Linear Algebra, Theory And Applications, 2012a

204 VECTOR SPACES AND FIELDS
Differentiate the above equation n − 1 times yielding the equations
⎛
⎞
a 1 f 1 + a 2 f 2 + ···+ a n f n =0
a 1 f 1 ′ + a 2 f 2 ′ + ···+ a n f n ′ =0
⎜
⎟
⎝
.
⎠
a 1 f (n−1)
1 + a 2 f (n−1)
2 + ···+ a n f n (n−1) =0
Now plug in t. Then the above yields
⎛
⎞ ⎛ ⎞ ⎛
f 1 (t) f 2 (t) ··· f n (t) a 1
f 1 ′ (t) f 2 ′ (t) ··· f n ′ (t)
a 2
⎜
⎝
.
.
⎟ ⎜ . ⎟
. ⎠ ⎝
⎠ = ⎜
⎝
f (n−1)
1 (t) f (n−1)
2 (t) ··· f n
(n−1) (t) a n
Since the determinant of the matrix on the left is assumed to be nonzero, it follows this
matrix has an inverse and so the only solution to the above system of equations is to have
each a k =0.
Here is a useful lemma.
Lemma 8.2.10 Suppose v /∈ span (u 1 , ··· , u k ) and {u 1 , ··· , u k } is linearly independent.
Then {u 1 , ··· , u k , v} is also linearly independent.
Proof: Suppose ∑ k
i=1 c iu i + dv =0. It is required to verify that each c i = 0 and that
d = 0. But if d ≠ 0, then you can solve for v as a linear combination of the vectors,
{u 1 , ··· , u k },
k∑ ( ci
)
v = − u i
d
i=1
contrary to assumption. Therefore, d =0. But then ∑ k
i=1 c iu i = 0 and the linear independence
of {u 1 , ··· , u k } implies each c i = 0 also.
Given a spanning set, you can delete vectors till you end up with a basis. Given a linearly
independent set, you can add vectors till you get a basis. This is what the following theorem
is about, weeding and planting.
Theorem 8.2.11 If V =span(u 1 , ··· , u n ) then some subset of {u 1 , ··· , u n } is a basis for
V. Also, if {u 1 , ··· , u k }⊆V is linearly independent and the vector space is finite dimensional,
then the set, {u 1 , ··· , u k }, can be enlarged to obtain a basis of V.
Proof: Let
S = {E ⊆{u 1 , ··· , u n } such that span (E) =V }.
For E ∈ S, let |E| denote the number of elements of E. Let
m ≡ min{|E| such that E ∈ S}.
Thus there exist vectors
{v 1 , ··· , v m }⊆{u 1 , ··· , u n }
such that
span (v 1 , ··· , v m )=V
and m is as small as possible for this to happen. If this set is linearly independent, it follows
it is a basis for V and the theorem is proved. On the other hand, if the set is not linearly
independent, then there exist scalars
c 1 , ··· ,c m
0
0
.
0
⎞
⎟
⎠