12 1. Linea r Algebr a Revie w
label Xi a s good, and otherwis e w e labe l i t a s bad. Le t B b e th e se t
of al l th e good elements. W e clai m tha t B i s a basi s fo r V .
B i s nonempty becaus e not al l the xi ar e zero and the first nonzer o
Xi will b e good. We must first verif y tha t B i s a linearl y independen t
set. Le t B = {61, 62,. •., b^} and assum e w e have scalar s ci , C2, •, Q
such that Ci6i+C2&2 H Vcgbi = 0 . The n ci&i+C2&2 H hQ_i^_ i =
—cibi, an d b y th e constructio n o f th e se t B, bi i s no t a linea r com -
bination o f th e previou s elements , whic h therefor e implie s tha t Q
must b e zero . Bu t w e ca n repea t thi s argument . W e no w kno w tha t
cibi + c 2b2 H h ci-\bt-\ = 0 , s o that ci& i + c 262 H h
Q _2 ^ _
2
=
Q _ I 6 ^ _I an d agai n thi s implie s tha t Q _ I = 0 . Repeatin g thi s wil l
prove tha t al l the coefficient s mus t b e zero . Therefore , B i s a linearl y
independent set .
We leav e i t a s Exercis e 3 to sho w tha t th e spa n o f th e element s
of B i s al l o f V .
Example 1.17 . I t i s eas y t o se e tha t th e proo f o f th e abov e propo -
sition i s actuall y a n algorith m fo r choosin g a subse t tha t i s a basis .
Such a proo f i s calle d constructive . Fo r a specifi c example , le t S b e
the spa n o f the polynomial s 1 + x , 1 -f- 2x, x, 1 + x
2
i n P2 . I n th e first
step of the proof, w e see 1-fx i s labelled good. Next, sinc e 1 + x, l + 2x
is linearl y independent , th e elemen t 1 + 2x i s als o labelle d good. We
see b y inspectio n tha t x ca n b e writte n a s x = ( 1 + 2x) (1 + x) ,
so x i s labelle d bad. Lastly , 1 + x
2
canno t b e writte n a s a linea r
combination o f th e first thre e 1 + x , 1 + 2x , x, becaus e i t i s o f degre e
2 an d th e first thre e ar e o f degre e 1, s o i t i s labelle d good. Thu s
B = { 1 + x , 1 + 2x , 1 + x 2 } i s ou r basi s fo r 5 .
Note: I t shoul d als o be noted tha t th e algorith m i n the proof doe s
not yiel d th e onl y subse t tha t i s a basis. I n particular , { 1 + 2x , x, 1 +
x2} i s als o readil y see n t o b e a basis .
Proposition 1.18 . Every linearly independent set in V can be ex-
tended to a basis ofV.
Proof. Le t {xi,X2,... , x^} b e a linearl y independen t subse t o f V .
We mus t sho w tha t eithe r {#i}f =1i s alread y a basi s fo r V , o r els e
there exis t additiona l vector s {x^
+
i,.. . ,x p} i n V suc h tha t {xi}^
=1
is a basi s fo r V .
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