Bases

DEFINITION 3.3.1 (Basis of a Vector Space)  

1. A non-empty subset ${\cal B}$ of a vector space $V$ is called a basis of $V$ if
   1. ${\cal B}$ is a linearly independent set, and
   2. $L({\cal B}) = V,$ i.e., every vector in $V$ can be expressed as a linear combination of the elements of ${\cal B}.$
2. A vector in ${\cal B}$ is called a basis vector.

Remark 3.3.2   Let $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_p \}$ be a basis of a vector space $V ({\mathbb{F}}).$ Then any ${\mathbf v}\in V$ is a UNIQUE LINEAR COMBINATION of the basis vectors, ${\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_p.$

Observe that if there exists a ${\mathbf v}\in W$ such that ${\mathbf v}= \alpha_1 {\mathbf v}_1 + \alpha_2 {\mathbf v}_2 + \cdots + \alpha_p {\mathbf v}_p$ and ${\mathbf v}= \beta_1 {\mathbf v}_1 + \beta_2 {\mathbf v}_2 + \cdots + \beta_p {\mathbf v}_p$ then

$${\mathbf 0}= {\mathbf v}- {\mathbf v}= (\alpha_1 - \beta_1) {\mat... ...alpha_2 - \beta_2) {\mathbf v}_2 + \cdots + (\alpha_p - \beta_p) {\mathbf v}_p.$$

But then the set $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_p \}$ is linearly independent and therefore the scalars $\alpha_i - \beta_i$ for $1 \leq i \leq p$ must all be equal to zero. Hence, for $1 \le i \le p,$ ${\alpha}_i = \beta_i$ and we have the uniqueness.

By convention, the linear span of an empty set is $\{{\mathbf 0}\}.$ Hence, the empty set is a basis of the vector space $\{{\mathbf 0}\}.$

EXAMPLE 3.3.3  

1. Check that if $\; V = \{(x,y,0) : x, y \in {\mathbb{R}} \} \subset {\mathbb{R}}^3,$ then ${\cal B}= \{(1,0,0), (0,1,0) \}$ or ${\cal B}= \{(1,0,0), (1,1,0) \}$ or ${\cal B}= \{(2,0,0), (1,3,0) \}$ or $\cdots$ are bases of $V.$
2. For $1 \leq i \leq n,$ let ${\mathbf e}_i = ( 0, \ldots, 0, \underbrace{1}_{i\mbox{ th place}}, 0, \ldots, 0) \in {\mathbb{R}}^n.$ Then, the set ${\cal B}=\{ {\mathbf e}_1, {\mathbf e}_2, \ldots, {\mathbf e}_n \}$ forms a basis of ${\mathbb{R}}^n.$ This set is called the standard basis of ${\mathbb{R}}^n.$

   That is, if $n=3,$ then the set $\{(1,0,0), (0,1,0), (0,0,1) \}$ forms an standard basis of ${\mathbb{R}}^3.$

3. Let $V = \{(x, y, z) : x+ y - z = 0, \; x,y,z \in {\mathbb{R}} \}$ be a vector subspace of ${\mathbb{R}}^3.$ Then $S = \{(1,1, 2), (2,1,3), (1,2,3) \} \subset V.$ It can be easily verified that the vector $(3,2,5) \in V$ and
   $$(3,2,5) = (1,1,2) + (2,1,3) = 4 (1,1,2) - (1,2,3).$$
   Then by Remark 3.3.2, $S$ cannot be a basis of $V.$

