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Mathematics
Linear and Abstract Algebra
Same vector space for arbitrary independent vectors?
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[QUOTE="jbunniii, post: 5485918, member: 81553"] Let ##R## denote the vector space spanned by ##k_1,k_2,\ldots,k_n##. Since each ##k_j## is in ##Q## and ##Q## is a vector space, every linear combination of the ##k_j##'s is in ##Q##, hence ##R \subseteq Q##. Now suppose that ##q## is an arbitrary element of ##Q##. Since ##k_1,k_2,\ldots,k_n## is a basis for ##Q##, we can write ##q = a_1 k_1 + a_2 k_2 + \cdots + a_n k_n## for some scalars ##a_1,a_2,\ldots a_n##. Therefore ##q## is contained in the subspace spanned by ##k_1,k_2,\ldots,k_n##, which is ##R##. This shows that ##Q \subseteq R##. Since we have shown both containments ##R \subseteq Q## and ##Q \subseteq R##, we conclude that ##Q = R##. [/QUOTE]
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Forums
Mathematics
Linear and Abstract Algebra
Same vector space for arbitrary independent vectors?
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