OpenLLM-France/wikimedia huggingface dataset:
# Bilinear form/Linear forms/Nondegenerate/Fact

Let $K$ be a field, and let $V$ denote a $K$-vector space, endowed with a bilinear form $\left\langle -,-\right\rangle$. Then the following statements hold.
* For every vector $u\in V$, the assignments and are $K$-linear.
    $V\longrightarrow K,v\longmapsto \left\langle u,v\right\rangle ,$
    $V\longrightarrow K,v\longmapsto \left\langle v,u\right\rangle ,$
* The assignment is $K$-linear.
    $V\longrightarrow {V}^{*},u\longmapsto \left\langle u,-\right\rangle ,$
* If $\left\langle -,-\right\rangle$ is nongenerate, then the assignment in (2) is injective. If, moreover, $V$ is finite-dimensional, then this assignment is bijective.
# Euclidean space/Linear form/Corresponding vector/Fact/Proof2

This follows immediately from fact (3). The extra statement is clear because of
    $\left\langle w,u_{i}\right\rangle =\left\langle \sum _{j=1}^{n}a_{j}u_{j},u_{i}\right\rangle =a_{i}=f(u_{i})\,.$
# Bilinear form/Gram matrix/Introduction/Section
## Definition

Bilinear form/Gram matrix/Definition
In example, the matrix ${\left(a_{ij}\right)}_{ij}$ is the Gram matrix with respect to the standard basis of $K^{n}$. If the Gram matrix of a bilinear form $\left\langle -,-\right\rangle$ with respect to a basis $v_{1},\ldots ,v_{n}$ is given, then one can compute $\left\langle v,w\right\rangle$ for arbitrary vectors. For this, just write $v=\sum _{i=1}^{n}b_{i}v_{i}$ and $w=\sum _{i=1}^{n}c_{i}v_{i}$; then we get, using the general distributive law,
    ${\begin{aligned}\left\langle v,w\right\rangle &=\left\langle \sum _{i=1}^{n}b_{i}v_{i},\sum _{j=1}^{n}c_{j}v_{j}\right\rangle \\&=\sum _{1\leq i,j\leq n}b_{i}c_{j}\left\langle v_{i},w_{j}\right\rangle \\&=\sum _{i=1}^{n}b_{i}{\left(\sum _{j=1}^{n}c_{j}\left\langle v_{i},w_{j}\right\rangle \right)}\\&=(b_{1},\ldots ,b_{n})G{\begin{pmatrix}c_{1}\\\vdots \\c_{n}\end{pmatrix}}.\end{aligned}}$
Thus we obtain the value of the bilinear form at two vectors in applying the Gram matrix to the coordinate tuple of the second vector, and multiplying the result (which is a column vector) with the coordinate tuple of the first vector, considered as a row tuple. Put briefly,
    $\left\langle v,w\right\rangle ={v^{\text{tr}}}Gw\,.$

## Lemma
### Proof

$\Box$
# Bilinear form/Gram matrix/Definition

Vector space/Finite dimensional/Bilinear form/Situation Let $v_{1},\ldots ,v_{n}$ be a basis of $V$. The the $n\times n$-matrix
    $[[:Template:Grammatrix]]$
is called the Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis.
# Bilinear form/Gram matrix under base change/Fact

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a bilinear form on $V$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ be two bases of $V$, and let $G$ and $H$ be the Gram matrices of $\left\langle -,-\right\rangle$ with respect to these bases. Suppose that we have the relations
    $w_{j}=\sum _{i=1}^{n}a_{ij}v_{i}\,$
between the basis elements, which we encode in the transformation matrix $A={\left(a_{ij}\right)}_{i,j}$.
    $H={A^{\text{tr}}}GA\,$
# Bilinear form/Gram matrix under base change/Fact/Proof

We have
    ${\begin{aligned}\left\langle w_{r},w_{s}\right\rangle &=\left\langle \sum _{i=1}^{n}a_{ir}v_{i},\sum _{k=1}^{n}a_{ks}v_{k}\right\rangle \\&=\sum _{1\leq i,k\leq n}a_{ir}a_{ks}\left\langle v_{i},v_{k}\right\rangle \\&=\sum _{1\leq i\leq n}a_{ir}{\left(\sum _{1\leq k\leq n}a_{ks}\left\langle v_{i},v_{k}\right\rangle \right)}\\&=\sum _{1\leq i\leq n}a_{ir}{\left(G\circ A\right)}_{is}\\&={\left({A^{\text{tr}}}\circ {\left(G\circ A\right)}\right)}_{rs}.\end{aligned}}$
# Bilinear form/Symmetric/Introduction/Section
## Definition

Bilinear form/Symmetric/Definition
As in the case of an inner product, we have again a polarization formula.

