# Rank–nullity theorem

Rank–nullity theorem "Rank theorem" redirige ici. For the rank theorem of multivariable calculus, see constant rank theorem. This article may require cleanup to meet Wikipedia's quality standards. The specific problem is: see talk:Rank–nullity theorem#Terminology. Please help improve this article if you can. (Avril 2022) (Découvrez comment et quand supprimer ce modèle de message) Rank–nullity theorem The rank–nullity theorem is a theorem in linear algebra, which asserts that the dimension of the domain of a linear map is the sum of its rank (the dimension of its image) and its nullity (the dimension of its kernel).[1][2][3][4] Contenu 1 Stating the theorem 1.1 Matrices 2 Preuves 2.1 Première preuve 2.2 Deuxième preuve 3 Reformulations and generalizations 4 Citations 5 References Stating the theorem Let {style d'affichage T:Vto W} be a linear transformation between two vector spaces where {style d'affichage T} 's domain {style d'affichage V} is finite dimensional. Alors {nom de l'opérateur de style d'affichage {Rank} (J)~+~operatorname {Nullity} (J)~=~dim V,} où {nom de l'opérateur de style d'affichage {Rank} (J)~:=~dim(nom de l'opérateur {Image} (J))qquad {texte{ et }}qquad operatorname {Nullity} (J)~:=~dim(nom de l'opérateur {Ker} (J)).} Autrement dit, {style d'affichage sombre(nom de l'opérateur {je suis} J)+dim(ker T)=dim(nom de l'opérateur {domain} J).} This theorem can be refined via the splitting lemma to be a statement about an isomorphism of spaces, not just dimensions. Explicitement, since T induces an isomorphism from {displaystyle V/operatorname {Ker} (J)} à {nom de l'opérateur de style d'affichage {Image} (J),} the existence of a basis for V that extends any given basis of {nom de l'opérateur de style d'affichage {Ker} (J)} implique, via the splitting lemma, ce {nom de l'opérateur de style d'affichage {Image} (J)oplus operatorname {Ker} (J)cong V.} Taking dimensions, the rank–nullity theorem follows.

Matrices Since {nom de l'opérateur de style d'affichage {Tapis} _{mtimes n}(mathbb {F} )cong operatorname {Lui} la gauche(mathbb {F} ^{n},mathbb {F} ^{m}droit),} [5] matrices immediately come to mind when discussing linear maps. In the case of an {displaystyle mtimes n} matrice, the dimension of the domain is {displaystyle n,} the number of columns in the matrix. Thus the rank–nullity theorem for a given matrix {displaystyle Min operatorname {Tapis} _{mtimes n}(mathbb {F} )} immediately becomes {nom de l'opérateur de style d'affichage {Rank} (M)+nom de l'opérateur {Nullity} (M)=n.} Proofs Here we provide two proofs. The first[2] operates in the general case, using linear maps. The second proof[6] looks at the homogeneous system {style d'affichage mathbf {Ax} = mathbf {0} } pour {style d'affichage mathbf {UN} in operatorname {Tapis} _{mtimes n}(mathbb {F} )} with rank {style d'affichage r} and shows explicitly that there exists a set of {displaystyle n-r} linearly independent solutions that span the kernel of {style d'affichage mathbf {UN} } .

While the theorem requires that the domain of the linear map be finite-dimensional, there is no such assumption on the codomain. This means that there are linear maps not given by matrices for which the theorem applies. Despite this, the first proof is not actually more general than the second: since the image of the linear map is finite-dimensional, we can represent the map from its domain to its image by a matrix, prove the theorem for that matrix, then compose with the inclusion of the image into the full codomain.

First proof Let {style d'affichage V,O} be vector spaces over some field {style d'affichage mathbb {F} } et {style d'affichage T} defined as in the statement of the theorem with {displaystyle dim V=n} .

Comme {nom de l'opérateur de style d'affichage {Ker} Tsubset V} est un sous-espace, there exists a basis for it. Supposer {displaystyle dim operatorname {Ker} T=k} et laissez {style d'affichage {mathématique {K}}:={v_{1},ldots ,v_{k}}nom de l'opérateur du sous-ensemble {Ker} (J)} be such a basis.

We may now, by the Steinitz exchange lemma, extend {style d'affichage {mathématique {K}}} avec {displaystyle n-k} linearly independent vectors {displaystyle w_{1},ldots ,w_{nk}} to form a full basis of {style d'affichage V} .

