Page 191 - Advanced engineering mathematics

6.4 The Vector Space R n 171

V , V , ..., V k
1
2
S
G , G , ..., G t
1
2

S
A : G , V , V , ..., V k
1
1
2
3

S
A : G , G , V , ..., V k
1
3
2
2

S
A : G , G j–1 , ..., G , A j+1 , ..., A t
j
j
1

FIGURE 6.15 The sets A 1 ,A 2 , ··· formed
in the proof of Theorem 6.4.

Now adjoin G 2 to this list of vectors to form
G 2 ,G 1 ,V 2 ,··· ,V k .
This set spans S and is linearly dependent because G 2 is a linear combination of the other vectors.
Arguing as we did for A 1 ,some V j is a linear combination of the other vectors in this list. Again
for notational ease, suppose this is V 2 . Deleting this vector from the list therefore yields a set of
vectors that still spans S. Denote this set A 2 :
A 2 : G 2 ,G 1 ,V 3 ,··· ,V k .
The vectors in A 2 span S and are linearly dependent. We can continue this process of replacing,
one by one, the vectors in V 1 ,··· ,V k with vectors in G 1 ,··· ,G t . Figure 6.15 illustrates this
interchange of vectors between the basis to the spanning set that we have been carrying out.
There are two possibilities for this process to end.
First, this process may exhaust the basis vectors G 1 ,··· ,G t with some vectors V j remaining.
Since we delete a V j from the list exactly when we adjoin some G i , this would imply that t ≤ k.
The other possibility is that at some stage we have removed all of the V j s, and still have

some G i s left (so we would have t > k). At this stage, we would have the list
A k : G k ,G k−1 ,··· ,G 1 .

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October 14, 2010 14:21 THM/NEIL Page-171 27410_06_ch06_p145-186