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Number of indie vectors ##\leq ## cardinality of spanning set
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[QUOTE="Mark44, post: 6065045, member: 147785"] Actually what you wrote is "Suppose that n < m..." which is correct for a proof by contradiction if your spanning set is ##\{u_1, u_2, \dots, u_n \}## Yes, your proof seems more complicated than it needs to be, in part because of the use of lots more letters than are needed. For example, instead of having coefficients of ##a_i, b_i,## and ##\eta_i##, just use two subscripts. So ##v_i = \sum_{j = 1}^n a_{i~j}u_j##. If you write the equations for ##v_1, v_2, \dots, v_n## in the form of a matrix product, you have ##\vec v = A \vec u##, where A is a matrix with m rows and n columns.I think you can assume without loss of generality that the u vectors are a minimal spanning set; i.e., a basis for the space. Since you need to prove the given statement for every spanning list, it doesn't matter if you prove it for the smallest spanning set. [/QUOTE]
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Number of indie vectors ##\leq ## cardinality of spanning set
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