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Finding a Basis for a set of vectors
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[QUOTE="HallsofIvy, post: 4414004, member: 637751"] What, exactly is your answer? Once you have row reduced, the only 4 dimensional vectors we can get from that matrix are (1, 0, 0, 0) and (-3, 0, 0, 1). That clearly cannot be a basis for this space because the second and third components would always be 0 which is not the case here. And what do you [b]mean[/b] by "Is a−3b,b a suitable basis for this?". a and b, and so a- 3b, are [b]numbers[/b], not vectors and so cannot be a "basis". Here is how I would do it. H is the set of all vectors of the form (a- 3b, b- a, a, b). (a- 3b, b- a, a, b)= (a, -a, a, 0)+ (-3b, b, 0, b)= a(1, -1, 1, 0)+ b(-3, 1, 0, 1). And now a basis, and the dimension, are obvious. [/QUOTE]
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Finding a Basis for a set of vectors
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