Page 154 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Thus h is 0 on the subspace spanned by N and – and that subspace is V. We conclude that
f
h 0, i.e. that g cf .

Theorem 4: Let , ,...,g f 1 f r be linear functionals on a vector space V with respective null space
N ,N ,...,N Then g is a linear combination of f ,..., f if and only if N contains the intersection
.
1 r 1 r
N  ... N .

1 r
Proof: If g c f ... c f and f i 0 for each i, then clearly g 0. Therefore, N contains
1 1
r r

N  ... N .
1 r
We shall prove the converse (the ‘if’ half of the theorem) by induction on the number r. The
preceding lemma handles the case r = 1. Suppose we know the result for r = k – 1, and let f 1 ,..., f r

be linear functionals with null spaces N 1 ,...,N such that N  ... N is contained in N, the pull
k
k
1
',
space of g. Let g f ' ,..., f ' be the restrictions of , ,...,g f f 1 to the subspace N . Then
1 k 1 1 k k
g ', f ' ,..., f ' are linear functionals on the vector space N . Furthermore, if is a vector in N and
1 k 1 k k

f i ' 0, i 1,...,k 1, then is in N  ... N and so 'g 0. By the induction hypothesis
1
k
(the case r k 1), there are scalars c such that
i
' g c f ' ... c f ' .
1 1 k 1 k 1
Now let
k 1
h g c . f ...(10)
i i
i 1
Then h is a linear functional on V and (10) tells us that h 0 for every in N . By the preceding
k
leema, h is a scalar multiple of f . If h c f then
,
k k k
k
g c . f
i i
i 1
Self Assessment

1. Let n be a positive integer and F a field. Let W be the set of all vectors x 1 ,...,x n

in F n such that x ... x 0.
1 n
(a) Prove that W 0 consists of all linear functionals f of the form

n
f x ,...,x c x .
1 n j
j 1

148 LOVELY PROFESSIONAL UNIVERSITY