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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 Fall '08
 Moss
 Linear Algebra, Vectors, Vector Space, Vector Motors, 9a+7b=y, class1.m

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