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## Section20.3Linear Independence

Let $S = \{v_1, v_2, \ldots, v_n\}$ be a set of vectors in a vector space $V\text{.}$ 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 $\{ \alpha_1, \alpha_2 \ldots \alpha_n \}\text{.}$

If

\begin{equation*} v = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = \beta_1 v_1 + \beta_2 v_2 + \cdots + \beta_n v_n, \end{equation*}

then

\begin{equation*} (\alpha_1 - \beta_1) v_1 + (\alpha_2 - \beta_2) v_2 + \cdots + (\alpha_n - \beta_n) v_n = {\mathbf 0}. \end{equation*}

Since $v_1, \ldots, v_n$ are linearly independent, $\alpha_i - \beta_i = 0$ for $i = 1, \ldots, n\text{.}$

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

Suppose that $\{ v_1, v_2, \dots, v_n \}$ is a set of linearly dependent vectors. Then there exist scalars $\alpha_1, \ldots, \alpha_n$ such that

\begin{equation*} \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = {\mathbf 0 }, \end{equation*}

with at least one of the $\alpha_i$'s not equal to zero. Suppose that $\alpha_k \neq 0\text{.}$ Then

\begin{equation*} v_k = - \frac{\alpha_1}{\alpha_k} v_1 - \cdots - \frac{\alpha_{k - 1}}{\alpha_k} v_{k-1} - \frac{\alpha_{k + 1}}{\alpha_k} v_{k + 1} - \cdots - \frac{\alpha_n}{\alpha_k} v_n. \end{equation*}

Conversely, suppose that

\begin{equation*} v_k = \beta_1 v_1 + \cdots + \beta_{k - 1} v_{k - 1} + \beta_{k + 1} v_{k + 1} + \cdots + \beta_n v_n. \end{equation*}

Then

\begin{equation*} \beta_1 v_1 + \cdots + \beta_{k - 1} v_{k - 1} - v_k + \beta_{k + 1} v_{k + 1} + \cdots + \beta_n v_n = {\mathbf 0}. \end{equation*}

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 $V\text{.}$

###### Example20.12

The vectors $e_1 = (1, 0, 0)\text{,}$ $e_2 = (0, 1, 0)\text{,}$ and $e_3 =(0, 0, 1)$ form a basis for ${\mathbb R}^3\text{.}$ The set certainly spans ${\mathbb R}^3\text{,}$ since any arbitrary vector $(x_1, x_2, x_3)$ in ${\mathbb R}^3$ can be written as $x_1 e_1 + x_2 e_2 + x_3 e_3\text{.}$ 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 ${\mathbb R}^3\text{:}$ the set $\{ (3, 2, 1), (3, 2, 0), (1, 1, 1) \}$ is also a basis for ${\mathbb R}^3\text{.}$

###### Example20.13

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

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. 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.

Since $\{ e_1, e_2, \ldots, e_m \}$ is a basis, it is a linearly independent set. By Proposition 20.11, $n \leq m\text{.}$ Similarly, $\{ f_1, f_2, \ldots, f_n \}$ is a linearly independent set, and the last proposition implies that $m \leq n\text{.}$ Consequently, $m = n\text{.}$

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