## Lemma
### Proof

$\Box$

## Definition

Symmetric bilinear form/Nondegenerate/Gradient/Definition

## Definition

Symmetric bilinear form/Orthogonal/Definition

## Definition

Symmetric bilinear form/Orthogonal basis/Definition
Far a symmetric bilinear form, it is possible, different to the case of an inner product, that a vector $\neq 0$ is orthogonal to itself. It is possible, at least in the degenerate case, that a vector $\neq 0$ is orthogonal to all vectors. Like in the case of an inner product, there exist orthogonal bases.

## Definition

Symmetric bilinear form/Degeneracy space/Definition
The degeneracy space is indeed a linear subspace of $V$, see exercise.

## Theorem
### Proof

$\Box$
# Bilinear form/Symmetric/Definition

Let $K$ be a field, let $V$ be a $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a bilinear form on $V$. The bilinear form is called symmetric, if
    $\left\langle v,w\right\rangle =\left\langle w,v\right\rangle \,$
holds for all $v,w\in V$.
# Symmetric bilinear form/Polarization formula/Fact

Let $K$ be a field, and suppose that its characteristic is not $2$. Let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on the $K$-vector space $V$.
    $\left\langle v,w\right\rangle ={\frac {1}{2}}{\left(\left\langle v+w,v+w\right\rangle -\left\langle v,v\right\rangle -\left\langle w,w\right\rangle \right)}\,$
# Long terminal repeats

Long terminal repeats (LTRs) are sequences of DNA that repeat hundreds or thousands of times. They are found in retroviral DNA and in retrotransposons, flanking functional genes. They are used by viruses to control of gene expression and integration. For example, a retrovirus genome might contain the following features:
    LTR -- gag -- pol -- env -- LTR
The LTRs are partially transcribed into an RNA intermediate, followed by reverse transcription into complementary DNA (cDNA) and ultimately dsDNA (double-stranded DNA) with full LTRs. The LTRs then mediate integration of the retroviral DNA via an LTR specific integrase into another region of the host chromosome. Retroviruses such as Human Immunodeficiency Virus (HIV) use this basic mechanism.
# Symmetric bilinear form/Nondegenerate/Gradient/Definition

Let $V$ be a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a nondegenerate symmetric bilinear form on $V$. For a given linear form
    $L\colon V\longrightarrow K,$
the uniquely determined vector $z\in V$ fulfilling
    $L(v)=\left\langle z,v\right\rangle =\left\langle v,z\right\rangle \,$
for all $v\in V$, is called the gradient of $L$ with respect to the bilinear form.
# Symmetric bilinear form/Orthogonal/Definition

Vector space/Symmetric bilinear form/Situation Two vectors $v,w\in V$ are called orthogonal, if
    $\left\langle v,w\right\rangle =0\,$
holds.
# Symmetric bilinear form/Degeneracy space/Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on $V$. The linear subspace
    ${\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in V\right\}}$
is called the degeneracy space of the bilinear form.
# Symmetric bilinear form/Orthogonal basis/Fact

Let $K$ be a field, let $V$ be a finite-dimensional $K$-vector space endowed with a symmetric bilinear form $\left\langle -,-\right\rangle$.
orthogonal basis
$V$
# Bilinear form/Vector space/Section

Let $V$ be a vector space over a field $K$, and let $\Psi _{1}$ and $\Psi _{2}$ denote bilinear forms on $V$. The sum of these two bilinear forms is defined pointwisely, that is,
    $(\Psi _{1}+\Psi _{2})(u,v):=\Psi _{1}(u,v)+\Psi _{2}(u,v)\,.$
In the same way, for a scalar $c\in K$, the form $c\Psi$ is defined via
    $(c\Psi )(u,v)=c\Psi (u,v)\,.$
These functions are again bilinear, see exercise. With these definition, we obtain a vector space structure on the set of all bilinear forms on $V$.

