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
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