Span.
Let \(S=\{v_1,v_2,\ldots,v_k\}\) be a subset of a vector space \(V\text{.}\) Define
\begin{equation*}
\operatorname{span}(S)=\{x_{1}v_{1}+x_{2}v_{2}+\ldots+x_{k}v_{k}|x_{i}\in \mathbb{R}\}
\end{equation*}
Section 5.3 Span
Example 5.3.1.
Let \(S=\left\{\left[\begin{array}{r}
1 \\
1 \\
\end{array}\right],\left[\begin{array}{r}
0 \\
1
\end{array}\right]\right\}\text{.}\)
\begin{equation*}
\operatorname{span}(S)=\{x_{1}v_{1}+x_{2}v_{2}|x_{i}\in \mathbb{R}\}=\mathbb{R}^{2}
\end{equation*}
Activity 5.3.1.
Show that the set \(S\) below spans \(\mathbb{R}^{4}\text{.}\)
\begin{equation*}
S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
-2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
2
\end{array}\right], \left[\begin{array}{r}
1 \\
1 \\
1 \\
1
\end{array}\right]\right\}.
\end{equation*}
That is, for any \(v=\left[\begin{array}{r}
a \\
b \\
c \\
d
\end{array}\right]\text{,}\) the vector equation
\begin{equation*}
v = x_{1}v_{1}+x_{2}v_{2}+x_{3}v_{3}+x_{4}v_{4}+x_{5}v_{5}
\end{equation*}
is consistent(has a solution).
Question: Is \(S\) a basis of \(\mathbb{R}^{4}?\)
Activity 5.3.2.
Let
\begin{equation*}
S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
-2
\end{array}\right]\right\}.
\end{equation*}
Suppose that \(v=\left[\begin{array}{r}
a \\
b \\
c \\
d
\end{array}\right]\) is in \(\operatorname{span}(S)\text{.}\) Find a relation of \(a,b,c,d\text{.}\)
