1.2. Base s fo r Vecto r Space s 11

set i s calle d a Hamel basis. W e wil l no t us e th e notio n o f a Hame l

basis i n thi s text ; w e simpl y includ e thi s fac t fo r studen t interest. )

If {xi}i

=1

i s a basi s fo r V , the n th e coefficient s ci,C2,...,Cf c

in th e linea r expansio n o f an y vecto r x mus t b e unique . I f not ,

then ther e woul d b e tw o way s o f writin g x a s a linea r combina -

tion o f #1, X2,..., Xfe . Bu t i f ^

i=

iCiXi = x — Yli=i^i xi wit h a t

least on e inde x valu e io fo r whic h Ci

0

^ c^

0

, the n w e woul d hav e

0 = x — x = ^2(ci — di)Xi an d no t al l coefficients woul d b e zero . Thi s

would contradic t th e linea r independenc e o f {xi}^ =1.

The followin g lemm a give s a genera l principl e whic h i s usefu l i n

constructing an d otherwis e workin g wit h bases .

Lemma 1.15 . If {#i , X2, •.., Xk} is a linearly independent subset of

a vector space V, and if Xk+i is a vector in V which is not contained

in span{xi,X2,.. . ,x^}

7

then {xi,X2,.. . ,Xfc?#fc+i} is a linearly inde-

pendent set.

Proof. Suppos e ci , C2,..., Ck+i are scalar s suc h tha t c\X\ + C2X2 +

\-Ck+iXk+i = 0 . Suppose , b y way o f contradiction, tha t c^+ i ^ 0 .

Then

k

Zfc+1 = ^[-(Ck+l^C^Xi

1 = 1

and s o Xk+i G span{#i , #2,...,£&}, a contradiction t o the hypothesis .

Hence w e mus t conclud e tha t c^+ i = 0 , whic h the n implie s c\X\ +

C2X2 H + CkXk — 0 . Sinc e {#1, #2? • • •

xk]

i s linearl y independent ,

this implie s tha t c i = C 2 = • • • = c & = 0 . Henc e {#i , X2,. •. , %k+i}

must b e linearl y independent . •

Proposition 1.16 . Let V be a nontrivial vector space which has a

spanning set {xi}!?

=1

. Then there is a subset of {xi}%

=1

which is a

basis for V.

Proof. W e wil l divid e th e se t {xi}*-

=1

int o tw o sets , whic h w e wil l

call good and bad. If x\ 7 ^ 0, then w e label x\ a s good and i f it i s zero,

we label i t a s bad. For eac h i 2 , if xi £ spanjrci,... , £;_i} , the n w e