## Definition

Bilinear form/Vector space/Definition

## Lemma
### Proof

$\Box$
# Bilinear form/Vector space/Definition

Let $V$ be a vector space over the field $K$. The set of all bilinear forms on $V$, endowed with pointwise addition and scalar multiplication, is called the vector space of bilinear forms. It is denoted by $[[:Template:Op:Bilin]]$.
# Vector space/Finite-dimensional/Gram matrices/Fact

Let $V$ be a finite-dimensional $K$-vector space.
basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$, the mapping
    $\operatorname {Bilin} _{\left(V\right)}\longrightarrow \operatorname {Mat} _{n}(K),\Psi \longmapsto G_{\mathfrak {v}}(\Psi ),$
which assigns to a bilinear form $\Psi$ its Gram matrix with respect to the given basis, is an isomorphism
# Vector space/Finite-dimensional/Gram matrices/Fact/Proof

The injectivity of the mapping follows from fact. The surjectivity follows from the fact that an arbitrary matrix might b interpret as a bilinear form in the sense of example. The linearity follows immediately from the pointwise definition of the vector space structure on $\operatorname {Bilin} _{\left(V\right)}$.
# Linear algebra (Osnabrück 2024-2025)/K^n/Bilinear forms/Arbitrary coefficients/Special cases/Example/Examplereferencenumber

Example 38.
# Vector space over C/Sesquilinearform/Vector space structure/Remark

The set of all sesquilinear forms on a $\mathbb {C}$-vector space $V$ form a $\mathbb {C}$-vector space. It is denoted by $[[:Template:Op:Sesqui]]$.
# Hermitian form/Definition

A sesquilinear form $\left\langle -,-\right\rangle$ on a complex vector space $V$ is called Hermitian, if
    $\left\langle v,u\right\rangle ={\overline {\left\langle u,v\right\rangle }}\,$
holds for all $u,v\in V$.
# Matrix/Hermitian/Definition

A complex square matrix
    $M={\left(a_{ij}\right)}_{ij}\,$
is called Hermitian, if
    $a_{ij}={\overline {a_{ji}}}\,$
holds for all $i,j$.
# Sesquilinear forms/Introduction/Section
## Definition

Vector spaces over C/Mapping/Antilinear/Definition
If we consider the complex vector spaces as a real vector space, then this is, in particular, a real-linear mapping. We have encountered this property already in the context of a complex inner product.

## Definition

Vector space over C/Sesquilinear form/Definition
We impose linearity in the first and antilinearity in the second component. There exists also the other convention.
Many concepts and statements carry over, with some minor changes, from the real to the complex situation.

## Definition

Sesquilinear form/Gram matrix/Definition
If the Gram matrix of a Sesquilinear form $\left\langle -,-\right\rangle$ with respect to a basis $v_{1},\ldots ,v_{n}$ is given, then we can compute $\left\langle v,w\right\rangle$ for arbitrary vectors. We write $v=\sum _{i=1}^{n}b_{i}v_{i}$ and $w=\sum _{i=1}^{n}c_{i}v_{i}$, and we obtain, using the general distributive law,
    ${\begin{aligned}\left\langle v,w\right\rangle &=\left\langle \sum _{i=1}^{n}b_{i}v_{i},\sum _{j=1}^{n}c_{j}v_{j}\right\rangle \\&=\sum _{1\leq i,j\leq n}b_{i}{\overline {c_{j}}}\left\langle v_{i},w_{j}\right\rangle \\&=\sum _{i=1}^{n}b_{i}{\left(\sum _{j=1}^{n}{\overline {c_{j}}}\left\langle v_{i},w_{j}\right\rangle \right)}\\&=(b_{1},\ldots ,b_{n})G{\begin{pmatrix}{\overline {c_{1}}}\\\vdots \\{\overline {c_{n}}}\end{pmatrix}}.\end{aligned}}$
Thus, we get the value of the sesquilinear form at two vectors by applying the Gram matrix to the coordinate tuple of the complex-conjugated second vector, and by multiplying the result (which is a column vector) with the coordinate tuple of the first vector, considered as a row tuple. Therefore,
    $\left\langle v,w\right\rangle ={v^{\text{tr}}}G{\overline {w}}\,.$

## Lemma
### Proof

$\Box$

## Remark

The set of all sesquilinear forms on a $\mathbb {C}$-vector space $V$ form a $\mathbb {C}$-vector space. It is denoted by $\operatorname {Sesq} _{\left(V\right)}$.
# Vector spaces over C/Mapping/Antilinear/Definition

Let $V$ and $W$ be vector spaces over the complex numbers $\mathbb {C}$. A mapping
    $\varphi \colon V\longrightarrow W$
is called antilinear, if
    $\varphi (u+v)=\varphi (u)+\varphi (v)\,$
holds for all $u,v\in V$, and if
    $\varphi (\lambda v)={\overline {\lambda }}\varphi (v)\,$
holds for all $v\in V$ and $\lambda \in \mathbb {C}$.
# Vector space over C/Sesquilinear form/Definition

