# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 39/refcontrol

    Definiteness of bilinear forms
We want to classify symmetric bilinear forms over the real numbers. For this, inner products play e key role, as they represent extreme cases.

## Definition

Let $V$ be a real vector space,MDLD/real vector space endowed with a symmetricMDLD/symmetric (bilinear) bilinear formMDLD/bilinear form $\left\langle -,-\right\rangle$. This bilinear form is called
* positive definite, if $\left\langle v,v\right\rangle >0$ holds for all $v\in V$, $v\neq 0$.
* negative definite, if $\left\langle v,v\right\rangle <0$ holds for all $v\in V$, $v\neq 0$.
* positive semidefinite, if $\left\langle v,v\right\rangle \geq 0$ holds for all $v\in V$.
* negative semidefinite, if $\left\langle v,v\right\rangle \leq 0$ holds for all $v\in V$.
* 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-dimensionalMDLD/finite-dimensional real vector space,MDLD/real vector space endowed with a symmetricMDLD/symmetric (bilinear) bilinear formMDLD/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.

## TheoremCreate referencenumber

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

### Proof

With respect to an orthogonal basisMDLD/orthogonal basis (bilinear) $u_{1},\ldots ,u_{n}$ of $V$ (which exists due to Lemma 38. ), 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;MDLD/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.MDLD/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 Corollary 9.8 . 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 $\mathbb {R} ^{n}$.
    Type criteria for symmetric bilinear forms
There are several methods to determine the type of a symmetric bilinear form. The first possibility is given by Sylvester's law of inertia . But this has the disadvantage that one has to construct an orthogonal basis. We discuss the minor criterion and the eigenvalue criterion. A minor is the determinantMDLD/determinant of a square submatrix of a matrix. We could call the following criterion also determinant criterion.

## TheoremTheorem 39. change

Let $\left\langle -,-\right\rangle$ be a symmetric bilinear formMDLD/symmetric bilinear form on a finite-dimensionalMDLD/finite-dimensional real vector spaceMDLD/real vector space $V$. Let $v_{1},\ldots ,v_{n}$ denote a basisMDLD/basis of $V$. Let $G$ be the Gram matrixMDLD/Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis. Suppose that the determinantMDLD/determinant $D_{k}$ of the squareMDLD/square (matrix) submatricesMDLD/submatrices
    $M_{k}={\left(\left\langle v_{i},v_{j}\right\rangle \right)}_{1\leq i,j\leq k}\,$
are not $0$ for $k=1,\ldots ,n$. Let $a$ be the number of changes of sign in the sequence
    $D_{0}=1,\,D_{1}=\det M_{1},\,D_{2}=\det M_{2},\ldots ,D_{n}=\det M_{n}=\det G.$
typeMDLD/type (bilinear)
$\left\langle -,-\right\rangle$
$(n-a,a)$

### Proof

Due to the condition, the determinant of the Gram matrix is not $0$. Therefore, by exercise *****, the bilinearform is nodegenerat.MDLD/nodegenerat Hence, its type has the form $(n-q,q)$. We have to show that $q=a$. We prove this statement by induction over the dimension of $V$, the base case is trivial. So suppose that the statement is proven up to dimension $n-1$, and that we have an $n$-dimensional vector space, together with a basis $v_{1},\ldots ,v_{n}$ and fulfilling the condition. The linear subspaceMDLD/linear subspace
    $U=\langle v_{1},\ldots ,v_{n-1}\rangle \,$
has dimension $n-1$, and the sequence of the determinants of the submatrices of the Gram matrixMDLD/Gram matrix of the restricted form $\left\langle -,-\right\rangle {|}_{U}$ coincides with the given sequence, only the last member
    $D_{n}=\det M_{n}=\det G\,$
is missing. By the induction hypothesis, the form $\left\langle -,-\right\rangle {|}_{U}$ has type $(n-1-b,b)$, where $b$ is the number of sign changes in the sequence
    $D_{0}=1,\,D_{1},\ldots ,D_{n-1}.$
Due to the definition of the type,MDLD/type (bilinear) we have
    $b\leq q\leq b+1\,,$
since a $q$-dimensional linear subspace $W\subseteq V$, where the bilinear form id negative definite,MDLD/negative definite yields a linear subspace
    $W'=U\cap W\subseteq U\,$
of dimension $q$ oder $q-1$, and where again the restricted form is negative definite. According to exercise *****, the sign of $D_{n-1}$ equals $(-1)^{b}$, and the sign of $D_{n}$ equals $(-1)^{q}$. This means that we have from $D_{n-1}$ to $D_{n}$ another sign change (and therefore $a=b+1$) if and only if
    $q=b+1\,.$
$\Box$

## CorollaryCreate referencenumber

Let $\left\langle -,-\right\rangle$ be a symmetric bilinear formMDLD/symmetric bilinear form on a finite-dimensionalMDLD/finite-dimensional real vector space,MDLD/real vector space and let $v_{1},\ldots ,v_{n}$ be a basisMDLD/basis (vs) of $V$. Let $G$ be the Gram matrixMDLD/Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis, and let $D_{k}$ denote the determinantsMDLD/determinants of the squareMDLD/square (matrix) submatricesMDLD/submatrices
    $M_{k}={\left(\left\langle v_{i},v_{j}\right\rangle \right)}_{1\leq i,j\leq k},\,k=1,\ldots ,n.$
* The form $\left\langle -,-\right\rangle$ is positive definiteMDLD/positive definite if and only if all $D_{k}$ are positive.
* The form $\left\langle -,-\right\rangle$ is negative definiteMDLD/negative definite if and only if the sign in the sequence $D_{0}=1,\,D_{1},\,D_{2},\ldots ,D_{n}$ changes at every step.

