SOLUTION: Let {a1,...,ap} be a set of vectors from R^n. If a1=x2a2+...+xpap where x2...xp are scalars, then span(a1...ap) = span(a2...ap). I need assistance showing that this statement is
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Question 107124: Let {a1,...,ap} be a set of vectors from R^n. If a1=x2a2+...+xpap where x2...xp are scalars, then span(a1...ap) = span(a2...ap). I need assistance showing that this statement is true.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Remember the span of a set of vectors is the set of all linear combinations of every vector in the set. In other words...
for some scalar 
But since 
, we can replace 
with the expanded notation and simplify.
The final result will be 
So here is one way you could go about solving this proof:
 +c_2\vec{a}_2+c_3\vec{a}_3+\cdots+c_p\vec{a}_p \mbox{...Replace \vec{a}_1 with x_1\vec{a}_2+\cdots+x_p\vec{a}_p} \\ &=&\left(c_1x_1\vec{a}_2+\cdots+c_1x_p\vec{a}_p\right)+c_2\vec{a}_2+c_3\vec{a}_3+\cdots+c_p\vec{a}_p \mbox{...Distribute} \\&=&\left(c_2+c_1x_1\right)\vec{a}_2+\left(c_3+c_1x_2\right)\vec{a}_3+\cdots+\left(c_{p}+c_1x_{p-1}\right)\vec{a}_p \mbox{...Collect the scalar terms} \\&=&s_1\vec{a}_2+s_2\vec{a}_3+\cdots+s_p\vec{a}_p \mbox{...Replace each sum c_{i}+c_{1}x_i with s_i (which are also scalar)} \\&=&\textrm{span}\left\{\vec{a}_2,\ldots,\vec{a}_p\right\} \mbox{...Now replace the linear combination with the span of the set of vectors}\end{eqnarray})
So this shows that 
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