Laisser {style d'affichage {mathématique {S}}:={w_{1},ldots ,w_{nk}}subset Vsetminus operatorname {Ker} (J)} tel que {style d'affichage {mathématique {B}}:={mathématique {K}}Coupe {mathématique {S}}={v_{1},ldots ,v_{k},w_{1},ldots ,w_{nk}}subset V} is a basis for {style d'affichage V} . De cela, we know that {nom de l'opérateur de style d'affichage {Je suis} T=operatorname {Span} J({mathématique {B}})=nomopérateur {Span} {J(v_{1}),ldots ,J(v_{k}),J(w_{1}),ldots ,J(w_{nk})}=nomopérateur {Span} {J(w_{1}),ldots ,J(w_{nk})}=nomopérateur {Span} J({mathématique {S}}).} We now claim that {style d'affichage T({mathématique {S}})} is a basis for {nom de l'opérateur de style d'affichage {Je suis} J} . The above equality already states that {style d'affichage T({mathématique {S}})} is a generating set for {nom de l'opérateur de style d'affichage {Je suis} J} ; it remains to be shown that it is also linearly independent to conclude that it is a basis.

Supposer {style d'affichage T({mathématique {S}})} is not linearly independent, et laissez {somme de style d'affichage _{j=1}^{nk}Alpha _{j}J(w_{j})=0_{O}} pour certains {style d'affichage alpha _{j}en mathbb {F} } .

Ainsi, owing to the linearity of {style d'affichage T} , il s'ensuit que {displaystyle Tleft(somme _{j=1}^{nk}Alpha _{j}w_{j}droit)=0_{O}implies left(somme _{j=1}^{nk}Alpha _{j}w_{j}droit)in operatorname {Ker} T=operatorname {Span} {mathématique {K}}subset V.} This is a contradiction to {style d'affichage {mathématique {B}}} being a basis, unless all {style d'affichage alpha _{j}} are equal to zero. This shows that {style d'affichage T({mathématique {S}})} is linearly independent, and more specifically that it is a basis for {nom de l'opérateur de style d'affichage {Je suis} J} .

To summarize, Nous avons {style d'affichage {mathématique {K}}} , a basis for {nom de l'opérateur de style d'affichage {Ker} J} , et {style d'affichage T({mathématique {S}})} , a basis for {nom de l'opérateur de style d'affichage {Je suis} J} .

Finally we may state that {nom de l'opérateur de style d'affichage {Rank} (J)+nom de l'opérateur {Nullity} (J)=dim operatorname {Je suis} T+dim operatorname {Ker} T=|J({mathématique {S}})|+|{mathématique {K}}|=(nk)+k=n=dim V.} This concludes our proof.

Second proof Let {style d'affichage mathbf {UN} in operatorname {Tapis} _{mtimes n}(mathbb {F} )} avec {style d'affichage r} linearly independent columns (c'est à dire. {nom de l'opérateur de style d'affichage {Rank} (mathbf {UN} )=r} ). Nous allons montrer que: There exists a set of {displaystyle n-r} linearly independent solutions to the homogeneous system {style d'affichage mathbf {Ax} = mathbf {0} } . That every other solution is a linear combination of these {displaystyle n-r} solutions.

To do this, we will produce a matrix {style d'affichage mathbf {X} in operatorname {Tapis} _{ntimes (n-r)}(mathbb {F} )} whose columns form a basis of the null space of {style d'affichage mathbf {UN} } .

Sans perte de généralité, assume that the first {style d'affichage r} columns of {style d'affichage mathbf {UN} } are linearly independent. Alors, nous pouvons écrire {style d'affichage mathbf {UN} ={commencer{pmatrice}mathbf {UN} _{1}&mathbf {UN} _{2}fin{pmatrice}},} où {style d'affichage mathbf {UN} _{1}in operatorname {Tapis} _{mtimes r}(mathbb {F} )} avec {style d'affichage r} linearly independent column vectors, et {style d'affichage mathbf {UN} _{2}in operatorname {Tapis} _{mtimes (n-r)}(mathbb {F} )} , each of whose {displaystyle n-r} columns are linear combinations of the columns of {style d'affichage mathbf {UN} _{1}} .

Cela signifie que {style d'affichage mathbf {UN} _{2}= mathbf {UN} _{1}mathbf {B} } pour certains {style d'affichage mathbf {B} in operatorname {Tapis} _{rtimes (n-r)}} (see rank factorization) et, Par conséquent, {style d'affichage mathbf {UN} ={commencer{pmatrice}mathbf {UN} _{1}&mathbf {UN} _{1}mathbf {B} fin{pmatrice}}.} Laisser {style d'affichage mathbf {X} ={commencer{pmatrice}-mathbf {B} \mathbf {je} _{n-r}fin{pmatrice}},} où {style d'affichage mathbf {je} _{n-r}} est le {style d'affichage (n-r)fois (n-r)} identity matrix. We note that {style d'affichage mathbf {X} in operatorname {Tapis} _{ntimes (n-r)}(mathbb {F} )} satisfait {style d'affichage mathbf {UN} mathbf {X} ={commencer{pmatrice}mathbf {UN} _{1}&mathbf {UN} _{1}mathbf {B} fin{pmatrice}}{commencer{pmatrice}-mathbf {B} \mathbf {je} _{n-r}fin{pmatrice}}=-mathbf {UN} _{1}mathbf {B} +mathbf {UN} _{1}mathbf {B} = mathbf {0} _{mtimes (n-r)}.} Par conséquent, each of the {displaystyle n-r} columns of {style d'affichage mathbf {X} } are particular solutions of {style d'affichage mathbf {Ax} = mathbf {0} _{mathbb {F} ^{m}}} .

