## 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 }\text{,} \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\text{,} \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}\text{.} \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 }\text{,} \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\text{.} \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\text{.} \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}\text{.} \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.