## Linear Dependence, Bases, and Dimension

Last update: 13 August 2013

## Linear Dependence, Bases, and Dimension

Let $R$ be a ring and let $M$ be an $R\text{-module.}$ Let $S$ be a subset of $M\text{.}$

• The submodule generated by $S$ is the smallest submodule of $M$ containing $S\text{;}$ i.e. $\text{span}\left(S\right)$ is a submodule of $M$ such that  (a) $S\subseteq \text{span}\left(S\right),$ (b) If $N$ is a submodule of $M$ such that $S\subseteq N$ then $\text{span}\left(S\right)\subseteq N\text{.}$
• A module $M$ is finitely generated if there is a finite subset $S$ of $M$ such that $\text{span}\left(S\right)=M\text{.}$
• A vector space $V$ is finite dimensional if there is a finite subset $S$ of $V$ such that $\text{span}\left(S\right)=V\text{.}$
• A linear combination of elements of $S$ is an element $v\in M$ of the form $v=∑s∈Srss,$ where ${r}_{s}\in R$ are such that ${r}_{s}=0$ for all but a finite number of $s\in S\text{.}$
• The set $S$ is linearly independent if it satisfies the condition, $If∑s∈S rss=0,then rs=0,for all s∈S.$
• The set $S$ is linearly dependent if it is not linearly independent.
• A basis of $M$ is a set $B\subseteq M$ such that  a) $\text{span}\left(B\right)=V,$ and b) $B$ is linearly independent.
• A free module is an $R\text{-module}$ $M$ that has a basis.

HW: Show that $\text{span}\left(S\right)$ is the set of linear combinations of elements in $S\text{.}$

HW: Let $R$ be a commutative ring. Give an example of a finitely generated $R\text{-module}$ $M$ that does not have a basis.

Let $V$ be a vector space over a field $𝔽\text{.}$ Then $V$ has a basis.

Example. Let $R$ be the ring of infinite matrices with rows and columns indexed by ${ℤ}_{>0},$ entries in $ℂ,$ and only a finite number of nonzero entries in each row,

$R= { ( a11a12a13⋯ a21a22a23⋯ a31a32⋱ ⋮⋱ ) | aij∈ℂ } ,and letM=R$

be the regular $R\text{-module.}$ Let

$b0= ( 1000⋯ 0100⋯ 0010⋯ 0001⋯ ⋮⋱ ) , b1= ( 10000⋯ 00100⋯ 00001⋯ 00000⋯ ⋮⋮⋱ ) , b2= ( 010000⋯ 000100⋯ 000001⋯ 000000⋯ ⋮0⋮⋱ ) .$

Then

$B1={b0}and B2={b1,b2}$

are both bases of $M\text{.}$

 (a) Let $R$ be a ring and let $M$ be a free $R\text{-module}$ with an infinite basis. Any two bases of $M$ have the same number of elements. (b) Let $V$ be a vector space over a field $𝔽\text{.}$ Any two bases of $V$ have the same number of elements. (c) Let $R$ be a commutative ring and let $M$ be a free $R\text{-module.}$ Any two bases of $M$ have the same number of elements.

Let $R$ be a commutative ring and let $M$ be a free $R\text{-module.}$ Let $V$ be a vector space over a field $F\text{.}$

• The rank of $M$ is the number of elements in a basis of $M\text{.}$
• The dimension of $V$ is the number of elements in a basis of $V\text{.}$

Let $B$ be a set and let $R$ be a ring.

• The free module on the set $B$ is the $R\text{-module}$ $RB= { ∑b∈Brbb | rb∈R, and all but a finite number of the rb are equal to 0 } ,$ with addition given by $(∑b∈Brbb)+ (∑b∈Bsbb)= ∑b∈B(rb+sb)b,$ and $R\text{-action}$ given by $r∑b∈Brbb= ∑b∈B(rrb) b.$

HW: Let $B$ be a set. Show that ${R}^{B}$ is a free $R\text{-module}$ with basis $B\text{.}$

Let $M$ be an $R\text{-module}$ and let $B$ be a subset of $M\text{.}$

 (a) There is an $R\text{-module}$ homomorphism $\varphi :{R}^{B}\to M$ such that $\varphi \left(b\right)=b\text{.}$ (b) $M$ is generated by $B$ if and only if $M$ is a quotient of ${R}^{B}\text{.}$ (c) $M$ is a free module with basis $B$ if and only if $M\cong {R}^{B}\text{.}$

Let $B$ and $C$ be sets. Let $R$ be a ring.

