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Section 5.2 Linear independent
Linear independent.
Let \(S=\{v_1,v_2,\ldots,v_k\}\) be a subset of a vector space \(V\text{.}\) The set \(S\) is called linear independent if
\begin{equation*}
x_{1}v_{1}+x_{2}v_{2}+\ldots+x_{k}v_{k}|x_{i}=\mathbf{0}
\end{equation*}
has only trivial solution. That is, \(x_1=x_2=\ldots=x_n=0\) is the only solution. Otherwise, the set is called linear dependent.
Activity 5.2.1 .
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], \left[\begin{array}{r}
1 \\
1 \\
1 \\
1
\end{array}\right]\right\}.
\end{equation*}
Show that \(S\) is linear dependent.
Activity 5.2.2 .
Find a maximum linear independent subset \(W\) of \(S\text{.}\) That is, \(W\) is linear independent, but for any \(W\subsetneq U\) (proper included), \(U\) is linear dependent.
Activity 5.2.3 .
Write the fifth vector \(v_5\) as a linear combination of the other four vectors \(v_1, v_2,v_3\) and \(v_4\text{.}\)
Activity 5.2.4 .
Find a maximal linear independent subset of the set
\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*}
