Let $v_1=\begin{bmatrix}1\\-2\\1\end{bmatrix}$, $v_2=\begin{bmatrix}2\\-1\\1\end{bmatrix}$, $v_3=\begin{bmatrix}-2\\-2\\0\end{bmatrix}$, $v_4=\begin{bmatrix}1\\-2\\-2\end{bmatrix}$
$S=\{v_1,v_2,v_3,v_4\}$
$T=\{v_1,v_2,v_4\}$
Show that the set of all linear combinations of vectors from $S$ is the same as the set of all linear combinations of vectors from $T$.
how to show that two stets of vectors have the same linear combinations?
$\quad$ Actually, the linear combination of vectors contained in T should be linear combination of that of S. What we want to prove is to get $ \vec{v}_3 $ could be wrote as a linear combination of {$\vec{v}_1, \vec{v}_2, \vec{v}_4$}. Because you have done this process. Therefore, we go ahead. We will use a example to illustrate the principle. Example: If $ \vec{v}_3 = \vec{v}_1 + \vec{v}_2 + \vec{v}_4$, then any vector expressed in term of {$\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4 $} could be expressed in term of {$\vec{v}_1, \vec{v}_2, \vec{v}_4 $}. Since $\vec{x} = x_1\vec{v}_1+x_2 \vec{v}_2+x_3 \vec{v}_3+x_4 \vec{v}_4 =x_1\vec{v}_1+x_2 \vec{v}_2+x_3( \vec{v}_1 + \vec{v}_2 + \vec{v}_4)+x_4 \vec{v}_4= (x_1+x_3)\vec{v}_1+(x_2+x_3) \vec{v}_2+(x_4+x_3) \vec{v}_4$.