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Mathematics 17 Online
OpenStudy (anonymous):

Could someone help me out with this: Explain why S is not a basis for R^3. S={(6,4,1),(3,-5,1),(8,13,6),(0,6,9)}

OpenStudy (anonymous):

A set of vectors B={v1, v2, v3,....,vN} is basis of a vector space E if it satisfies the following conditions: 1) All of the vectors in B are linearly independent 2) They span the vector space E For R^n, we need n linearly independent vectors to span that vector space, as soon as we have one more vector, then it will be a linear combination of the other three and it won't satisfy condition 1 anymore, so in your example: S={(6,4,1),(3,-5,1),(8,13,6),(0,6,9)} is not a basis, because the first three vectors are already linearly independent and can span R^3, so a basis for R^3 could be S={(6,4,1),(3,-5,1),(8,13,6)} For more information: http://en.wikibooks.org/wiki/Linear_Algebra/Basis

OpenStudy (anonymous):

"then it will be a linear combination of the other three" I meant of the other "n" (Was thinking about R^3 in my head)

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