page 19.
Fig. 7.
(0,0,0)
(0,-1,0)
(0,1,0)
(0,4,0)
Fig. 8
(5,-4-0)
(0,1,0)
(1, 0, 0)
(6, 0, 0)
The vector (0,1,0) determines a space which includes any 3-vector whose first and last components are zero, but does not include any other vector. Exactly the same space is determined by the two 3-vectors (0,1,0) & (0, 58,0). (0,1,0) & (0,0,0) also determine this space. See Fig 7; this figure, and those which follow it, are really views of 3-dimensional figures.
(1,0,0) & (0,1,0) determine a space which includes any vector whose last component is zero. For example, (5,-4,0) is included, since
5(1,0,0) -4(0,1,0)-(5,-4,0) =
= (0, 0, 0)
and (6, 0, 0) also, since
6(1,0,0) - (6,0,0) = (0,0,0)
But no vector with last component different from zero is included; because no sum of multiples of the determining vectors will have its last component different from zero, and therefore any such sum added to any multiple of the vector in question will give a sum with a non-zero final component, hence cannot give (0,0,0) as required. See Fig. 8.
Now observe that all four of (1,0,0), (0,1,0), (5,-4,0), and (6,0,0), taken together, determine the same space as do the first two taken by themselves. the same space is determined by any three of them; and by any two of them, except (1,0,0) & (6,0,0). The space determined by this pair would not, for example, include (0,1,0).
The space determined by (1,0,0), (0,1,0), & (0,0,1) includes all 3-vectors. (1,3,0), (2,-1,0), & (1,3,4) will serve just as well. So only three vectors are necessary to determine this space. Yet we can easily find sets of more than three vectors which do not

page 19.
Fig. 7.
(0,0,0)
(0,-1,0)
(0,1,0)
(0,4,0)
Fig. 8
(5,-4-0)
(0,1,0)
(1, 0, 0)
(6, 0, 0)
The vector (0,1,0) determines a space which includes any 3-vector whose first and last components are zero, but does not include any other vector. Exactly the same space is determined by the two 3-vectors (0,1,0) & (0, 58,0). (0,1,0) & (0,0,0) also determine this space. See Fig 7; this figure, and those which follow it, are really views of 3-dimensional figures.
(1,0,0) & (0,1,0) determine a space which includes any vector whose last component is zero. For example, (5,-4,0) is included, since
5(1,0,0) -4(0,1,0)-(5,-4,0) =
= (0, 0, 0)
and (6, 0, 0) also, since
6(1,0,0) - (6,0,0) = (0,0,0)
But no vector with last component different from zero is included; because no sum of multiples of the determining vectors will have its last component different from zero, and therefore any such sum added to any multiple of the vector in question will give a sum with a non-zero final component, hence cannot give (0,0,0) as required. See Fig. 8.
Now observe that all four of (1,0,0), (0,1,0), (5,-4,0), and (6,0,0), taken together, determine the same space as do the first two taken by themselves. the same space is determined by any three of them; and by any two of them, except (1,0,0) & (6,0,0). The space determined by this pair would not, for example, include (0,1,0).
The space determined by (1,0,0), (0,1,0), & (0,0,1) includes all 3-vectors. (1,3,0), (2,-1,0), & (1,3,4) will serve just as well. So only three vectors are necessary to determine this space. Yet we can easily find sets of more than three vectors which do not