Page 192 - Advanced engineering mathematics
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172 CHAPTER 6 Vectors and Vector Spaces
But each time we form such a list by replacing a spanning set vector with a basis vector, we
obtain a new set of vectors that spans S. This would make G k+1 a linear combination of the first
k basis vectors, and then these basis vectors would be linearly dependent, a contradiction.
This proves that this possibility cannot occur, leaving the first possibility, and t ≤ k.
n
This theorem has a profound consequence—all bases for a given subspace of R have the
same number of vectors in them.
COROLLARY 6.1
n
Let G 1 ,··· ,G m and H 1 ,··· ,H k be bases for a subspace S of R . Then m = k.
Proof Each basis is a spanning set, so two applications of Theorem 6.4 gives us m ≤k and also
k ≤ m.
n
The number of vectors in a basis for a subspace S of R is called the dimension of S.For
n
3
example, R has dimension n, and the subspace of R in Example 6.17 has dimension 2.
n
Now suppose S is a k-dimensional subspace of R , and v 1 ,v 2 ,··· ,v k form a basis for S.If
X is in S, then there are numbers c 1 ,c 2 ,··· ,c k such that
k
X = c 1 v 1 + c 2 v 2 + ··· + c k v k = c j v k .
j=1
The numbers c 1 ,··· ,c k are called the coordinates of X with respect to this basis. These
coordinates are unique to X and to this basis.
For, if
X = d 1 v 1 + ··· + d k v k
then
k
X − X = O = (c 1 − d 1 )v 1 + ··· + (c k − d k )v k = (c j − d j )v j .
j=1
Since the vectors v 1 ,··· ,v k are linearly independent, each c j − d j = 0, and therefore each
c j = d j .
A nontrivial subspace of R has many bases, and each n-vector X has unique coordinates
n
with respect to each basis. However, on a practical level, some bases are more convenient to work
with in the sense that coordinates of vectors with respect to these bases are easier to determine.
4
To illustrate, let S be the subspace of R consisting of all vectors < x, y,0,0>, with x and y any
real numbers.
This is a two-dimensional subspace with e 1 =< 1,0,0,0 > and e 2 =< 0,1,0,0 > forming a
basis B 1 for S. The vectors
w 1 =< 2,−6,0,0 > and w 2 =< 2,4,0,0 >
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October 14, 2010 14:21 THM/NEIL Page-172 27410_06_ch06_p145-186