Linear dependence and independence.

We consider a vector space V and a set I (this will be a set of indices, generally of the form $I= \{ 1,2, \dots , n \}$, but not necessarily).

Definition 5.2.1   A family

$$F= \{ \overrightarrow{u_i} ; \; i \in I \}$$

of vectors in V is linearly independent if, for any scalars $\alpha_i, \; i \in I$, we have:

$$\underset{i \in I}{\sum}\alpha_i \overrightarrow{u_i} = \overrightarrow{0}\Longrightarrow \forall i \in I, \alpha_i =0.$$

If I is a finite set, e.g. $I= \{ 1,2, \dots , n \}$, we can write Eq.  7 as follows:

 

$$\alpha_1 \overrightarrow{u_1} + \alpha_1 \overrightarrow{u_2}... ...{0}\Longrightarrow \alpha_1 = \alpha_2 = \dots = \alpha_n = 0.$$

Example 5.2.2   We prove that the vectors $\overrightarrow{u_1} = \begin{pmatrix}1 \\ 2 \end{pmatrix}$and $\overrightarrow{u_2} = \begin{pmatrix}2 \\ 3 \end{pmatrix}$are linearly independent in $\mathbb{R} ^2$.

Let $\alpha_1, \alpha_2$be two scalars (i.e. here two real numbers). Then:

 

$$\alpha_1 \overrightarrow{u_1} + \alpha_1 \overrightarrow{u_2}... ...}\alpha_1 + 2 \alpha_2 \\ 2 \alpha_1 + 3 \alpha_2 \end{pmatrix}$$

Therefore:

$$\alpha_1 \overrightarrow{u_1} + \alpha_1 \overrightarrow{u_2}... ...pha_2 = 0 \end{cases}\Longleftrightarrow \alpha_1=\alpha_2 =0.$$

Proposition 5.2.3   Let F be a family of vectors in V. If F is linearly independent, then any non empty subset of F is linearly independent too.

Definition 5.2.4   If a family F of vectors in V is not linearly independent, it is linearly dependent.

This means that there exists a family of scalars $\alpha_i$such that at least one of them is not zero and $\underset{i \in I}{\sum}\alpha_i \overrightarrow{u_i} = \overrightarrow{0}$.

Example 5.2.5  

1.

The family of one vector $\{ \overrightarrow{0}\}$is linearly dependent, as $2 \overrightarrow{0} = \overrightarrow{0}$.

2.

Every family containing the zero vector is linearly dependent (take 1 as the coefficient of $\overrightarrow{0}$and zeros as the other coefficients).

3.

The vectors $\overrightarrow{u} =\begin{pmatrix}1 \\ 2 \\ 3 \end{pmatrix}$and $\overrightarrow{v} =\begin{pmatrix}2 \\ 4 \\ 6 \end{pmatrix}$are linearly dependent in $\mathbb{R} ^3$, as $\overrightarrow{v} = 2 \overrightarrow{u}$.