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also a basis. If (-1, 0, 0) were replaced by (1, 0, 0), it would also form the canonical basis of .

Another spanning set for the same space is given by {(1, 2, 3), (0, 1, 2), (-1, 1/2, 3), (1, 1, 1)}, but this set is not a basis, because it is linearly dependent.

The set } is not a spanning set of , since its span is the space of all vectors in whose last component is zero. That space is also

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