### Proof

(1). If the bilinear form is positive definite,MDLD/positive definite then, because of exercise *****, the sign of the determinantMDLD/determinant of the Gram matrixMDLD/Gram matrix equals $(-1)^{0}=1$, and is positive. Since the restriction of the form to the linear subspaces $U_{i}=\langle v_{1},\ldots ,v_{i}\rangle$ is also positive definite, the determinants of all relevant submatrices are positive. If, the other way round, all determinants are positive, then Theorem 39. , implies that the bilinear form is positive definite.
(2) follows from (1) by considering the negative bilinear form, that is, $-\left\langle -,-\right\rangle$.
$\Box$
We will proof the following criterion in lecture 42.

## TheoremCreate referencenumber

Let $\left\langle -,-\right\rangle$ be a symmetric bilinear formMDLD/symmetric bilinear form on a finite-dimensionalMDLD/finite-dimensional real vector space,MDLD/real vector space and let $v_{1},\ldots ,v_{n}$ eine BasisMDLD/Basis (vs) of $V$. Let $G$ denote the Gram matrixMDLD/Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis. Then the typeMDLD/type (bilinear) $(p,q)$ of the form has the following interpretation: $p$ is the sum of the dimensions of the eigenspacesMDLD/eigenspaces of $G$ for positive eigenvalues,MDLD/eigenvalues
$q$
$G$

### Proof

Theorem 42.11 .
$\Box$

## Remark

For a function
    $f\colon \mathbb {R} ^{n}\longrightarrow \mathbb {R} ,$
one would like to know, like in the case $n=1$, where the function has local extrema, for example maxima, that is, points $P\in \mathbb {R} ^{n}$ with the property that in a small nighborhood of $P$, the function is bounded from above by $f(P)$. In case $n=2$, $f$ describes a mountain range over a plane and we are looking for the summits. If the function is twice continuously differentiable, then there exist, like in the one-dimensional situation, necessary and sufficient differential criteris for the existence of local maxima and minima. The necessary criterion is that $P$ is a critical point, that is, the partial derivativesMDLD/partial derivatives ${\frac {\partial f}{\partial x_{i}}}(P)$ vanish for $i=1,\ldots ,n$. In this case, we consider the second partial derivatives, and form the so-called Hesse-matrix
    ${\left({\frac {\partial }{\partial x_{i}}}{\frac {\partial f}{\partial x_{j}}}(P)\right)}_{ij}.$
The corresponding symmetric bilinear form determines often whether we have a local maximum or a local minimum. If this form is positive definite,MDLD/positive definite then we have an isolated local minimum; if it is negative definite,MDLD/negative definite then we have an isolated local maximum; if it is indefinite,MDLD/indefinite then we do not have a local extremum (see Fact *****). In the remaining cases, e.g., in the case of the zero matrix, we need further considerations.
    Perfect pairings
Bilinear forms can also be defined for two different vector spaces. The following property is a variant of the property of not being degenerate.

## Definition

Let $K$ be a field,MDLD/field let $V$ and $W$ denote vector spacesMDLD/vector spaces over $K$, and let
    $\Psi \colon V\times W\longrightarrow K$
denote a bilinear form.MDLD/bilinear form (multi) We say that $\Psi$ defines a perfect pairing, if the mapping
    $V\longrightarrow {W}^{*},v\longmapsto {\left(w\mapsto \Psi (v,w)\right)},$
is
If the spaces are finite-dimensional, then a perfect pairing can only occur for vector spaces of the same dimension. For a finite-dimensional vector space and a nondegenerate bilinear form, we get a perfect pairing between the space and its dual space, which is given by the evaluation; see exercise *****.
    Footnotes
* ↑ In mathematics, the term classification means to describe a set of mathematical objects clearly and completely, to give criteria in order to determine when two objects are essentially equal (or equivalent), to distinguish different objects by numerical invariants, and to represent each object by a simple representative. For example, finite-dimensional vector spaces are classified by their dimensions, vector spaces of the same dimension are isomorphic. Linear mappings from $\mathbb {C} ^{n}$ to itself are classified by their Jordan normal forms. The main question is which Jordan blocks occur with what length and what eigenvalues. Here, we discuss the type of a real-symmetric bilinear form. Other classification results in linear algebra refer to quadratic forms, and to finite groups of motions in space.