Par ailleurs, la {displaystyle n-r} columns of {style d'affichage mathbf {X} } are linearly independent because {style d'affichage mathbf {Xu} = mathbf {0} _{mathbb {F} ^{n}}} will imply {style d'affichage mathbf {tu} = mathbf {0} _{mathbb {F} ^{n-r}}} pour {style d'affichage mathbf {tu} en mathbb {F} ^{n-r}} : {style d'affichage mathbf {X} mathbf {tu} = mathbf {0} _{mathbb {F} ^{n}}implique {commencer{pmatrice}-mathbf {B} \mathbf {je} _{n-r}fin{pmatrice}}mathbf {tu} = mathbf {0} _{mathbb {F} ^{n}}implique {commencer{pmatrice}-mathbf {B} mathbf {tu} \mathbf {tu} fin{pmatrice}}={commencer{pmatrice}mathbf {0} _{mathbb {F} ^{r}}\mathbf {0} _{mathbb {F} ^{n-r}}fin{pmatrice}}implies mathbf {tu} = mathbf {0} _{mathbb {F} ^{n-r}}.} Par conséquent, the column vectors of {style d'affichage mathbf {X} } constitute a set of {displaystyle n-r} linearly independent solutions for {style d'affichage mathbf {Ax} = mathbf {0} _{mathbb {F} ^{m}}} .

We next prove that any solution of {style d'affichage mathbf {Ax} = mathbf {0} _{mathbb {F} ^{m}}} must be a linear combination of the columns of {style d'affichage mathbf {X} } .

For this, laisser {style d'affichage mathbf {tu} ={commencer{pmatrice}mathbf {tu} _{1}\mathbf {tu} _{2}fin{pmatrice}}en mathbb {F} ^{n}} be any vector such that {style d'affichage mathbf {Au} = mathbf {0} _{mathbb {F} ^{m}}} . Note that since the columns of {style d'affichage mathbf {UN} _{1}} are linearly independent, {style d'affichage mathbf {UN} _{1}mathbf {X} = mathbf {0} _{mathbb {F} ^{m}}} implique {style d'affichage mathbf {X} = mathbf {0} _{mathbb {F} ^{r}}} .

Par conséquent, {style d'affichage {commencer{déployer}{rcl}mathbf {UN} mathbf {tu} &=&mathbf {0} _{mathbb {F} ^{m}}\implique {commencer{pmatrice}mathbf {UN} _{1}&mathbf {UN} _{1}mathbf {B} fin{pmatrice}}{commencer{pmatrice}mathbf {tu} _{1}\mathbf {tu} _{2}fin{pmatrice}}&=&mathbf {UN} _{1}mathbf {tu} _{1}+mathbf {UN} _{1}mathbf {B} mathbf {tu} _{2}&=&mathbf {UN} _{1}(mathbf {tu} _{1}+mathbf {B} mathbf {tu} _{2})&=&mathbf {0} _{mathbb {F} ^{m}}\implies mathbf {tu} _{1}+mathbf {B} mathbf {tu} _{2}&=&mathbf {0} _{mathbb {F} ^{r}}\implies mathbf {tu} _{1}&=&-mathbf {B} mathbf {tu} _{2}fin{déployer}}} {displaystyle implies mathbf {tu} ={commencer{pmatrice}mathbf {tu} _{1}\mathbf {tu} _{2}fin{pmatrice}}={commencer{pmatrice}-mathbf {B} \mathbf {je} _{n-r}fin{pmatrice}}mathbf {tu} _{2}= mathbf {X} mathbf {tu} _{2}.} This proves that any vector {style d'affichage mathbf {tu} } that is a solution of {style d'affichage mathbf {Ax} = mathbf {0} } must be a linear combination of the {displaystyle n-r} special solutions given by the columns of {style d'affichage mathbf {X} } . And we have already seen that the columns of {style d'affichage mathbf {X} } are linearly independent. Ainsi, the columns of {style d'affichage mathbf {X} } constitute a basis for the null space of {style d'affichage mathbf {UN} } . Par conséquent, the nullity of {style d'affichage mathbf {UN} } est {displaystyle n-r} . Depuis {style d'affichage r} equals rank of {style d'affichage mathbf {UN} } , il s'ensuit que {nom de l'opérateur de style d'affichage {Rank} (mathbf {UN} )+nom de l'opérateur {Nullity} (mathbf {UN} )=n} . This concludes our proof.

