Page 187 - Advanced engineering mathematics

6.4 The Vector Space R n 167

EXAMPLE 6.12
3
The vectors i,j,k span all of R . But so do
3i,2j,−k.
The vectors
F 1 = i + k,F 2 = i + j,F 3 = j + k
3
also span R . To see this, let V = ai + bj + ck be any 3− vector. Then
a + b − c a − b + c 3a − b − c
V = F 1 + F 2 + F 3 .
2 2 2
In this example these spanning sets all have three vectors in them. But a spanning set for R 3
may have more than three vectors. For example, the vectors
√
i,j,k,−4i, 97k
3
also span R because we can write any 3-vector as
√
< x, y, z >= xi + yj + zk + 0(−4i) + 0( 97kj).
3
This set of five vectors spans R , but does so inefficiently in the sense that two of the vectors are
not needed to have a spanning set for R .
3
n
More generally, if vectors V 1 ,··· ,V k span a subspace S of R , we can adjoin any number
m of other vectors of S to these k vectors, and the resulting m + k vectors will still span S.
n
Going the other way, if V 1 ,··· ,V k span a subspace S of R ,it may be possible to remove
some vectors from this set and have the smaller set of vectors still span S. This occurs when
V 1 ,··· ,V k contain redundant information and not all of them are needed to completely specify S.
The efficiency of a spanning set (the idea of whether it contains unnecessary vectors) is addressed
through the notions of linear dependence and independence.

n
A (finite) set of vectors in R is called linearly dependent if one of the vectors is a linear
combination of the others. Otherwise, if no one of the vectors is a linear combination of
the others, then these vectors are linearly independent.

EXAMPLE 6.13
The vectors

F =< 3,−1,0,4 >,G =< 3,−2,−1,10 >,H =< 6,−1,1,2 >
are linearly dependent in R because G = 3F − H. The two vectors F and G are linearly
4
independent, because neither is a scalar multiple of the other.
Think of linear independence in terms of information. Suppose F 1 ,··· ,F k are vectors in R .
n
If these vectors are linearly dependent, then at least one of them, say F k for convenience, is a
linear combination of F 1 ,··· ,F k−1 . This means that any linear combination of these k vectors
is really a linear combination of just the first k − 1 of them. Put another way, the subspace S
spanned by all k of these vectors is the same as the subspace space spanned by just the first k − 1
of them, and F k is not needed in specifying S.

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October 14, 2010 14:21 THM/NEIL Page-167 27410_06_ch06_p145-186