14 1. Linea r Algebr a Revie w
is not a linear combination of the vectors {x\,x
2l
... ,x
n
-i}. Then
{x\,x2l... ,x
n
-i,z} is also a basis for V '.
Proof. Sinc e {xi}f
=1
i s a basi s fo r V and z G V, we can writ e
z = cixi + c 2x2 H h cn x
n
for som e set o f scalars ci , C2,..., cn . Sinc e 2 ^ span^}™!"/ , th e valu e
of c
n
canno t b e zero . Hence ,
xn
= C "
1^
~
C
l ^ l ~
c
2 # 2 C
n
_ i X
n
_ i ) .
This show s #
n
6 span{xi,X2,.. . , x
n
_ i , 2:}. Therefore , an y vecto r
which i s a linea r combinatio n o f {£;}™
=1
ca n b e writte n a s a linea r
combination o f th e vector s {x\,x
2
,... , £n _i,2:}, whic h mean s tha t
{x\,X2, , #
n
_ i , 2} is als o a spannin g se t fo r V .
Referencing Lemm a 1.15 , z £ s p a n ^ i } ^ 1 implie s tha t th e se t
{xi, x2, •, xn-i,z} i s linearly independent , an d therefor e a basis fo r
V.
Theorem 1.21. Let V be a finite-dimensional vector space. Suppose
that V has a basis containing n elements.
(i) Any other basis of V also has n elements.
(ii) Any linearly independent set ofn vectors in V is also a basis
forV.
Proof, (i ) Suppos e {x{}2
=
i an d {yi}i
=1
ar e bot h base s fo r V . With -
out los s o f generality , w e ca n assum e n k. B y th e definitio n o f
linear independence , {#2 , •, xn} doe s no t spa n al l o f V . I n particu -
lar, ther e exist s a t leas t on e o f the vectors , sa y y^ , i n {2/1,2/2, •, Vk]
such that yi
x
£ span{a^ } ™=2- B y Lemm a 1.20, {y^,x 2,... ,x n}
1S a
l
s o
a basi s fo r V .
We repea t thi s argumen t fo r x 2. Sinc e {y^ , # 3 , . . . , x n} doe s no t
span V , we ca n selec t yi
2
£ spanjy^ , £ 3 , . . . , x n}. Not e tha t thi s cer -
tainly require s i
2
7 ^ i\. Lemm a 1.20 agai n prove s {yi 2, y^ , £ 3 , . . . , x n}
is a basi s fo r V .
We procee d inductively . Assum e fo r som e p , 2 p n , w e hav e
selected distinc t {ii,i
2
, •, iP} fro m amon g {1,2,... , k} s o tha t th e
set {yi
x
, yi
2
,..., yi
v
, x
p
+\,..., x
n
} i s a basis fo r V . A t leas t on e of th e
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