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
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