• A $C×B$ matrix with entries in $R$ is a collection $F=\left({F}_{cb}\right)$ of elements ${F}_{cb}\in R$ indexed by the elements of $C×B$ and such that for each $b\in B$ all but a finite number of the ${F}_{cb}$ are $0\text{.}$
• ${M}_{C×B}\left(R\right)=\left\{C×B \text{matrices with entries in} R\right\}\text{.}$
• The sum of two matrices ${F}_{1},{F}_{2}\in {M}_{C×B}\left(R\right)$ is the matrix ${F}_{1}+{F}_{2}$ given by $(F1+F2)cb =(F1)cb+ (F2)cb, for all b∈B,c∈C.$
• The product of matrices ${F}_{1}\in {M}_{D×C}\left(R\right)$ and ${F}_{2}\in {M}_{C×B}\left(R\right)$ is the matrix ${F}_{1}{F}_{2}\in {M}_{D×B}\left(R\right)$ given by $(F1F2)db= ∑c∈C (F1)dc (F2)cb, for all b∈B and d ∈D.$

Let $M$ and $N$ be free $R\text{-modules}$ with bases $B$ and $C,$ respectively. Let $f:M\to N$ be a homomorphism.

• The matrix of $f:M\to N$ with respect to $B$ and $C$ is the matrix ${}^{\circ }f\in {M}_{C×B}\left({R}^{\text{op}}\right)$ given by $(∘f)cb =(fcb)op where fcb∈R are given byf(b)= ∑c∈Cfcb c,$ for all $b\in B\text{.}$

Let $M$ and $N$ be free $R\text{-modules}$ with bases $B$ and $C,$ respectively.

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

HW: Discuss the difficulties in trying to make the map in Proposition ??? (a) into an $R\text{-modules}$ homomorphism, or into an ${R}^{\text{op}}\text{-module}$ isomorphism.

Let $M$ be a free module and let ${B}_{o}$ and ${B}_{n}$ be bases of $M\text{.}$

• The change of basis matrix is the matrix ${T}_{o\to n}$ given by $b∈∑c∈Bn (To→n)cb c,for all b∈Bo.$

Let $M$ be a free module with bases ${B}_{o}$ and ${B}_{n}$ and let $N$ be a free module with bases ${C}_{o}$ and ${C}_{n}\text{.}$ Then

$∘fn= To→n (∘fo) Tn→o$

Let $V$ be a vector space. Let $L,S$ be subsets of $V$ such that $L\subseteq S\subseteq$ such that $L$ is linearly independent and $\text{span}\left(S\right)=V\text{.}$ Then there exists a basis $B$ of $V$ such that $L\subseteq B\subseteq S\text{.}$

Proof.

Let $𝒞$ be the set of linearly independent subsets $L\subseteq C\subseteq S,$ partially ordered by inclusion. Let $B$ be a maximal element of $𝒞\text{.}$

To show: $B$ is a basis of $V\text{.}$

To show:

 (a) $\text{span}\left(B\right)=V\text{.}$ (b) $B$ is linearly independent.

(a) Suppose that $\text{span}\left(B\right)\ne V\text{.}$ Let $s\in S$ such that $s\notin \text{span}\left(B\right)\text{.}$ Then $L\subseteq \left(B\cup \left\{s\right\}\right)\subseteq S,$ and if there is a linear combination

$ξss+∑b∈B ξbb=0,with ξs≠0,then s=ξs-1 (∑b∈Bξbb) ∈span(B).$

Since $s\notin \text{span}\left(B\right),$ ${\xi }_{s}=0,$ but then since $B$ is linearly independent ${\xi }_{b}=0$ for all $b\in B\text{.}$ So $B\cup \left\{s\right\}$ is linearly independent. This is a contradiction to the maximality of $B\text{.}$ So $\text{span}\left(B\right)=V\text{.}$

(b) $B$ is linearly independent by definition.

$\square$

Let $R$ be a ring and let $M$ be a free $R\text{-module.}$ If $M$ has an infinite basis, or $R$ is a field, or $R$ is a commutative ring, then any two bases of $M$ have the same number of elements.

