Chapter 6 : Linear dependence

1. Linear combination

If a vector can be expressed as a linear combination of other vectors, they are said to be linearly dependent.

Example I

Figure 6.1 : vectors linearly dependent

We’d like to know if one vector in Figure 6.1 depends upon the others, as the following :

$$\begin{equation} \vec{v_1} = \lambda{_2} \vec{v_2} + \lambda{_3} \vec{v_3} \end{equation}$$

We have thus :

$$\begin{equation} \underbrace{ \begin{pmatrix} 8 \\ -4 \\ -3 \end{pmatrix}}_{\color{red}{\vec{v_1}}} = \lambda{_2} \underbrace{ \begin{pmatrix} 2 \\ -4 \\ 0 \end{pmatrix}}_{\color{orange}{\vec{v_2}}} + \lambda{_3} \underbrace{ \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}}_{\color{green}{\vec{v_3}}} \end{equation}$$

Let’s factorize the right hand-side of $(2)$ :

$$\begin{pmatrix} 8 \\ -4 \\ -3 \end{pmatrix} = \begin{pmatrix} 2 \lambda{_2} \\ -4 \lambda{_2} \\ 0 \lambda{_2} \end{pmatrix} + \begin{pmatrix} -2 \lambda{_3} \\ 0 \lambda{_3} \\ 1 \lambda{_3} \end{pmatrix} = \begin{pmatrix} 2 \lambda{_2} + -2 \lambda{_3} \\ -4 \lambda{_2} + 0 \lambda{_3} \\ 0 \lambda{_2} + 1 \lambda{_3} \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ -4 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \lambda{_2} \\ \lambda{_3} \end{pmatrix}$$

Let’s proceed with the matrix elimination as we have seen in the dedicated chapter :

$$\left( \begin{array}{ c c| c} 2 & -2 & 8 \\ -4 & 0 & -4 \\ 0 & 1 & -3 \end{array} \right) \overset{2R{_1}+R{_2}\rightarrow{R_2}}{\iff} \left( \begin{array}{ c c | c} 2 & -2 & 8 \\ 0 & -4 & 12 \\ 0 & 1 & -3 \end{array} \right) \overset{ 4R{_3}+R{_2}\rightarrow{R_3}}{\iff} \left( \begin{array}{c c | c} 2 & -2 & 8 \\ 0 & -4 & 12 \\ 0 & 0 & 0 \end{array} \right)$$

From the matrix in echelon form, we get :

$ -4\lambda_3 = 12 \iff$ $\lambda_3 = -3$

Let’s replace $\lambda_3$ in the first row :

$2\lambda_2 + -2 \cdot \underbrace{-3}_{\lambda_3} = 8 \iff$ $\lambda_2 = 1$

The solution is :

Therefore any vector $\vec{v_i} \in \{\vec{v_1}, \vec{v_2}, \vec{v_3}\}$ can be expressed as the linear combination of the others.

2. Generalization

$(1)$ is obviously equivalent to $(4)$ :

$$\begin{equation} \vec{0} = -\vec{v_1} + \lambda{_2} \vec{v_2} + \lambda{_3} \vec{v_3} \end{equation}$$

More generally, we have :

If the unique solution to $(5)$ is $\lambda{_1}=\lambda{_2}=\lambda{_3}=0$ then none of the vectors can be expressed as the linear combination of the other. Otherwise the vectors are linearly dependent.

Recapitulation