Section 20.3 Linear 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
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
implies that
for any set of scalars \(\{ \alpha_1, \alpha_2 \ldots \alpha_n \}\text{.}\)
Proposition 20.9.
Let \(\{ v_1, v_2, \ldots, v_n \}\) be a set of linearly independent vectors in a vector space. Suppose that
Then \(\alpha_1 = \beta_1, \alpha_2 = \beta_2, \ldots, \alpha_n = \beta_n\text{.}\)
Proof.
If
then
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.
Proposition 20.10.
A set \(\{ v_1, v_2, \dots, v_n \}\) of vectors in a vector space \(V\) is linearly dependent if and only if one of the \(v_i\)'s is a linear combination of the rest.
Proof.
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
with at least one of the \(\alpha_i\)'s not equal to zero. Suppose that \(\alpha_k \neq 0\text{.}\) Then
Conversely, suppose that
Then
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 endofchapter exercises.
Proposition 20.11.
Suppose that a vector space \(V\) is spanned by \(n\) vectors. If \(m \gt n\text{,}\) then any set of \(m\) vectors in \(V\) must be linearly dependent.
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 \(V\text{.}\)
Example 20.12.
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 \({\mathbb R}^3\text{.}\)
Example 20.13.
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.
Proposition 20.14.
Let \(\{ e_1, e_2, \ldots, e_m \}\) and \(\{ f_1, f_2, \ldots, f_n \}\) be two bases for a vector space \(V\text{.}\) Then \(m = n\text{.}\)
Proof.
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.
Theorem 20.15.
Let \(V\) be a vector space of dimension \(n\text{.}\)
If \(S = \{v_1, \ldots, v_n \}\) is a set of linearly independent vectors for \(V\text{,}\) then \(S\) is a basis for \(V\text{.}\)
If \(S = \{v_1, \ldots, v_n \}\) spans \(V\text{,}\) then \(S\) is a basis for \(V\text{.}\)

If \(S = \{v_1, \ldots, v_k \}\) is a set of linearly independent vectors for \(V\) with \(k \lt n\text{,}\) then there exist vectors \(v_{k + 1}, \ldots, v_n\) such that
\begin{equation*} \{v_1, \ldots, v_k, v_{k + 1}, \ldots, v_n \} \end{equation*}is a basis for \(V\text{.}\)