# famlles génératrices

## English translation: generating families

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GLOSSARY ENTRY (DERIVED FROM QUESTION BELOW)
 French term or phrase: famlles génératrices English translation: generating families Entered by:

 10:20 Feb 15, 2018
French to English translations [PRO]
Mathematics & Statistics / Algèbre linéaire
 French term or phrase: famlles génératrices Algèbre 2 Etudier l'algèbre linéaire Espaces vectoriels, sous-espaces, somme direct. Familles libres, /////familles génératrices,///// bases, théorème de la base incomplète. Dimension. From document describing the courses for a Bachelor's degree "licence de sciences, technologies, santé - Mention mathématiques" Thanks for your help with reference if possible.
 Ghyslaine LE NAGARDNew Caledonia Local time: 20:07
 generating families Explanation:Famille de vecteurs pouvant généré un espace de dimension finie. Il s'agit d'une notion d'algèbre linéaire de base qui est associée à "famille libre".--------------------------------------------------Note added at 44 minutes (2018-02-15 11:04:49 GMT)--------------------------------------------------famille libre = linear independent family
Selected response from:

Frederic Rosard
France
Local time: 11:07
 Merci4 KudoZ points were awarded for this answer

5generating families
 Frederic Rosard
5set of basis vectors
 Francois Boye
4linear span
 herbalchemist

18 mins   confidence:
generating families

Explanation:
Famille de vecteurs pouvant généré un espace de dimension finie.

Il s'agit d'une notion d'algèbre linéaire de base qui est associée à "famille libre".

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Note added at 44 minutes (2018-02-15 11:04:49 GMT)
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famille libre = linear independent family

Example sentence(s):
• Prove that the familly of vectors of XXX is a generating familly of vectors of XXX
 Frederic RosardFranceLocal time: 11:07Specializes in fieldNative speaker of: FrenchPRO pts in category: 4
 Merci
 Asker: Merci. J'ai justement poster une question pour "families libres" juste avant celle-ci mais elle ne semble pas apparaître dans le listing des questions.

5 hrs   confidence:
set of basis vectors

Explanation:
In mathematics, a set of elements (vectors) in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set.

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Note added at 8 hrs (2018-02-15 18:53:55 GMT)
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basis vectors are generating vectors because they are the components of the basis; as the result, they can generate any vector of the vector space.

This is linear algebra 101

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Note added at 1 day 19 hrs (2018-02-17 06:20:06 GMT)
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https://www.cmoncours.com/cours/175/

 Francois BoyeUnited StatesLocal time: 05:07Specializes in fieldNative speaker of: FrenchPRO pts in category: 4

neutral  philgoddard: This is just lifted from Wikipedia. It's not an explanation.
 48 mins
-> Any person that has a math background understands this!

2 days 23 hrs   confidence:
famille génératrice
linear span

Explanation:
https://fr.wikipedia.org/wiki/Famille_génératrice
En algèbre linéaire, une famille génératrice est une famille de vecteurs d'un espace vectoriel dont les combinaisons linéaires permettent de construire tous les autres vecteurs de l'espace...
Soit un corps K, et soit E un espace vectoriel sur K. Une famille finie ( f 1 , f 2 , … , f n ) {\displaystyle (f_{1},f_{2},\ldots ,f_{n})} (f_{1},f_{2},\ldots ,f_{n}) d'éléments de E (vecteurs) est dite génératrice de E si...
En bref, la famille est génératrice de E si tous les vecteurs de l'espace E s'expriment comme combinaisons linéaires des vecteurs de la famille ( f i ) i ∈ I {\displaystyle (f_{i})_{i\in I}} (f_{i})_{{i\in I}}...
Si en plus la famille est libre, alors c'est une base de E.

https://fr.wikipedia.org/wiki/Famille_(mathématiques)
En mathématiques, la notion de famille est une généralisation de celle de suite, suite finie ou suite indexée par tous les entiers naturels. Ainsi on pourra parler, en algèbre linéaire, de la famille de vecteurs (u1, u2, …, un), qui est une famille finie, ou de la famille dénombrable (un)n ∈ N.

https://en.wikipedia.org/wiki/Basis_(linear_algebra)
In mathematics, a set of elements (vectors) in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set.[1] In more general terms, a basis is a linearly independent spanning set.

https://en.wikipedia.org/wiki/Linear_span
The real vector space R3 has {(-1,0,0), (0,1,0), (0,0,1)} as a spanning set. This particular spanning set is also a basis. If (-1,0,0) were replaced by (1,0,0), it would also form the canonical basis of R3.
Another spanning set for the same space is given by {(1,2,3), (0,1,2), (−1,1/2,3), (1,1,1)}, but this set is not a basis, because it is linearly dependent...
Suppose that X is a normed vector space and let E be any non-empty subset of X. The closed linear span of E, denoted by Sp ¯ ( E ) or Span ¯ ( E ), is the intersection of all the closed linear subspaces of X which contain E.

https://fr.wikipedia.org/wiki/Sous-espace_vectoriel_engendré
Dans un espace vectoriel E, le sous-espace vectoriel engendré par une partie A de E est le plus petit sous-espace vectoriel de E contenant A. C'est aussi l'ensemble des combinaisons linéaires de vecteurs de A. Le sous-espace vectoriel engendré par une famille de vecteurs est le plus petit sous-espace contenant tous les vecteurs de cette famille.
Une famille de vecteurs ou une partie est dite génératrice de E si le sous-espace qu'elle engendre est l'espace entier E.

https://www.math.ksu.edu/~nagy/lin-alg/notes.pdf
Definition. Let V be a k-vector space, and let M ⊂ V be an arbitrary subset
of V . Consider the family F = {X : X k-linear subspace of V , and X ⊃ M}.
The set Spank(M) =∩(X∈F) X, which is a linear subspace of V by the preceding exercise, is called the k-linear span of M in V...
8) Let V be a k-vector space, and M be a subset of V . For an element v ∈ V ,
prove that the following are equivalent:
(i) v ∈ Spank(M);
(ii) there exists an integer n ≥ 1, elements x1, . . . , xn ∈ M, and scalars
λ1, . . . , λn ∈ k such that v = λ1x1 + · · · + λnxn.
Hint: First prove that the set of elements satisfying property (ii) is a linear subspace. Second, prove that the linear span of M contains all elements satisfying (ii).

 herbalchemistGermanyLocal time: 11:07Native speaker of: English

neutral  Francois Boye: You have confirmed the concept of 'set of basis vectors'. So why do you reject it?
 2 days 5 hrs

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