(y)-p\cdot h-\tfrac{1}{2}\,h\cdot(Bh)\Bigr|\le\varepsilon\lVert h\rVert^{2}$$

and the same inequality with $p'$ and $B'$ in place of $p$ and $B$. Then $p=p'$ and $B=B'$.