1.2. Base s fo r Vecto r Space s
15
vectors yi i s not i n the spa n o f the se t {y
ix
, y
i2
,..., y
ip
, x
p + 2
, . . . , xn } ,
so w e ca n selec t i
p
+i £ {1, 2 , . . ., k} suc h tha t
Vip+1 $ span{y
il
, y*2 ,..., y
ip
, x
p + 2
, . . . , x
n
}.
In particular, w e see that thi s requires ip
+
i ^ {^i,...,i
p
}. B y Lemm a
1.20, th e se t {y
i]L
, y;2 ,... , y
ip+1
, xp
+ 2
, . . . , x
n
} i s a basi s fo r V .
We ca n repea t thi s algorith m unti l al l th e origina l Xi hav e bee n
replaced, an d w e hav e a basi s {yi x, yi 2,..., yi
n
} for V . I f k n, the n
there i s a j G {1,2,... , A;} such tha t j £ {zi , i2 , . . . , i n}- Bu t fo r
this j , yj V ; henc e i t mus t b e i n th e spa n o f th e basi s element s
{yii^yi2- - iVin}- Thi s contradict s th e assumptio n tha t {yi}f_
1
i s
linearly independent , s o we must hav e n = k.
(ii) Repea t th e abov e argument , replacin g th e basi s {yi}i =i wit h
a linearl y independen t se t o f n element s {yi}™
=1
. Th e inductio n ar -
gument prove s tha t {y^,yi 2,... ,yi n} i s a basi s fo r V . Sinc e th e se t
{ii, i2 , . . . , i n} i s a permutatio n o f {1,2,... , n}, w e hav e {yi}^
=1
i s
the sam e set , an d thu s i s a basi s fo r V .
Prom th e abov e theorem , ever y basi s o f a vecto r spac e V has th e
same cardinalit y (numbe r o f elements) .
Definition 1.22. Th e dimension o f a vector space V is the cardinalit y
of a basis , whic h w e denote b y di m V .
Example 1.23. Le t F b e eithe r R o r C . Then , a s w e hav e seen , F n
is a vecto r spac e unde r th e usua l additio n an d scala r multiplication .
One basi s fo r V is {ei , e2 , . . . , e
n
} wit h
ri]
0
0
L0.
,e
2
=
roi
I
0
.0.
» * * *
i^-n

r°i
0
0
. i .
This i s usually calle d th e standar d basi s fo r F
n
. W e then se e that th e
dimension o f F
n
i s n .
Lemma 1.24. Suppose V is a nontrivial vector space, let {xi}\
=1
be
any finite subset ofV, and let E be the linear span of {xi} l
i=1
. Then
dimE I
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