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Spanning1

# Spanning1 - Spanning1.nb Alternative Basis and Spanning...

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Alternative Basis and Spanning Spaces Derivation of an alternative set of spanning vectors As demonstrated in the MatLab program class1.m, the vector (14,39,54)' can be written as a linear combination of vectors (1,9,12)' and (4,7,10)'. However, the two spanning vectors are somewhat unappealing. For example, a graph of the space spanned by the two vectors is somewhat difficult to draw. Hence, we are interested in deriving two vectors which span the same space with somewhat more desirable characteristics. As a first step in this process, let us assume that we want to derive a vector (x,y,z)' as a linear combination of (1,9,12)' and (4,7,10)'. Specifically, for a given (x,y,z)' we want to derive that a and b such that a(1,9,12)'+b(4,7,10)'=(x,y,z)'. First note that a+4b=x. Solving for this relationship in terms of a we get: step1 = Solve @ a + 4b == x,a D {{a -> -4 b + x}} This is what is known as a mathematica rule. We can use this rule in the second step which is to solve for b using the information in 9a+7b=y, or step2 = Solve @ 9a + 7b == y .step1, b D 9 x - y {{b -> -------}} ------ 29 The "/." denotes "such that" or "substitute". It instructs Mathematica to substitute the rule from step 1 into the expression.

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