66 I The Expanding Universe of Numbers
for someα 1 ,...,αm ∈D.Thecoefficientsα 1 ,...,αmneed not be uniquely deter-
mined. Evidently a vectorvis a linear combination ofv 1 ,...,vm if it is a linear
combination of some proper subset, sincewe can add the remaining vectors with zero
coefficients.
IfSis any nonempty subset ofV, then the set〈S〉of all vectors inVwhich are
linear combinations of finitely many elements ofSis a subspace ofV, the subspace
‘spanned’ orgeneratedbyS. ClearlyS⊆〈S〉and〈S〉is contained in every subspace
ofVwhich containsS.
A finite subset{v 1 ,...,vm}ofVis said to belinearly dependent(overD)ifthere
existα 1 ,...,αm∈D, not all zero, such that
α 1 v 1 +···+αmvm= 0 ,and is said to belinearly independentotherwise.
For example, inR^3 the vectors
v 1 =( 1 , 0 , 1 ), v 2 =( 1 , 1 , 0 ), v 3 =( 1 , 1 / 2 , 1 / 2 )are linearly dependent, sincev 1 +v 2 − 2 v 3 =0. On the other hand, the vectors
e 1 =( 1 , 0 , 0 ), e 2 =( 0 , 1 , 0 ), e 3 =( 0 , 0 , 1 )are linearly independent, sinceα 1 e 1 +α 2 e 2 +α 3 e 3 =(α 1 ,α 2 ,α 3 ), and this is 0 only
ifα 1 =α 2 =α 3 =0.
In any vector spaceV,theset{v}containing the single vectorvis linearly indepen-
dent ifv=0 and linearly dependent ifv=0. Ifv 1 ,...,vmare linearly independent,
then any vectorv∈〈v 1 ,...,vm〉has a unique representation as a linear combination
ofv 1 ,...,vm, since if
α 1 v 1 +···+αmvm=β 1 v 1 +···+βmvm,then
(α 1 −β 1 )v 1 +···+(αm−βm)vm= 0and hence
α 1 −β 1 =···=αm−βm= 0.Evidently the vectorsv 1 ,...,vmare linearly dependent if some proper subset is
linearly dependent. Hence any nonempty subset of a linearly independent set is again
linearly independent.
A subsetSof a vector spaceVis said to be abasisforVifSis linearly indepen-
dent and〈S〉=V. In the previous example, the vectorse 1 ,e 2 ,e 3 are a basis forR^3 ,
since they are not only linearly independent but also generateR^3.
Any nontrivial finitely generated vector space has a basis. In fact if a vector space
Vis generated by a finite subsetT,thenVhas a basisB⊆T. Moreover, any linearly
independent subset ofVis also finite and its cardinality does not exceed that ofT.It
follows that any two bases contain the same number of elements.