The Key Ingredients of Vector Spaces

Let $V$ be a vector space over a field $\mathbb K$ . We have previously seen that for vectors $\mathbf v_1, \mathbf v_2 \in V$ ,

$\mathrm{span}(\{\mathbf v_1, \mathbf v_2\}) = \mathbb K\{\mathbf v_1\} + \mathbb K \{\mathbf v_2\}.$

Furthermore, if $\mathbf v \neq \mathbf 0$ , we require at least one vector, namely $\mathbf v$ , to cook up the space $\mathbb K\{\mathbf v\}$ .

We are tempted then to think that $\mathrm{span}(\{\mathbf v_1, \mathbf v_2\})$ requires two minimum ingredients. However, we have seen that this does not always hold. For instance, suppose $V$ contains some $\mathbf v \neq \mathbf 0$ . Define $\mathbf v_1 = \mathbf v$ and $\mathbf v_2 = 2\mathbf v$ . Then $\mathbb K\{\mathbf v_1\} + \mathbb K \{\mathbf v_2\} = \mathbb K\{\mathbf v\}$ yields

$\mathrm{span}(\{\mathbf v_1, \mathbf v_2\}) = \mathrm{span}(\{\mathbf v\}),$

which requires only one ingredient. The key observation is that $\mathbf v_2 = 2 \mathbf v_1 \in \mathrm{span}(\{\mathbf v_1\})$ . That is, $\mathbf v_2$ doesn’t truly increase $\mathrm{span}(\{\mathbf v_1\})$ at all. We can consolidate by making the following observation.

Lemma 1. Let $W_1,W_2 \subseteq V$ be sets of vectors. Suppose $W_1 \subseteq W_2$ . Then

$\mathrm{span}(W_1) \subseteq \mathrm{span}(W_2)$

as subspaces. Furthermore, if $W_2 \backslash W_1 \subseteq \mathrm{span}(W_1)$ , then

$\mathrm{span}(W_1) = \mathrm{span}(W_2).$

Proof. Firstly, $W_1 \subseteq W_2 \subseteq \mathrm{span}(W_2)$ , so that $\mathrm{span}(W_2)$ contains $W_1$ . Since $\mathrm{span}(W_1)$ is the smallest subspace containing $W_1$ , we must have

$\mathrm{span}(W_1) \subseteq \mathrm{span}(W_2).$

If $W_2 \backslash W_1 \subseteq \mathrm{span}(W_1)$ , then $W_2 = (W_1 \cap W_2) \cup W_2 \backslash W_1$ implies

$\begin{aligned} \mathrm{span}(W_2) &\subseteq \mathrm{span}(W_1 \cap W_2) + \mathrm{span}(W_2 \backslash W_1) \\ &\subseteq \mathrm{span}(W_1) + \mathrm{span}(W_1) \\ &\subseteq \mathrm{span}(W_1). \end{aligned}$

Hence, $\mathrm{span}(W_1) = \mathrm{span}(W_2)$ .

The main point of the equality is that for any excess vectors $W_2 \backslash W_1$ , if these vectors belong to $\mathrm{span}(W_1)$ , then $\mathrm{span}(W_2)$ doesn’t add any new information to $\mathrm{span}(W_1)$ , and thus does not increase the minimum required number of ingredients to generate the space.

Recall our example $\{\mathbf v_1, \mathbf v_2\}$ where both vectors are nonzero. We have seen that if $\mathbf v_2 \in \mathrm{span}(\{ \mathbf v_1 \})$ , then

$\mathrm{span}(\{\mathbf v_1, \mathbf v_2\}) = \mathrm{span}(\{\mathbf v_1\}).$

What if $\mathbf v_2 \notin \mathrm{span}(\{ \mathbf v_1 \}) = \mathbb K( \{\mathbf v_1\} )$ ? Then there are no scalars $c \in \mathbb K$ such that $\mathbf v_2 = c\mathbf v_1 \iff c\mathbf v_1 +(-1)\mathbf v_2 = \mathbf 0$ . In fact, something stronger happens.

Lemma 2. Let $\mathbf v_1,\mathbf v_2 \in V$ be nonzero vectors. Then $\mathbf v_2 \notin \mathrm{span}(\{\mathbf v_1\})$ if and only if for any $c_1,c_2 \in \mathbb K$ ,

$c_1 \mathbf v_1 + c_2 \mathbf v_2 = \mathbf 0\quad \Rightarrow \quad c_1 = c_2 = 0.$

Proof. We first prove $(\Rightarrow)$ . Fix scalars $c_1,c_2 \in \mathbb K$ such that

$c_1 \mathbf v_1 + c_2 \mathbf v_2 = \mathbf 0.$

By algebra, $c_1 \mathbf v_1 = -c_2 \mathbf v_2$ . If $c_2 = 0$ , then we have $c_1 = 0$ and we are done. Otherwise, $\mathbf v_2 = -(c_1/c_2) \mathbf v_1 \in \mathrm{span}(\{\mathbf v_1\})$ , a contradiction. Therefore, $c_1 = c_2 = 0$ necessarily.

We next prove $(\Leftarrow)$ by contrapositive. Suppose $\mathbf v_2 \in \mathrm{span}(\{\mathbf v_1\})$ . Then there exists some $c \in \mathbb K$ such that $\mathbf v_2 = c\mathbf v_1$ . If $c = 0$ then $\mathbf v_2 = \mathbf 0$ , a contradiction. Hence, $c \neq 0$ . Therefore,

$c\mathbf v_1 + (-1)\mathbf v_2 = \mathbf 0.$

Setting $c_1 = c$ and $c_2 = -1$ ,

$c_1 \mathbf v_1 + c_2\mathbf v_2 = \mathbf 0,$

and yet $c_1 = c_2 = 0$ is not true.