Let $V$ be a $\mathbb {C}$-vector space. A mapping
    $V\times V\longrightarrow \mathbb {C} ,(v,w)\longmapsto \left\langle v,w\right\rangle ,$
is called a sesquilinear form, if, for all $v\in V$, the induced mappinga
    $V\longrightarrow \mathbb {C} ,w\longmapsto \left\langle v,w\right\rangle ,$
are $\mathbb {C}$-antilinear, and, for all $w\in V$, the induced mappings
    $V\longrightarrow \mathbb {C} ,v\longmapsto \left\langle v,w\right\rangle ,$
are $\mathbb {C}$-linear.
# Sesquilinear form/Gram matrix/Definition

Let $V$ be a finite-dimensional $\mathbb {C}$-vector space, endowed with a sesquilinear form $\left\langle -,-\right\rangle$. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$. The $n\times n$-matrix
    $\left\langle v_{i},v_{j}\right\rangle _{1\leq i,j\leq n}$
is called the Gram matrix of $\left\langle -,-\right\rangle$ with respect to this basis.
# Sesquilinear form/Gram matrix under base change/Fact

Let $V$ be a finite-dimensional $\mathbb {C}$-vector space, endowed with a sesquilinear form $\left\langle -,-\right\rangle$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote two bases of $V$, and let $G$ and $H$ denote the Gram matrices of $\left\langle -,-\right\rangle$ with respect to these bases. Suppose that the basis elements are related by
    $w_{j}=\sum _{i=1}^{n}a_{ij}v_{i}\,,$
which we encode in the transformation matrix $A={\left(a_{ij}\right)}_{i,j}$.
    $H={A^{\text{tr}}}G{\overline {A}}\,.$
# Sesquilinear form/Gram matrix under base change/Fact/Proof

We have
    ${\begin{aligned}\left\langle w_{r},w_{s}\right\rangle &=\left\langle \sum _{j=1}^{n}a_{rj}v_{j},\sum _{k=1}^{n}a_{sk}v_{k}\right\rangle \\&=\sum _{1\leq j,k\leq n}a_{rj}{\overline {a_{sk}}}\left\langle v_{j},v_{k}\right\rangle \\&=\sum _{1\leq j\leq n}a_{rj}{\left(\sum _{1\leq k\leq n}{\overline {a_{sk}}}\left\langle v_{j},v_{k}\right\rangle \right)}\\&={\left({A^{\text{tr}}}\circ {\left(G\circ {\overline {A}}\right)}\right)}_{rs}.\end{aligned}}$
# Symmetric bilinear form/Orthogonal basis/Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on $V$. A basis $v_{i}$, $i\in I$, of $V$ is called orthogonal basis, if
    $\left\langle v_{i},v_{j}\right\rangle =0\,$
holds for all $i\neq j$.
# Korean/Words/미르

* ㄱ ㄲ ㄴ ㄷ ㄸ ㄹ ㅁ ㅂ ㅃ ㅅ ㅆ ㅇ ㅈ ㅉ ㅊ ㅋ ㅌ ㅍ ㅎ -ㄱ-ㄲ-ㄴ-ㄷ-ㄸ-ㄹ-ㅁ-ㅂ-ㅃ-ㅅ-ㅆ-ㅇ-ㅈ-ㅉ-ㅊ-ㅋ-ㅌ-ㅍ-ㅎ- 말+/말씀+/물+/불+ a2z/Basics/News/Notes/Table/Legend/Sandbox

## 미르
### Noun

* ‘용’의 옛말.
    (obsolete) The oriental dragon with a very long body.

### Synonyms

* 용 (龍, yong, North Korean ryong, Chinese long cf. English long)

### Compounds

* 미리내 (miri-nae, "Milky Way," literally "Dragon River")

### Relatives

* 물 (mul, "water, body of water, esp. river")

### Comparatives

    (The mysterious Eurasian uses)
* mirus #Latin "marvelous, wonderful, amazing, awesome"
* miror #Latin "to admire, wonder at"
* mirer #French "to watch, stare"
* mirage #French "mirage"
* miroir #French "mirror"
* admirer #French "to admire"
* amiral #French "admiral"
* emir #French "emir"
* मीर (mir) # Hindi "(historical) Mir: title used by Muslim rulers of princely states"
* मीर (mira) # Sanskrit "sea, ocean"
* कश्मीर (kashmir) # Hindi : See Etymology through!

