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For unit $u$, put $A=I-2u{u}^T$. Then $A^T=A$ and $A^2=I$, so $A$ is an orthogonal matrix. The affine reflection in a hyperplane is $R(x)=Ax+2cu$. Hence $R(x)-R(y)=A(x-y)$ preserves the Euclidean norm and distance, and $R(x)=x$ whenever $u\cdot x=c$. Thus it is an Euclidean isometry fixing the hyperplane pointwise.

For $p\ne q$, any reflection exchanging them must have normal parallel to $p-q$, and its fixed hyperplane must contain their midpoint. These requirements determine the perpendicular bisector uniquely:Changing both signs of $u,c$ leaves the same reflection. It belongs to the orthogonal group $O(n)$ exactly when its translation term vanishes, that is, $c=0$. Since $c=(\Vert p{\Vert }^2-\Vert q{\Vert }^2)/(2\Vert p-q\Vert )$,To prove the finite reflection decomposition of a Euclidean isometry, first note that a distance-preserving map $g$ fixing zero preserves inner products by the polarization identity. Its values on an orthonormal basis $e_i$ form an orthonormal basis, and $g(x)\cdot g(e_i)=x_i$ shows $g(x)={\sum }_i{x}_{i}g(e_i)$. Thus $g$ is orthogonal.

Every orthogonal transformation is a product of at most $n$ linear reflections: if $g(e_1)\ne e_1$, reflect $g(e_1)$ to $e_1$ using the hyperplane normal $g(e_1)-e_1$, which passes through zero. The resulting orthogonal transformation fixes $e_1$, and its restriction to $e_1^{\perp }$ is handled inductively in dimension $n-1$. If $g(e_1)=e_1$, no initial reflection is necessary. Reflections on the complement extend by fixing $e_1$; the induction starts in dimension zero.

For an arbitrary isometry $f$, if $f(0)\ne 0$, first reflect $f(0)$ to zero across its perpendicular bisector with zero. Composing this one affine reflection with $f$ gives an orthogonal transformation. If $f(0)=0$, skip the initial reflection. ThereforeThe glide reflection $(x,y)\mapsto (x+1,-y)$ cannot use fewer than three: its determinant sign is negative, excluding zero or two reflections, while absence of fixed points excludes a single reflection. Three suffice by reflecting successively in $x=0$, $x=1/2$, and $y=0$.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

Use polarization identity of the quadratic variations:This is continuous and adapted, starts at zero, and has finite variation on every compact interval because it is a difference of two nondecreasing processes. The defining quadratic variation property gives that $(M+N{)}^2-[M+N]$ and $(M-N{)}^2-[M-N]$ are continuous local martingales; their difference divided by four is $MN-[M,N]$. This proves existence without assuming a product formula involving an as yet undefined quadratic covariation.

If $C$ and $D$ both satisfy the requirements, $C-D$ is a continuous finite-variation process and a local martingale. By continuous finite-variation local martingale is constant, it is constant; its initial value is zero, so $C=D$ up to indistinguishability of stochastic processes.

There is a minor initial-value convention in this characterization. The displayed uncentred local martingale identity is directly valid when $M_0=N_0=0$, or when the initial terms have the requisite integrability. With arbitrary finite initial values, centre the quadratic variation construction and use $MN-M_0{N}_0-[M,N]$, which starts at zero. Adding back $M_0{N}_0$ requires it to be integrable under the usual definition of a local martingale; the quadratic covariation itself is unaffected by centring.

By the Itô formula,To compute the Stratonovich integral correction, apply the Itô formula also to $f^{\prime }$, which is twice continuously differentiable because $f$ is $C^3$. The continuous local martingale part of $f^{\prime }(X)$ is $\int f^{\prime \prime }(X_s)d{L}_s$, where $L$ is the continuous local martingale part of $X$. A finite-variation process has zero quadratic covariation with $X$, and the quadratic variation of a stochastic integral together with polarization identity givesAll these statements can be localized to compact ranges of $X$, so unbounded derivatives create no global integrability requirement. Substitute this identity into the Stratonovich integral to obtain the Stratonovich chain rule:This is the integrated meaning of the requested differential identity; there is no extra second-order term after the Stratonovich integral correction has been included.

The symmetric bilinear form associated with a quadratic form is obtained by polarization:To diagonalize it, argue by induction on $\dim V$. If $\varphi (v,v)=0$ for every $v$, polarization gives $\varphi =0$, so every basis works. Otherwise choose $e$ with $\varphi (e,e)\ne 0$. Thenand diagonalize the restriction to $e^{\perp }$ inductively. Thus some basis gives a diagonal matrix with positive, negative, and zero entries. By Sylvester's law of inertia, their counts $n_{+},n_{-},n_0$ are basis-independent. In the convention relevant here, the signature of a quadratic form is

Suppose $R$ lies in the radical of a bilinear form, so $\varphi (r,v)=0$ for all $r\in R,v\in V$. Then $q(r)=0$ andHence $q^{\prime }(v+R)=q(v)$ is a well-defined quadratic form on the quotient vector space $V/R$. In an adapted diagonal basis, quotienting by $R$ merely removes zero diagonal directions, so $n_{+}$ and $n_{-}$, and therefore the signature, are unchanged.

Now let $W=\mathrm{span}(e,f)$. Its Gram matrix iswhich has determinant $-1$. Thus $W$ is nondegenerate and has one positive and one negative square: it is a hyperbolic plane (quadratic form) and has signature zero. Nondegeneracy gives the orthogonal direct sumSignature is additive under orthogonal direct sums, so

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

A Hermitian form is conjugate-linear in its first argument, linear in its second, and satisfies $\varphi (w,v)=\overline{\varphi (v,w)}$. For $H=\varphi -\psi$, the hypothesis says $H(v,v)=0$. The complex polarization identitytherefore gives $H(v,w)=0$ for all $v,w$.

The displayed matrix is Hermitian. By Sylvester's criterion, its leading principal minors areHence the form is

Choose any tangent vector $V^a\in T_p\Sigma$ and let the affinely parametrized geodesic with initial tangent $V^a$ start at $p$. If $\Sigma$ is totally geodesic, then $U^a{n}_a$ remains zero. Its initial derivative is therefore zero. Part iii, with $U^a{n}_a=0$ at $p$, givesThis holds for every tangent $V$. Since $K$ is symmetric, the polarization identity implies

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

The quadratic covariation of two continuous semimartingales is the limit in probabilityThe polarization identity gives $[M,N]=\frac{1}{4}([M+N]-[M-N])$.