1.2. Base s fo r Vector Space s

13

Since V is finite-dimensional, i t has a spanning set {s±, 52,..., s r}.

If w e tak e th e union wit h ou r linearly independen t set , we still hav e

a spannin g se t {x\, ..., X£, s i , . . ., s

r

}. W e apply th e good/bad argu -

ment fro m th e proof o f Proposition 1.1 6 to this list , extractin g a basis

for V . Sinc e we considered th e elements i n order, w e see that eac h of

the element s Xi, i = 1,2,... ,£ are good, an d therefore ar e present i n

the ne w basis. Thus , th e new basis extend s {xi}f =1. •

Lemma 1.19. Let E and F be subspaces ofV with EnF = {0 } and

span(F, F) = V . Suppose that £ = {e\,...,e k} is a basis for E and

T — {/1,..., fm} is a basis for F. Then £ U T is a basis for V.

Proof. Le t

B = £ UT = { e i , e2 , . . . , e f c , / i , / 2 , . . . , /

m

} .

Then span(K ) contain s bot h th e set s E = spanjei , e2, ..., e k} an d

F = s p a n { / i , /

2

, . . . , /

m

} , s o span(# ) 2 spem(E,F) = V . Thu s

span(#) = V.

Thus w e need onl y prove that B is linearly independent . Fo r this,

suppose {ci,.. . , Ck, d\, . . . , dm } are scalars suc h tha t

ciei H h ckek + di/i H h im/m = 0.

We hav e

ciei H h cfcefc = - ( d i / i H h dm /

m

) .

Let z = cie i + • • • -I- Cfeefe. The n z £ E becaus e E i s close d

under takin g o f linear combination s o f its members. Similarly , z =

- ( d i / i + • • • + 4 / m ) an d s o z G F. Therefor e 2 : e E n F . Bu t

E C\ F = {0} which implie s tha t z = cie i + • • • + c kek = 0 . So , since

{e±,..., e/c } are linearly independent , w e must hav e

ci = c

2

= • • • = Cf c = 0.

Similarly, b y using the fact tha t {/1,..., f m} i s a basis for F, we

can sho w tha t d\ = d

2

= • • • = d

m

= 0 as well. Thi s prove s tha t B is

linearly independent . Sinc e a linearl y independen t spannin g se t is a

basis, w e have prove n tha t B is a basis fo r V. L I

Lemma 1.20. Let V be a finite-dimensional vector space, and sup-

pose {xi}f

=1

is a basis for V . Suppose z is any vector in V which