   A basis of $V$ can be obtained by the following method:
   The condition $x + y - z = 0$ is equivalent to $z = x + y.$ we replace the value of $z$ with $x + y$ to get

   $$(x,y,z) = (x, y, x+y) = (x, 0, x) + (0, y, y) = x (1, 0, 1)+ y (0, 1, 1).$$
   Hence, $\{(1,0,1), (0,1,1)\}$ forms a basis of $V.$
4. Let $V = \{a + i b : a,b \in {\mathbb{R}} \}$ and ${\mathbb{F}} = {\mathbb{C}}.$ That is, $V$ is a complex vector space. Note that any element $a+ i b \in V$ can be written as $a + i b = (a+ i b) {\mathbf 1}.$ Hence, a basis of $V$ is $\{{\mathbf 1} \}.$
5. Let $V = \{a + i b : a,b \in {\mathbb{R}} \}$ and ${\mathbb{F}} = {\mathbb{R}}.$ That is, $V$ is a real vector space. Any element $a+ i b \in V$ is expressible as $a \cdot {\mathbf 1} + b \cdot {\mathbf i}.$ Hence a basis of $V$ is $\{{\mathbf 1}, {\mathbf i}\}.$

   Observe that $i$ is a vector in ${\mathbb{C}}.$ Also, $i \not\in {\mathbb{R}}$ and hence $i \cdot(1+ 0 \cdot i )$ is not defined.

6. Recall the vector space ${\cal P}({\mathbb{R}}),$ the vector space of all polynomials with real coefficients. A basis of this vector space is the set
   $$\{1, x, x^2, \ldots, x^n, \ldots\}.$$
   This basis has infinite number of vectors as the degree of the polynomial can be any positive integer.

DEFINITION 3.3.4 (Finite Dimensional Vector Space)   A vector space $V$ is said to be finite dimensional if there exists a basis consisting of finite number of elements. Otherwise, the vector space $V$ is called infinite dimensional.

In Example 3.3.3, the vector space of all polynomials is an example of an infinite dimensional vector space. All the other vector spaces are finite dimensional.

Remark 3.3.5   We can use the above results to obtain a basis of any finite dimensional vector space $V$ as follows:

Step 1: Choose a non-zero vector, say, ${\mathbf v}_1 \in V.$ Then the set $\{{\mathbf v}_1\}$ is linearly independent.

Step 2: If $V = L({\mathbf v}_1),$ we have got a basis of $V.$ Else there exists a vector, say, ${\mathbf v}_2 \in V$ such that ${\mathbf v}_2 \not\in L({\mathbf v}_1).$ Then by Corollary 3.2.6, the set $\{{\mathbf v}_1, {\mathbf v}_2\}$ is linearly independent.

Step 3: If $V = L({\mathbf v}_1, {\mathbf v}_2),$ then $\{{\mathbf v}_1, {\mathbf v}_2\}$ is a basis of $V.$ Else there exists a vector, say, ${\mathbf v}_3 \in V$ such that ${\mathbf v}_3 \not\in L({\mathbf v}_1, {\mathbf v}_2).$ So, by Corollary 3.2.6, the set $\{{\mathbf v}_1, {\mathbf v}_2, {\mathbf v}_3\}$ is linearly independent.

At the $i^{\mbox{th}}$ step, either $V = L({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_i),$ or $L({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_i) \ne V.$
In the first case, we have $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_i\}$ as a basis of $V.$

In the second case, $L({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_i) \subset V$ . So, we choose a vector, say, ${\mathbf v}_{i+1} \in V$ such that ${\mathbf v}_{i+1} \not\in L({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_i).$ Therefore, by Corollary 3.2.6, the set $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_{i+1}\}$ is linearly independent.

This process will finally end as $V$ is a finite dimensional vector space.

EXERCISE 3.3.6  

1. Let $S = \{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_p \}$ be a subset of a vector space $V ({\mathbb{F}}).$ Suppose $L(S) = V$ but $S$ is not a linearly independent set. Then prove that each vector in $V$ can be expressed in more than one way as a linear combination of vectors from $S.$
2. Show that the set $\{(1,0,1), (1, i ,0), (1,1,1- i ) \}$ is a basis of ${\mathbb{C}}^3 ({\mathbb{C}}).$
3. Let $A$ be a matrix of rank $r.$ Then show that the $r$ non-zero rows in the row-reduced echelon form of $A$ are linearly independent and they form a basis of the row space of $A.$