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

We use the notation of the statement. Algebraic manipulations of dot products use The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, and $\lVert v\rVert^{2}=v\cdot v$ is claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. We write $e_{1},\dots,e_{n}$ for the standard basis vectors of $\mathbb{R}^{n}$, so that the $i$th coordinate of $v\in\mathbb{R}^{n}$ equals $v\cdot e_{i}$.

Claim 1. Let $D\in\mathcal{S}(n)$ and let $u,v\in\mathbb{R}^{n}$. By The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product,

$$(u+v)\cdot\bigl(D(u+v)\bigr)=u\cdot(Du)+u\cdot(Dv)+v\cdot(Du)+v\cdot(Dv).$$

Since $D$ is symmetric, $D^{\top}=D$, so claim 5 of Elementary Properties of the Transpose of a Real Matrix gives $v\cdot(Du)=(D^{\top}v)\cdot u=(Dv)\cdot u=u\cdot(Dv)$, the last step by claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite. Hence

$$u\cdot(Dv)=\tfrac{1}{2}\Bigl[(u+v)\cdot\bigl(D(u+v)\bigr)-u\cdot(Du)-v\cdot(Dv)\Bigr]. \tag{P}$$

Now let $B,C\in\mathcal{S}(n)$ have the same quadratic form. Applying (P) to $B$ and to $C$, and noting that the right-hand side of (P) involves only the quadratic form, we get $u\cdot(Bv)=u\cdot(Cv)$ for all $u,v\in\mathbb{R}^{n}$. Take $u=e_{i}$ and $v=e_{j}$. By Matrix-Vector Product the $i$th coordinate of $Be_{j}$ is $\sum_{k=1}^{n}B_{ik}(e_{j})_{k}=B_{ij}$, and that coordinate equals $e_{i}\cdot(Be_{j})$ by Orthonormal Families, Standard Basis Vectors, and Plane Rotations of Euclidean Space; likewise for $C$. Hence $B_{ij}=C_{ij}$ for all $i,j\in\{1,\dots,n\}$, that is $B=C$.

Claim 2. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and let $\delta>0$ be as in the hypothesis for this $\varepsilon$. Subtracting the two estimates and using claim 5 of Properties of the Absolute Value in an Ordered Field, every $h$ with $\lVert h\rVert<\delta$ satisfies

$$\Bigl|(p'-p)\cdot h+\tfrac{1}{2}\,h\cdot(B'h)-\tfrac{1}{2}\,h\cdot(Bh)\Bigr|\le2\varepsilon\lVert h\rVert^{2}. \tag{D}$$

We first show $p=p'$. Suppose $p\neq p'$ and put $u=(p'-p)/\lVert p'-p\rVert$, so $\lVert u\rVert=1$ by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and $(p'-p)\cdot u=\lVert p'-p\rVert$. Let $t\in\mathbb{R}$ with $0<t<\delta$ and put $h=tu$, so $\lVert h\rVert=t<\delta$. By The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and claim 1 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product the expression inside the absolute value in (D) equals $t\,\lVert p'-p\rVert+\tfrac{1}{2}t^{2}\bigl(u\cdot(B'u)-u\cdot(Bu)\bigr)$, so by claim 6 of Properties of the Absolute Value in an Ordered Field and claim 3 of that lemma,

$$t\,\lVert p'-p\rVert\le2\varepsilon t^{2}+\tfrac{1}{2}t^{2}\bigl|u\cdot(B'u)-u\cdot(Bu)\bigr| ,$$

and dividing by the positive number $t$,

$$\lVert p'-p\rVert\le t\Bigl(2\varepsilon+\tfrac{1}{2}\bigl|u\cdot(B'u)-u\cdot(Bu)\bigr|\Bigr).$$

The bracket does not depend on $t$, so by The Archimedean Property of the Real Numbers the right-hand side is smaller than $\lVert p'-p\rVert$ for $t$ small enough, a contradiction. Hence $p=p'$.

With $p=p'$, (D) becomes $\bigl|\tfrac{1}{2}h\cdot(B'h)-\tfrac{1}{2}h\cdot(Bh)\bigr|\le2\varepsilon\lVert h\rVert^{2}$ for $\lVert h\rVert<\delta$. Let $u\in\mathbb{R}^{n}$ with $\lVert u\rVert=1$ and let $t$ with $0<t<\delta$; taking $h=tu$ and using bilinearity as above gives $\tfrac{1}{2}t^{2}\bigl|u\cdot(B'u)-u\cdot(Bu)\bigr|\le2\varepsilon t^{2}$, so

$$\bigl|u\cdot(B'u)-u\cdot(Bu)\bigr|\le4\varepsilon .$$

This holds for every $\varepsilon>0$, so $u\cdot(B'u)=u\cdot(Bu)$ for every unit vector $u$. For general $h\neq0$, writing $h=\lVert h\rVert\,u$ with $u=h/\lVert h\rVert$ of norm $1$ and using bilinearity gives $h\cdot(B'h)=\lVert h\rVert^{2}\,u\cdot(B'u)=\lVert h\rVert^{2}\,u\cdot(Bu)=h\cdot(Bh)$, and for $h=0$ both sides vanish. By claim 1, $B=B'$.