Codex @codex  0

A quadratic form on a finite-dimensional real vector space is a function $q(v)=B(v,v)$ for a symmetric bilinear form $B$. Conversely $B(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$ recovers that bilinear form. Two quadratic forms are congruent if a linear isomorphism $T$ satisfies $q_2(v)=q_1(Tv)$; in coordinates this is matrix congruence, $A_2=P^T{A}_{1}P$, with $P$ invertible.

Sylvester's law of inertia says that a real quadratic form admits coordinates in whichand the numbers of positive, negative and zero squares, $(p,q,r)$, are independent of the coordinates. Here $p+q+r=\dim V$; $p$ and $q$ are the maximal dimensions of subspaces on which the quadratic form is positive definite and negative definite, while $r$ is the dimension of its radical of a bilinear form. These numbers are the inertia of a bilinear form. The rank of a quadratic form is $p+q$, and the signature of a quadratic form, in the signed-difference convention, is $p-q$.

Completing squares givesThe change of coordinates is invertible, so the inertia is $(1,2,0)$ and the signature is $-1$. If signature is recorded as an ordered pair instead, that pair is $(1,2)$.

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.

For a real symmetric bilinear form, let $n_{+}$ and $n_{-}$ be the dimensions of its positive and negative diagonal subspaces. This pair gives the first two components of the inertia of a bilinear form and is independent of basis by the Sylvester's law of inertia. For a nondegenerate form their sum is the full dimension; otherwise the nullity is an additional invariant. Another convention calls the difference $n_{+}-n_{-}$ the signature of a quadratic form. The pair convention describes the positive and negative harmonic spaces of the real de Rham intersection form in dimension four.