Bases of vector spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 14 August 2011

Bases and dimension

Let 𝔽 be a field and let V be a vector space over 𝔽. Let { v1, v2, …, vk} be a subset of V.

Let B be a subset of V. The following are equivalent

(a)   B is a basis of V.
(b)   B is a minimal element of {S⊆V | span(S)=V} , ordered by inclusion.
(c)   B is a maximal element of {L⊆V | Lis linearly independent} , ordered by inclusion.

Let V be a vector space over a field 𝔽.

(a)   Then V has a basis.
(b)   Any two bases of V have the same number of elements.

Let R be a ring and let M be an R-module. Let S be a subset of M.

  • The submodule generated by S is the smallest submodule of M containing S, i.e. the submodule generated by S is the submodule span(S) of M such that
    (a)   S⊆span(S),
    (b)   If N is a submodule of M such that S ⊆N then span(S)⊆N.
  • The R-module M is finitely generated if there is a finite subset S of M such that span(S)=M.
  • A linear combination of elements of S is an element v ∈M of the form v= ∑s∈S rss,
  • where rs∈R and rs=0 for all but a finite number of s∈S.
  • The set S is linearly independent if it satisfies the condition: if rs∈R and rs=0 for all but a finite number of s∈S, and ∑s∈S rss=0 then rs=0, for all s∈S.
  • A basis of M is a subset B ⊆M such that
    (a)   span(B)=M,
    (b)   B is linearly independent.
  • A free module is an R-module M that has a basis.

HW: Let M be an R-module and let S ⊆M be a subset of M. Show that span(S) exists and is unique by showing that span(S) is the intersection of all the submodules that contain S.

HW: Show that span(S) is the set of linear combinations of S.

HW: Let M be an R-module. Show that a subset S⊆M is linearly independent if and only if S satisfies the following property:

HW: If rm∈R such that rm=0 for all but a finite number of m∈M and ∑m∈S rmm=0 then rm for all m∈M.

HW: Let V be a vector space over a field 𝔽. Show that if v∈V and {v1, v2,…, vn} is a basis of V then there exist unique ci∈𝔽 such that v= c1v1+ c2v2+ ⋯+ cnvn .

HW: Let R be a commutative ring. Give an example of a finitely generated R-module that does not have a basis.

HW: Give an example of a ring R and a finitely generated module over R that has two different bases with different numbers of elements.

HW: Give an example of a finitely generated module that is not free.

HW: Give an example of a free module that is not finitely generated.

HW: Show that every module over a field is free.

HW: Show that R is a division ring if and only if every R-module is free.

(a)   Let R be a ring and let M be a free R-module with an infinite basis. Any two bases of M have the same number of elements.
(b)   Let R be a commutative ring and let M be a free R-module. Any two bases of M have the same number of elements.

Let B,C and D be sets. Let R be a ring.

  • A C×B matrix with entries in R is a collection F= (Fcb) of elements Fcb ∈R indexed by the elements of C×B and such that for each b∈B all but a finite number of the Fcb are equal to 0. MC×B( R)= {C×B matrices with entries inR} .
  • The sum of two matrices F1, F2∈ MC×B( R) is the matrix F1+ F2 given by ( F1+F2) cb = ( F1) cb + (F2) cb, for b∈B and c∈C.
  • The product of matrices F1 ∈ MD×C( R) and F2 ∈ MC×B( R) is the matrix F1 F2 ∈ MD×B( R) given by ( F1F2) db = ∑c∈C ( F1) dc (F2) cb, for b∈B and d∈D.

Let M and N be free R-modules with bases B and C, respectively. Let f:M→N be a homomorphism.

  • The matrix of f:M→N with respect to the bases B and C is the matrix fCB ∈ MC×B( Rop) given by ( fCB ) cb = fcb, where fcb∈ Rop are given by f(b) = ∑c∈C fcbc, for b∈B.

Let M be a free R-module and let B and C be bases of M.

  • The change of basis matrix from B to C is the matrix PCB ∈ MC×B( Rop) given by ( PCB ) cb = Pcb, where Pcb∈ Rop are given by b = ∑c∈C Pcbc, for b∈B.

Let M and N be free R-modules with bases B and C, respectively.

(a)   The function HomR(M,N) → MC×B( Rop) f ⟼ fCB is an isomorphism of abelian groups.
(b)   EndR(M) → MB×B( Rop) f ⟼ fCB is a ring isomorphism.

HW: Discuss the difficulties in trying to make the map in Proposition ??? (a) into an R-module homomorphism, or into an Rop-module homomorphism.

HW: Let M be a free R-module. Let B and C be bases of M. Let P be the change of basis matrix from B to C and let Q be the change of basis matrix from C to B. Show that Q=P-1.

(a)   Let f:M→N be an R-module homomorphism. Let B1 and B2 be bases of M and let P be the change of basis matrix from B2 to B1. Let C1 and C2 be bases of N and let Q be the change of basis matrix from C1 to C2. Then fC2B2 = Q fC1B1 P.
(b)   Let f:M→M be an R-module homomorphism. Let B1 and B2 be bases of M and let P be the change of basis matrix from B2 to B1. Then fB2 = P-1 fB1 P.

Proofs

Let M and N be free R-modules with bases B and C, respectively.

(a)   The function HomR(M,N) → MC×B( Rop) f ⟼ fCB is an isomorphism of abelian groups.
(b)   EndR(M) → MB×B( Rop) f ⟼ fCB is a ring isomorphism.

Proof.

In fact we shall show that if M,N and P are free modules with bases B,C and D, respectively, and if f∈HomR (M,N) and g∈HomR (N,P) then (fg) DB = fDC gCB . f1( f2(b)) = ∑c∈C (f2) cb f1(c) = ∑c∈C ∑d∈D (f2) cb (f1) dc d = ∑d∈D ( ∑c∈C (f2) cb (f1) dc ) d. So (f1f2) db = ( ∑c∈C (f2) cb (f1) dc ) op = (f1) dc (f2) cb = ( (f1) (f2) ) db .

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

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