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

Problems and solutions in Linear Algebra

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.
\[\left\{\, \begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}, \begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}\, \right\}.\]

 
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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
\[ \begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}
=
-\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}+2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}+3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}.\]


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