Page 190 - Advanced engineering mathematics

170 CHAPTER 6 Vectors and Vector Spaces

The vectors < 1,0,−1 > and < 0,1,−1 > span M. These vectors are also linearly independent,
since neither is a scalar multiple of the other. These vectors therefore form a basis for M.Both
vectors are needed to specify all vectors in S.
There is nothing unique about a basis for a subspace. For example,
< 2,0,−2 > and < 0,2,−2 >
also form a basis for M,asdo
< 1,0,−1 > and < 0,4,−4 >.

We will need some additional facts about bases. The first is that any spanning set for a
n
subspace S of R contains a basis.

THEOREM 6.3

n
Let S be a subspace of R that is spanned by F 1 ,··· ,F k . Then a basis for S can be formed from
some or all of the vectors F 1 ,··· ,F k .
We will sketch the idea of a proof. Suppose we have a set of vectors F 1 ,··· ,F k that span a
n
n
given subspace S of R (perhaps all of R ). If these vectors are also linearly independent, then
they form a basis for S.
If these spanning vectors are linearly dependent, then at least one F j is a linear combination
of others. Remove F j , and the remaining set (one vector smaller) spans S. If these vectors are
linearly dependent, then one is a linear combination of the others, and we can remove this one to
obtain a still smaller spanning set for S. Continuing in this way, we eventually reach a spanning
set for S that is linearly independent, with no one vector a linear combination of the others.
A spanning set for S is a basis if the vectors are linearly independent. If we are willing to
forego linear independence, however, then we can adjoin as many vectors from S as we like to
this spanning set and still have a spanning set for S. This suggests that a basis is limited in size,
while a spanning set is not. The next theorem is a careful statement of this idea, and says that
any spanning set for S has at least as many vectors in it as any basis for S. It is in this sense that
a basis for a subspace is a “smallest possible” spanning set for this subspace.

THEOREM 6.4

Suppose V 1 ,··· ,V k span a subspace S of R , and let G 1 ,··· ,G t be a basis for S. Then t ≤k.
n
Proof Since V 1 ,··· ,V k span S and G 1 is in S, then

G 1 = c 1 V 1 + ··· + c k V k
for some numbers c 1 ,··· ,c k . Then

G 1 − c 1 V 1 − ··· − c k V k = O.
If each c j = 0 then G 1 = O, impossible since G 1 is a basis vector. Therefore some c j is nonzero.
As a notational convenience, suppose c 1 = 0. Then
1 c 2 c k
V 1 =− G 1 − V 2 − ··· − V k .
c 1 c 1 c 1
Further, G 1 ,V 2 ,··· ,V k span S. Denote this set of vectors as A 1 :
A 1 : G 1 ,V 2 ,··· ,V k .

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October 14, 2010 14:21 THM/NEIL Page-170 27410_06_ch06_p145-186