This chapter defines vocabulary used to give a precise way to describe linear equations.
Definition 2.4.1(Linear combination). If $\vec v_1 , \dots, \vec v_k$ is a collection of vectors in $\R^n$, then a linear combination of the $\vec v_i$ is a vector $\vec w$ of the form
$$ \vec w = \sum_{i=1}^k a_i\vec v_i \text{\space\space\space\space\space for any scalars }a_i $$
Example. The vector $\begin{bmatrix}3\\4\end{bmatrix}$ is a linear combination of $\vec e_i$ and $\vec e_2$:
$$ \begin{bmatrix}3\\4\end{bmatrix} = 3\begin{bmatrix}1\\0\end{bmatrix} + 4\begin{bmatrix}0\\1\end{bmatrix}. $$
Linear independence is a way to talk about uniqueness of solutions to linear equations; span is a way to talk about the existence of solutions.
Definition 2.4.2(Linear independence). A set of vectors $\vec v_1,\dots \vec v_k$ is linearly independent if the only solution to
$$ a_1\vec v_1 + a_2\vec v_2 + \dots + a_k\vec v_k = \vec 0 \text{ \space\space is \space \space }a_1=a_2=\dots=a_k = 0. $$
An equivalent definition is that vectors $\vec v_1, \dots, \vec v_k \in \R^n$ are linearly independent if and only if a vector $\vec w \in \R^n$ can be written as a linear combination of those vectors in at most one way:
$$ \sum_{i=0}^k y_i \vec v_i = \sum_{i=0}^k x_i \vec v_i \text{ \space \space implies\space\space } y_1 =x_1, \dots,y_k = x_k. $$
It is also equivalent to say that vectors are linearly independent if none of $\vec v_i$ is a linear combination of the others. If the columns of $[T]$ are linearly independent, then the transformation $T$ is injective.
Geometrically, one vector is linearly independent if it isn't the zero vector. Two vectors are linearly independent if they don't both lie on the same line. Three vectors are linearly independent if they don't all lie in the same plane.
Example. The vectors $\vec e_1, \vec e_2 \in \R^2$ are linearly independent. There is only one way to write $\begin{bmatrix}3\\4\end{bmatrix}$ in terms of $\vec e_i$ and $\vec e_2$:
$$ 3\begin{bmatrix}1\\0\end{bmatrix} + 4\begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}3\\4\end{bmatrix}. $$
However, $\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}3\\2\end{bmatrix}$ are not linearly independent because
$$ \begin{bmatrix}3\\2\end{bmatrix} + 2\begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}3\\4\end{bmatrix}. $$