Bilinear form/Symmetric/Definiteness/Introduction/Section

Definition

Let ${}V$ be a real vector space, endowed with a symmetric bilinear form ${}\left\langle -,-\right\rangle$. This bilinear form is called

1. positive definite, if ${}\left\langle v,v\right\rangle >0$ holds for all ${}v\in V$, ${}v\neq 0$.
2. negative definite, if ${}\left\langle v,v\right\rangle <0$ holds for all ${}v\in V$, ${}v\neq 0$.
3. positive semidefinite, if ${}\left\langle v,v\right\rangle \geq 0$ holds for all ${}v\in V$.
4. negative semidefinite, if ${}\left\langle v,v\right\rangle \leq 0$ holds for all ${}v\in V$.
5. indefinite, if ${}\left\langle -,-\right\rangle$ is neither positive semidefinite nor negative semidefinite.

Positive definite symmetric bilinear forms are the the real inner products. We have an indefinite if there exist vectors ${}v$ and ${}w$ such that ${}\left\langle v,v\right\rangle >0$ and ${}\left\langle w,w\right\rangle <0$. The zero form is positive semidefinite and negative semidefinite at the same time, but neither positive definite nor negative definite.

It is possible to restrict a bilinear form on ${}V$ to a linear subspace ${}U\subseteq V$, yielding a bilinear form on ${}U$. If the original form is positive definite, then so is the restriction. However, an arbitrary form might become positive definite when restricted to certain linear subspaces, and negative definite when restricted to other linear subspaces. This leads us to the following definition.

Definition

Let ${}V$ be a finite-dimensional real vector space, endowed with a symmetric bilinear form ${}\left\langle -,-\right\rangle$. We say that this bilinear form has the type

$(p,q),$

where

${}p:={\max {\left(\dim _{\mathbb {R} }{\left(U\right)},U\subseteq V,\,\left\langle -,-\right\rangle {|}_{U}{\text{ positive definite}}\right)}}\,,$

and

${}q:={\max {\left(\dim _{\mathbb {R} }{\left(U\right)},U\subseteq V,\,\left\langle -,-\right\rangle {|}_{U}{\text{ negative definite}}\right)}}\,.$

For an inner product on a ${}n$-dimensional real vector space, the type is ${}(n,0)$. Because of exercise, we always have

${}p+q\leq \dim _{K}{\left(V\right)}\,.$

The matrix

${\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}}$

is the Gram matrix of a symmetric bilinear form on ${}\mathbb {R} ^{3}$, say with respect to the standard basis. The restriction of the form to ${}\mathbb {R} e_{1}$ is positive definite, the restriction to ${}\mathbb {R} e_{2}$ is negative definite, the restriction to ${}\mathbb {R} e_{3}$ is the zero form. Therefore, we get immediately ${}p,q\geq 1$. However, it is not immediately clear whether there exist some two-dimensional linear subspace such that the restriction to this space is positive definite. An investigation of "all“ linear subspaces is not really feasible. However, there are several possibilities to determine the type of a symmetric bilinear, without checking all linear subspaces of ${}V$. The following statement is called Sylvester's law of inertia.

James Joseph Sylvester (1814-1897)

Theorem

Let ${}V$ be a finite-dimensional real vector space, endowed with a symmetric bilinear form ${}\left\langle -,-\right\rangle$ of type ${}(p,q)$. Then the Gram matrix of ${}\left\langle -,-\right\rangle$ with respect to any orthogonal basis is a diagonal matrix

with ${}p$ positive and ${}q$ negative entries.

Proof

With respect to an orthogonal basis ${}u_{1},\ldots ,u_{n}$ of ${}V$ (which exists due to fact), the Gram matrix is in diagonal form. Let ${}p'$ be the number of positive diagonal entries, and let ${}q'$ be the number of negative diagonal entries. We may order the basis in such a way that the first ${}p'$ diagonal entries are positive, the following ${}q'$ diagonal entries are negative, and the remaining entries are ${}0$. On the linear subspace ${}U=\langle u_{1},\ldots ,u_{p'}\rangle$ of dimension ${}p'$, the restricted bilinear form is positive definite; therefore, ${}p'\leq p$ holds. Let ${}W=\langle u_{p'+1},\ldots ,u_{n}\rangle$; on this linear subspace, the restricted bilinear form is negative semidefinite. We have ${}V=U\oplus W$, and these spaces are orthogonal to each other.

Assume now that there exists a linear subspace ${}U'$, such that the bilinear form restricted to is positive definite, and such that its dimension ${}p$ is larger than ${}p'$.

The dimension of ${}W$ is ${}n-p'$, therefore, ${}W\cap U'\neq 0$ because of fact. For a vector ${}w\in W\cap U'$, ${}w\neq 0$, we get directly the contradiction ${}\left\langle w,w\right\rangle >0$ and ${}\left\langle w,w\right\rangle \leq 0$.

$\Box$

By scaling the orthogonal vectors, we can even achieve that only the values ${}1,-1,0$ occur in the diagonal. The form given on ${}\mathbb {R} ^{n}$ by the diagonal matrix with ${}p$ times ${}1$, ${}q$ times ${}-1$ and ${}n-p-q$ times ${}0$ shows that every type fulfilling

${}p+q\leq n\,$

can be realized. We talk about the standard form of type ${}(p,q)$ on ${}\mathbb {R} ^{n}$.