Bases

Definition 3.61 (Related exercises: Exercise 3.12, Exercise 3.22, Exercise 3.17, Exercise 4.6, Exercise 3.24, Exercise 3.21, Exercise 3.25, Exercise 3.27, Exercise 3.26, Exercise 3.29, Exercise 3.23)

A collection of vectors in a vector space

\[ v_1, \dots, v_m \in V \]

is called a basis if they span \(V\) and if they are linearly independent.

Example 3.62 (Related exercises: Exercise 4.3, Exercise 4.21)

The vectors

\[ e_1 = (1, 0, \dots, 0), e_2 = (0, 1, 0, \dots 0), \dots, e_n = (0, \dots, 0, 1) \in {\bf R}^n \]

are a basis, called the standard basis. Indeed, we have observed in Example 3.43 and Example 3.51 that they span \({\bf R}^n\) and that they are linearly independent.

We try and modify this basis a little bit and see what happens. If we omit one of the vectors and only consider, say

\[ e_2 = (0, 1, 0, \dots 0), \dots, e_n = (0, \dots, 0, 1) \in {\bf R}^n \]

these do not form a basis: while they are still linearly independent, they do not span \({\bf R}^n\).

On the other hand, we now consider

\[ e_1, \dots, e_n, v, \]

for an arbitrary vector \(v \in {\bf R}^n\). These do not form a basis: while they span \({\bf R}^n\) (even without the \(v\)), they are not linearly independent. Indeed, since \(e_1, \dots, e_n\) span \({\bf R}^n\), this means that

\[ v = a_1 e_1 + \dots + a_n e_n \]

for appropriate \(a_1, \dots, a_n \in {\bf R}\). According to Lemma 3.54, this means that \(e_1, \dots, e_n, v\) are linearly dependent.

Example 3.63 (Related exercises: Exercise 3.12)

The vectors

\[ v_1 = (0, 2, 1), v_2 = (1, 0, 2), v_3 = (-1, 1, 1) \]

form a basis of \({\bf R}^3\). To see this, we apply Method 3.56 and Method 3.46:

\[ \begin{align*} \left ( \begin{array}{ccc} 0 & 2 & 1 \\ 1 & 0 & 2 \\ -1 & 1 & 1 \end{array} \right ) \leadsto & \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ -1 & 1 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 1 & 3 \end{array} \right ) \\ \leadsto & \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & \frac 52 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & \frac 1 2 \\ 0 & 0 & 1 \end{array} \right ). \end{align*} \]

This matrix has three leading ones, so the vectors are linearly independent and span \({\bf R}^3\), so they form a basis.

Note that this is a different basis than \(e_1, e_2, e_3\) considered above.

The following result, which is simply a combination of the definition of generating systems and Proposition 3.60, is often described by saying that a basis gives rise to a coordinate system in a vector space.

Proposition 3.64 (Related exercises: Exercise 4.11)

Let \(v_1, \dots, v_m\) be a basis of a vector space \(V\). Then each vector \(v \in V\) can be written in a unique way as a linear combination

\[ v=a_1 v_1 + \dots + a_m v_m. \]

For some other vector \(w = b_1 v_1 + \dots+ b_m v_m\), we have

\[ v+w = (a_1+b_1) v_1 + \dots + (a_m + b_m) v_m. \]