= Solution
The symmetric bilinear form associated with a real <quadratic form> $Q$ is obtained by <polarization identity>:
$$
phi(u,v)=\frac12\bigl(Q(u+v)-Q(u)-Q(v)\bigr).
$$
In coordinates $Q(x)=x^TAx$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^TAv$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\geq0$ for every $v$, and positive definite when the inequality is strict for every $v\ne0$.
The diagonalization theorem for real quadratic forms says that some <basis> puts any symmetric form into
$$
x_1^2+\cdots+x_p^2-y_1^2-\cdots-y_q^2,
$$
with $r$ 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 $P'$ is positive definite for another diagonalization and $\dim P'>p$, then
$$
\dim P'+\dim(N\oplus Z)>(p+q+r),
$$
so the two spaces intersect nontrivially. A <vector> in the intersection would have both positive and nonpositive square, a contradiction. Thus $p'\leq p$; symmetry gives $p=p'$. Applying the same argument to $-\phi$ gives $q=q'$, and then $r=r'$.
For the <nondegenerate bilinear form>[nondegenerate form] on $V$, write its <inertia of a bilinear form>[inertia] as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: <polarization identity>[polarization] gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is <injective function>[injective], since a <vector> with zero positive projection cannot be <isotropic vector>[isotropic] unless it is zero. Hence $k\leq p$; projection to the negative space likewise gives $k\leq q$. Therefore $k\leq\min(p,q)\leq n$.
The <matrix> of $l^2$ is $ll^T$. If $l\ne0$, 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> $tt^T$, where $t_i=l_is_i$, so it has the same rank-one conclusion when $t\ne0$ 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 gives
$$
(f,g)=\sum_{a,b}(l_a^2,s_b^2),
$$
a sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.
Back to article page
