Linear Algebra, Theory And Applications, 2012a

58 MATRICES AND LINEAR TRANSFORMATIONS
Proof 2:
that
Define span{y 1 , ··· , y s }≡V, it follows there exist scalars c 1 , ··· ,c s such
s∑
x 1 = c i y i . (2.23)
i=1
Not all of these scalars can equal zero because if this were the case, it would follow that
x
∑ 1 = 0 and so {x 1 , ··· , x r } would not be linearly independent. Indeed, if x 1 = 0, 1x 1 +
r
i=2 0x i = x 1 = 0 and so there would exist a nontrivial linear combination of the vectors
{x 1 , ··· , x r } which equals zero.
Say c k ≠0. Then solve ((2.23)) for y 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 } by
{z 1 , ··· , z s−1 }≡{y 1 , ··· , y k−1 , y k+1 , ··· , y s }
Therefore, span {x 1 , z 1 , ··· , z s−1 } = V because if v ∈ V, there exist constants c 1 , ··· ,c s
such that
∑s−1
v = c i z i + c s y k .
i=1
Now replace the y k in the above with a linear combination of the vectors, {x 1 , z 1 , ··· , z s−1 }
to obtain v ∈ span {x 1 , z 1 , ··· , z s−1 } . The vector y k , in the list {y 1 , ··· , y s } , has now been
replaced with the vector x 1 and the resulting modified list of vectors has the same span as
the original list of vectors, {y 1 , ··· , y s } .
Now suppose that r>sand that span {x 1 , ··· , x l , z 1 , ··· , z p } = V where the vectors,
z 1 , ··· , z p are each taken from the set, {y 1 , ··· , y s } and l + p = s. This has now been done
for l = 1 above. Then since r>s,it follows that l ≤ s