Question 107124
Remember the span of a set of vectors is the set of all linear combinations of every vector in the set. In other words...


*[Tex \LARGE \textrm{span}\left\{\vec{a}_1,\ldots,\vec{a}_p\right\}=c_1\vec{a}_1+c_2\vec{a}_2+c_3\vec{a}_3+\cdots+c_p\vec{a}_p] for some scalar *[Tex \Large \textrm{c_i's}]


But since *[Tex \Large \vec{a}_1=x_1\vec{a}_2+\cdots+x_p\vec{a}_p], we can replace *[Tex \Large \vec{a}_1] with the expanded notation and simplify. 

The final result will be *[Tex \Large  \textrm{span}\left\{\vec{a}_2,\ldots,\vec{a}_p\right\}]




So here is one way you could go about solving this proof:



*[Tex \LARGE \begin{eqnarray} \textrm{span}\left\{\vec{a}_1,\ldots,\vec{a}_p\right\}&=&c_1\vec{a}_1+c_2\vec{a}_2+c_3\vec{a}_3+\cdots+c_p\vec{a}_p \\ &=&c_1\left(x_1\vec{a}_2+\cdots+x_p\vec{a}_p\right)   +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 *[Tex \LARGE  \textrm{span}\left\{\vec{a}_1,\ldots,\vec{a}_p\right\}=\textrm{span}\left\{\vec{a}_2,\ldots,\vec{a}_p\right\}]