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Matrix Theory: Linear Algebra and Beyonds

Peter Tan

Section 4.2 Linear independent

An important type of problem in linear algebra involves writing one vector \(\mathbf{x}\) as the sum of scalar multiples of other vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots\text{,}\) and \(\mathbf{v}_{n}\text{.}\) That is, for scalars \(c_{1}\text{,}\) \(c_{2}, \ldots, c_{n}\)
\begin{equation*} \mathbf{x}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{n} \mathbf{v}_{n} \end{equation*}
The vector \(\mathbf{x}\) is called a linear combination of the vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) with weight \(c_1,c_2,\ldots,c_n\text{.}\)

Linear Independent Set.

A set of vectors \(\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\}\) is called linear independent if the linear combination
\begin{equation*} x_{1} \mathbf{v}_{1}+x_{2} \mathbf{v}_{2}+\cdots+x_{n} \mathbf{v}_{n}=\mathbf{0} \end{equation*}
has only trivial solution, that is, \(x_1=x_2=\ldots =x_n=0\) is the only solution.

Activity 4.2.1.

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 4.2.2.

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 4.2.3.

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 4.2.4.

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*}
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