# Determine Linearly Independent or Linearly Dependent. Express as a Linear Combination

## Problem 277

Determine whether the following set of vectors is linearly independent or linearly dependent. If the set is linearly dependent, express one vector in the set as a linear combination of the others.
$\left\{\, \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix}\, \right\}.$

## Solution.

Consider the linear combination
$x_1\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}+x_2 \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}+x_3\begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}+x_4 \begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix}=\mathbf{0} \tag{*}$ with variables $x_1, x_2, x_3, x_4$.
We determine whether there is $(x_1, x_2, x_3, x_4)\neq (0,0,0,0)$ satisfying the linear combination (*).

The linear combination (*) is written as the matrix equation
$\begin{bmatrix} 1 & 1 & -1 & -2 \\ 0 &2 & -2 & -2 \\ -1 & 3 & 0 & 7 \\ 0 & 4 & 1 & 11 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}=\mathbf{0}.$

To find the solutions of the equation, we apply the Gauss-Jordan elimination.
The augmented matrix $[A\mid \mathbf{0}]$ can be reduced by elementary row operations as follows.
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 &2 & -2 & -2 & 0 \\
-1 & 3 & 0 & 7& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right] \xrightarrow{\substack{R_3+R_1 \\ \frac{1}{2}R_2}}
\left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 4 & -1 & 5& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right]\10pt] \xrightarrow{\substack{R_1-R_2\\ R_3-4R_2\\ R_4-4R_2}} \left[\begin{array}{rrrr|r} 1 & 0 & 0 & -1 & 0\\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 3 & 9& 0 \\ 0 & 0 & 5 & 15 & 0 \\ \end{array} \right] \xrightarrow{\substack{\frac{1}{3}R_3\\ \frac{1}{5}R_4}} \left[\begin{array}{rrrr|r} 1 & 0 & 0 & -1 & 0\\ 0 & 1 & -1 & -1 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ \end{array} \right]\\[10pt] \xrightarrow{\substack{R_2+R_3\\R_4-R_3}} \left[\begin{array}{rrrr|r} 1 & 0 & 0 & -1 & 0\\ 0 & 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]. \end{align*} Thus, the general solution is given by \begin{align*} x_1&=x_4\\ x_2&=-2x_4\\ x_3&=-3x_4, \end{align*} where x_4 is a free variable. If we let x_4=1, then we have a nonzero solution x_1=1, x_2=-2, x_3=-3, x_4=1. Thus the set is linearly dependent. Substituting these values into (*), we have \[\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}-2 \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}-3\begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}+ \begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix}=\mathbf{0}. Solving this for the last vector we obtain the linear combination
$\begin{bmatrix} -2 \\ -2 \\ 7 \\ 11 \end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix}+2 \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}+3\begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix}.$

Let $V$ be the vector space of all $2\times 2$ real matrices and let $P_3$ be the vector space of...