### From Wikipedia

* Amir al-Mu'minin (...) is an Arabic title that is usually translated "Commander of the Faithful" or "Leader of the Faithful".
* The Gur-e-Amīr or Guri Amir (...‎) is a mausoleum of the Turco-Mongol conqueror Timur (also known as Tamerlane) in Samarkand, Uzbekistan. [...] Gur-e Amir is Persian for "Tomb of the King".
* Mir Osman Ali Khan, ..., was the last Nizam (ruler) of the Princely State of Hyderabad, the largest princely state in British India.
* Kashmir would be derived from either kashyapa-mir (Kashyapa's Lake) or kashyapa-meru (Kashyapa's Mountain).
* Suleiman the Magnificent, Padishah of the Ottoman Empire.
* Mohammad Reza Pahlavi, Shahanshah of Iran from 1941 to 1979
* Deng Xiaoping
* Kim Jong-un
* ↑ https://ko.dict.naver.com/#/search?query=미르
* ↑ https://en.wiktionary.org/wiki/미르
* ↑ Hence, suggestive of the river as a long body of water.
* ↑ 龍 용(龍: 상상의 동물) dragon, the imaginary creature. 임금, 천자(天子) king, the son of the heaven ...
* ↑ late 14c. loan-translation of Latin Via lactea
* ↑ Etymology From Old French amirail, amiral, from Arabic أَمِير اَلْبَحْر‎ (ʾamīr al-baḥr, “commander of the fleet”).
* ↑ emir a prince, commander or other leader or ruler in an Islamic nation. a descendant of the prophet Muhammad. Etymology From Old French emir, from Arabic أَمِير‎ (ʾamīr, “commander, prince”). Akin to amir, Amir and admiral. Doublet of amira. Derived terms emirate
* ↑ EtymologyBorrowed from Persian میر‎ (mir), from Arabic أَمِير‎ (ʾamīr). Doublet of wikt: अमीर (amīr).
* ↑ Portrait attributed to Titian (circa 1530)
* ↑ The last ruler to hold the title of shah
# Identi.ca

Identi.ca (launched 2008) is an open source social networking and micro-blogging service. It uses the software StatusNet (licence: Affero GPLv3+).
Some advantages to use identi.ca are:
* all notices are published with Creative Commons Attribution 3.0 license
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* one can forward messages also to twitter
* FOAF (Friend of a friend) support: a machine-readable ontology describing persons, their activities and their relations to other people and objects. example:
* when you install your own StatusNet instance you have control over your data (e.g. you decide with whom to share or not, ...) + can connect with other network systems (e.g. twitter, ...) via federation

## Wikiversity related info on identi.ca

* tag #wikiversity: you can generate these when you add in your message #wikiversity
* group !wikiversity: you can post here when you add in your message !wikiversity

## some Wikiversity users who use identi.ca

(alphabetical)
* Cary Bass
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* add yourself
# Bilinear form/Symmetric/Definiteness/Introduction/Section
## Definition

Bilinear form/Symmetric/Definiteness/Definition
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

Bilinear form/Symmetric/Type/Definition
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.

## Theorem
### Proof

$\Box$
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 $\mathbb {R} ^{n}$.
# Classification/Linear algebra/Type/Remark

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.
# Bilinear form/Symmetric/Definiteness/Definition

Let $V$ be a real vector space, endowed with a symmetric 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.
# Bilinear form/Symmetric/Type/Inertia law/Fact

Let $V$ be a finite-dimensional real vector space, endowed with a symmetric bilinear form $\left\langle -,-\right\rangle$ of type $(p,q)$.
Gram matrix of $\left\langle -,-\right\rangle$ with respect to any orthogonal basis is a diagonal matrix
$p$
$q$
# Bilinear form/Symmetric/Type/Inertia law/Fact/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 $\left\langle w,w\right\rangle \leq 0$.
# Linear algebra (Osnabrück 2024-2025)/Symmetric bilinear form/Orthogonal basis/Fact/Factreferencenumber

Lemma 38.
# Bilinear form/Symmetric/Type criteria/Introduction/Section

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 determinant of a square submatrix of a matrix. We could call the following criterion also determinant criterion.

## Theorem
### Proof

$\Box$

## Corollary
### Proof

$\Box$

## Theorem
### Proof

fact.
$\Box$