Reformulations and generalizations This theorem is a statement of the first isomorphism theorem of algebra for the case of vector spaces; it generalizes to the splitting lemma.

In more modern language, the theorem can also be phrased as saying that each short exact sequence of vector spaces splits. Explicitement, given that {displaystyle 0rightarrow Urightarrow Vmathbin {overset {J}{rightarrow }} Rrightarrow 0} is a short exact sequence of vector spaces, alors {displaystyle Uoplus Rcong V} , Par conséquent {style d'affichage sombre(tu)+dim(R)=dim(V).} Here R plays the role of im T and U is ker T, c'est à dire. {displaystyle 0rightarrow ker Tmathbin {hookrightarrow } Vmathbin {overset {J}{rightarrow }} nom de l'opérateur {je suis} Trightarrow 0} Dans le cas de dimension finie, this formulation is susceptible to a generalization: si 0 → V1 → V2 → ⋯ → Vr → 0 is an exact sequence of finite-dimensional vector spaces, alors[7] {somme de style d'affichage _{je=1}^{r}(-1)^{je}dim(V_{je})=0.} The rank–nullity theorem for finite-dimensional vector spaces may also be formulated in terms of the index of a linear map. The index of a linear map {displaystyle Tin operatorname {Lui} (V,O)} , où {style d'affichage V} et {style d'affichage W.} are finite-dimensional, est défini par {nom de l'opérateur de style d'affichage {index} T=dim operatorname {Ker} (J)-dim operatorname {Coker} T} Intuitivement, {displaystyle dim operatorname {Ker} J} is the number of independent solutions {style d'affichage v} of the equation {displaystyle Tv=0} , et {displaystyle dim operatorname {Coker} J} is the number of independent restrictions that have to be put on {style d'affichage w} to make {displaystyle Tv=w} soluble. The rank–nullity theorem for finite-dimensional vector spaces is equivalent to the statement {nom de l'opérateur de style d'affichage {index} T=dim V-dim W.} We see that we can easily read off the index of the linear map {style d'affichage T} from the involved spaces, without any need to analyze {style d'affichage T} in detail. This effect also occurs in a much deeper result: the Atiyah–Singer index theorem states that the index of certain differential operators can be read off the geometry of the involved spaces.

Citations ^ Axler (2015) p. 63, §3.22 ^ Jump up to: a b Friedberg, Insel & Spence (2014) p. 70, §2.1, Théorème 2.3 ^ Katznelson & Katznelson (2008) p. 52, §2.5.1 ^ Valenza (1993) p. 71, §4.3 ^ Friedberg, Insel & Spence (2014) pp. 103-104, §2.4, Théorème 2.20 ^ Banerjee, Sudipto; Roy, Anindya (2014), Linear Algebra and Matrix Analysis for Statistics, Texts in Statistical Science (1st ed.), Chapman and Hall/CRC, ISBN 978-1420095388 ^ Zaman, Ragib. "Dimensions of vector spaces in an exact sequence". Mathematics Stack Exchange. Récupéré 27 Octobre 2015. References Axler, Sheldon (2015). Algèbre linéaire bien faite. Undergraduate Texts in Mathematics (3e éd.). Springer. ISBN 978-3-319-11079-0. Banerjee, Sudipto; Roy, Anindya (2014), Linear Algebra and Matrix Analysis for Statistics, Texts in Statistical Science (1st ed.), Chapman and Hall/CRC, ISBN 978-1420095388 Friedberg, Stephen H.; Insel, Arnold J.; Spence, Lawrence E. (2014). Linear Algebra (4e éd.). Pearson Education. ISBN 978-0130084514. Meyer, Carl D. (2000), Matrix Analysis and Applied Linear Algebra, SIAM, ISBN 978-0-89871-454-8. Katznelson, Yitzhak; Katznelson, Yonatan R. (2008). UN (Terse) Introduction to Linear Algebra. Société mathématique américaine. ISBN 978-0-8218-4419-9. Valenza, Robert J. (1993) [1951]. Linear Algebra: An Introduction to Abstract Mathematics. Undergraduate Texts in Mathematics (3e éd.). Springer. ISBN 3-540-94099-5. Catégories: Theorems in linear algebraIsomorphism theorems

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