 Proof. (a) Let $B$ be an infinite basis of $M\text{.}$ Let $C$ be a basis of $M\text{.}$ Define ${r}_{cb}\in R$ for $c\in C,$ $b\in B,$ by $b=∑c∈Crcbc and letSb= {c∈C | rcb≠0},$ for each $b\in B\text{.}$ Then each ${S}_{b}$ is a finite subset of $C$ and $C\subseteq {\bigcup }_{b\in B}{S}_{b}$ since $C$ is a minimal spanning set. So $\text{Card}\left(C\right)\le {\aleph }_{0}\text{Card}\left(B\right)=\text{Card}\left(B\right)\text{.}$ The same argument shows that $\text{Card}\left(B\right)\le {\aleph }_{0}\text{Card}\left(C\right)\text{.}$ So $\text{Card}\left(B\right)=\text{Card}\left(C\right)\text{.}$ (b) By part (a) we may assume that $V$ has a finite basis $B\text{.}$ Let $C$ be another basis of $B\text{.}$ Let $b\in B\text{.}$ Then there is an element $c\in C$ such that $c\notin \text{span}\left(B-\left\{b\right\}\right)\text{.}$ Then ${B}_{1}=\left(B-\left\{b\right\}\right)\cup \left\{c\right\}$ is a basis with the same cardinality as $B\text{.}$ By repeating this process, we can create a basis $B\prime$ of $V$ which contains all the elements of $C$ and which has the same cardinality as $C\text{.}$ Thus $\text{Card}\left(B\right)\ge \text{Card}\left(C\right)\text{.}$ A similar argument with $C$ in place of $B$ give that $\text{Card}\left(B\right)\le \text{Card}\left(C\right)\text{.}$ (c) The case when $\text{Card}\left(B\right)$ is infinite is covered by part (a). We will show that if $B$ is a basis of $M$ then $\stackrel{‾}{B}=\left\{b+𝔪M | b\in B\right\}$ is a basis of $M/𝔪M$ as a vector space over $R/𝔪\text{.}$ Then the result will follow from part (b). Let $∑ir‾ib‾i =0$ where ${\stackrel{‾}{r}}_{i}\in R/𝔪$ and ${\stackrel{‾}{b}}_{i}\in M/𝔪M\text{.}$ Then $0=∑ir‾i b‾i=∑i (ri+𝔪) (bi+𝔪M)= (∑iribi) +𝔪M.$ So ${\sum }_{i}{r}_{i}{b}_{i}\in 𝔪M\text{.}$ So there are elements ${\ell }_{j}\in 𝔪,$ ${m}_{j}\in M$ and elements ${c}_{{j}_{k}}\in R$ such that $∑iribi= ℓ1m1+⋯ ℓsms=∑j ℓj∑kcjk bk=∑j,k ℓjmjkbk.$ Since $B$ is linearly independent $rk=∑jℓj cjk∈𝔪, for each k.$ So ${\stackrel{‾}{r}}_{k}=0$ for each $k\text{.}$ So $\stackrel{‾}{B}$ is linearly independent. $\square$

Let $M$ and $N$ be free $R\text{-modules}$ with bases $B$ and $C,$ respectively.

 (a) The function $HomR(M,N) ⟶ MB×C(Rop) f⟼∘f$ is a bijection. (b) The function $EndR(M) ⟶ MB×B(Rop) f⟼∘f$ is a ring isomorphism.

 Proof. (b) In fact we shall show that if $M,$ $N$ and $P$ are free modules with bases $B,$ $C$ and $D$ respectively and if ${f}_{1}\in {\text{Hom}}_{R}\left(M,N\right)$ and ${f}_{2}\in {\text{Hom}}_{R}\left(N,P\right)$ then ${}^{\circ }\left({f}_{1}{f}_{2}\right)={{}^{\circ }f}_{1}{{}^{\circ }f}_{2}\text{.}$ $f1(f2(b))= ∑c∈C(f2)cb cf1(c)=∑c∈C ∑d∈D(f2)cb (f1)dcd= ∑d∈D ( ∑c∈C(f2)cb (f1)dc ) d.$ So $(∘(f1f2)) db = ( ∑c∈C (f2)cb (f1)dc ) op =∑c∈C (f1)dcop (f2)cbop= ∑c∈C (∘f1)dc (∘f2)cb =(∘f1∘f2)db$ $\square$

HW: Discuss the difficulties in trying to make the map in Proposition ??? (a) into an $R\text{-modules}$ homomorphism, or into an ${R}^{\text{op}}\text{-module}$ isomorphism.

Let $M$ be a free module and let $B=\left\{{b}_{i}\right\}$ and $B\prime =\left\{{b}_{j}^{\prime }\right\}$ be bases of $M\text{.}$

• The change of basis matrix $P\left(B\prime ,B\right)$ is the matrix given by $bi=∑jP (B,B′)ij bj′, for all bi∈B.$

HW: Let $M$ be an $R\text{-module}$ and let $S\subseteq M$ be a subset of $M\text{.}$ Show that $\text{span}\left(S\right)$ exists and is unique by showing that $\text{span}\left(S\right)$ is the intersection of all of the submodules that contain $S\text{.}$

HW: Let M be an $R\text{-module.}$ Show that

$span(S)= { ∑m∈Srmm | rm=0 for all but a finite number of m∈S } .$

HW: Let $M$ be an $R\text{-module.}$ Show that a subset $S\subseteq M$ is linearly independent if and only if $S$ satisfies the following property:
If ${r}_{m}\in R$ such that ${r}_{m}=0$ for all but a finite number of $m\in M$ and ${\sum }_{m\in S}{r}_{m}m=0$ then ${r}_{m}=0$ for all $m\in M\text{.}$

HW: Let $V$ be a vector space over a field $F\text{.}$ Show that if $v\in V$ and $\left\{{v}_{1},{v}_{2},\cdots ,{v}_{n}\right\}$ is a basis of $V$ then there exist unique ${c}_{i}\in F$ such that $v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\cdots +{c}_{n}{v}_{n}\text{.}$

HW: Give an example of a finitely generated module over a commutative ring 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 finite bases with different numbers of elements.