A First Course in Linear Algebra, 2017a

482 Vector Spaces
Say c k ≠ 0. Then solve 9.1 for⃗y k and obtain
⎧
⎫
s-1 vectors here
⎨
⃗y k ∈ span
⎩ ⃗x { }} { ⎬
1, ⃗y 1 ,···,⃗y k−1 ,⃗y k+1 ,···,⃗y s
⎭ .
Define {⃗z 1 ,···,⃗z s−1 } to be
{⃗z 1 ,···,⃗z s−1 } = {⃗y 1 ,···,⃗y k−1 ,⃗y k+1 ,···,⃗y s }
Now we can write
⃗y k ∈ span{⃗x 1 ,⃗z 1 ,···,⃗z s−1 }
Therefore, span{⃗x 1 ,⃗z 1 ,···,⃗z s−1 } = V . To see this, suppose ⃗v ∈ V. Then there exist constants c 1 ,···,c s
such that
s−1
⃗v = ∑ c i ⃗z i + c s ⃗y k .
i=1
Replace this⃗y k with a linear combination of the vectors {⃗x 1 ,⃗z 1 ,···,⃗z s−1 } to obtain⃗v ∈ span{⃗x 1 ,⃗z 1 ,···,⃗z s−1 }.
The vector⃗y k ,inthelist{⃗y 1 ,···,⃗y s }, has now been replaced with the vector⃗x 1 and the resulting modified
list of vectors has the same span as the original list of vectors, {⃗y 1 ,···,⃗y s }.
We are now ready to move on to the proof. Suppose that r > s and that
span { ⃗x 1 ,···,⃗x l ,⃗z 1 ,···,⃗z p
}
= V
where the process established above has continued. In other words, the vectors⃗z 1 ,···,⃗z p are each taken
from the set {⃗y 1 ,···,⃗y s } and l + p = s. This was done for l = 1 above. Then since r > s, it follows that
l ≤ s < r and so l + 1 ≤ r. Therefore, ⃗x l+1 is a vector not in the list, {⃗x 1 ,···,⃗x l } and since
there exist scalars, c i and d j such that
span { ⃗x 1 ,···,⃗x l ,⃗z 1 ,···,⃗z p
}
= V
⃗x l+1 =
l
∑
i=1
c i ⃗x i +
p
∑ d j ⃗z j . (9.2)
j=1
Not all the d j can equal zero because if this were so, it would follow that {⃗x 1 ,···,⃗x r } would be a linearly
dependent set because one of the vectors would equal a linear combination of the others. Therefore, 9.2
can be solved for one of the⃗z i ,say⃗z k ,intermsof⃗x l+1 and the other⃗z i and just as in the above argument,
replace that⃗z i with⃗x l+1 to obtain
Continue this way, eventually obtaining
⎧
⎫
p-1 vectors here
⎨
span
⎩ ⃗x { }} { ⎬
1,···⃗x l ,⃗x l+1 , ⃗z 1 ,···⃗z k−1 ,⃗z k+1 ,···,⃗z p
⎭ = V
span{⃗x 1 ,···,⃗x s } = V.