Let $S = \{v_1, v_2, \ldots, v_n\}$ be a set of vectors in a vector space $V$. If there exist scalars $\alpha_1, \alpha_2 \ldots \alpha_n \in F$ such that not all of the $\alpha_i$'s are zero and \begin{equation*}\alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = {\mathbf 0 },\end{equation*} then $S$ is said to be linearly dependent. If the set $S$ is not linearly dependent, then it is said to be linearly independent. More specifically, $S$ is a linearly independent set if \begin{equation*}\alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = {\mathbf 0 }\end{equation*} implies that \begin{equation*}\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0\end{equation*} for any set of scalars $\{ \alpha_1, \alpha_2 \ldots \alpha_n \}$.

The definition of linear dependence makes more sense if we consider the following proposition.

The following proposition is a consequence of the fact that any system of homogeneous linear equations with more unknowns than equations will have a nontrivial solution. We leave the details of the proof for the end-of-chapter exercises.

A set $\{ e_1, e_2, \ldots, e_n \}$ of vectors in a vector space $V$ is called a basis for $V$ if $\{ e_1, e_2, \ldots, e_n \}$ is a linearly independent set that spans $V$.

##### Example20.12

The vectors $e_1 = (1, 0, 0)$, $e_2 = (0, 1, 0)$, and $e_3 =(0, 0, 1)$ form a basis for ${\mathbb R}^3$. The set certainly spans ${\mathbb R}^3$, since any arbitrary vector $(x_1, x_2, x_3)$ in ${\mathbb R}^3$ can be written as $x_1 e_1 + x_2 e_2 + x_3 e_3$. Also, none of the vectors $e_1, e_2, e_3$ can be written as a linear combination of the other two; hence, they are linearly independent. The vectors $e_1, e_2, e_3$ are not the only basis of ${\mathbb R}^3$: the set $\{ (3, 2, 1), (3, 2, 0), (1, 1, 1) \}$ is also a basis for ${\mathbb R}^3$.

##### Example20.13

Let ${\mathbb Q}( \sqrt{2}\, ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}$. The sets $\{1, \sqrt{2}\, \}$ and $\{1 + \sqrt{2}, 1 - \sqrt{2}\, \}$ are both bases of ${\mathbb Q}( \sqrt{2}\, )$.

From the last two examples it should be clear that a given vector space has several bases. In fact, there are an infinite number of bases for both of these examples. \emph{In general, there is no unique basis for a vector space}. However, every basis of ${\mathbb R}^3$ consists of exactly three vectors, and every basis of ${\mathbb Q}(\sqrt{2}\, )$ consists of exactly two vectors. This is a consequence of the next proposition.

If $\{ e_1, e_2, \ldots, e_n \}$ is a basis for a vector space $V$, then we say that the dimension of $V$ is $n$ and we write $\dim V =n$. We will leave the proof of the following theorem as an exercise.