It is this latter condition that we will define as linear independence.

Definition 2. The finite set $\{\mathbf v_1, \dots, \mathbf v_n\} \subseteq V$ is called linearly independent if for any $c_1, \dots, c_n \in \mathbb K$ ,

$c_1 \mathbf v_1 + \dots + c_n \mathbf v_n = \mathbf 0\quad \Rightarrow \quad c_1 = \cdots = c_n.$

It is not hard to see that nonempty finite subsets of $\{\mathbf v_1, \dots, \mathbf v_n\}$ are linearly independent as well. A single set is linearly independent as well too.

We use this property to generalise to infinite sets. For any set $W \subseteq V$ , we say that $W$ is linearly independent if every nonempty finite subset of $W$ is linearly independent. We say that $W$ is linearly dependent if it is not linearly independent.

Corollary 1. Let $\mathbf v_1,\mathbf v_2 \in V$ be nonzero vectors. Then $\mathbf v_2 \notin \mathrm{span}(\{\mathbf v_1\})$ if and only if $\{\mathbf v_1,\mathbf v_2\}$ is linearly independent.

Roughly speaking, a set $\{\mathbf v_1,\dots,\mathbf v_n\}$ is linearly independent if it contains $n$ pieces of information to its span. This is what we define to be the dimension of a span.

However, let’s first return to mathematical earth and discuss $\mathbb R^n = \mathcal F(\{1,\dots,n\}, \mathbb R)$ . Intuitively, $\mathbb R^n$ ought to have $n$ dimensions. This is true.

Example 3. For each $i$ , define $\mathbf e_i := \mathbb I_{\{i\}}$ , which means

$\mathbf e_i(j) = \begin{cases}1, & i = j,\\ 0, & i \neq j.\end{cases}$

Then $\{ \mathbf e_1, \dots, \mathbf e_n \}$ is linearly independent, and

$\mathbb R^n = \mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \}).$

Proof. For linear independence, fix scalars $c_1,\dots,c_n \in \mathbb K$ such that

$c_1 \mathbf e_1 + \dots + c_n \mathbf e_n = \mathbf 0.$

Then for any $i$ ,

$\begin{aligned} 0 = \mathbf 0(i) &= (c_1 \mathbf e_1 + \dots + c_n \mathbf e_n)(i) \\ &= c_1 \mathbf e_1(i) + \dots + c_i \mathbf e_i(i) + \cdots + c_n \mathbf e_n(i) \\ &= c_1 \cdot 0+ \dots + c_i \cdot 1 + \cdots + c_n \cdot 0 = c_i.\end{aligned}$

Therefore, $c_1 = \dots = c_n = 0$ , so that $\{ \mathbf e_1, \dots, \mathbf e_n \}$ is linearly independent. For the spanning property, fix $\mathbf v \in \mathbb R^n$ , where $\mathbf v(i) = v_i$ for each $I$ . Then

$\mathbf v = v_1 \mathbf e_1 + \dots + v_n \mathbf e_n \in \mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \}).$

This yields $\mathbb R^n \subseteq \mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \})$ . On the other hand, since $\{ \mathbf e_1, \dots, \mathbf e_n \} \subseteq \mathbb R^n$ and $\mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \})$ is the smallest vector space containing $\{ \mathbf e_1, \dots, \mathbf e_n \}$ , we automatically have $\mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \}) \subseteq \mathbb R^n$ . Therefore,

$\mathbb R^n = \mathrm{span}(\{ \mathbf e_1, \dots, \mathbf e_n \}).$

We call $\{ \mathbf e_1, \dots, \mathbf e_n \}$ a basis for $\mathbb R^n$ , and can generalise the idea to other vector spaces.

Definition 3. We call $K \subseteq V$ a basis for $V$ if $K$ is linearly independent and $\mathrm{span}(K) = V$ .

Theorem 1. Suppose $\{ \mathbf v_1, \dots, \mathbf v_n \}$ is a basis for $V$ . Then for any vector $\mathbf v \in V$ , there exists unique scalars $c_1,\dots,c_n \in \mathbb K$ such that

$\mathbf v = c_1 \mathbf v_1 + \dots + c_n \mathbf v_n.$

Proof. Existence is immediate from $V = \mathrm{span}(\{ \mathbf v_1, \dots, \mathbf v_n \})$ . For uniqueness, suppose there are two representations

$c_1 \mathbf v_1 + \dots + c_n \mathbf v_n = d_1 \mathbf v_1 + \dots + d_n \mathbf v_n.$

Subtracting on both sides,

$(c_1-d_1) \mathbf v_1 + \dots + (c_n-d_n) \mathbf v_n = \mathbf 0.$

Since $\{ \mathbf v_1, \dots, \mathbf v_n \}$ is linearly independent, $c_i - d_i = 0\iff c_i = d_i$ for each $i$ , so that the “recipe” that creates $\mathbf v$ is unique.

Corollary 2. Suppose $\{ \mathbf v_1, \dots, \mathbf v_n \}$ is a basis for $V$ . Define the map $T : V \to \mathbb K^n$ in the following manner: $T(\mathbf v_i) = \mathbf e_i$ for any $i$ and for any $\mathbf u,\mathbf v \in V$ and $c,d \in \mathbb K$ ,

$T(c \mathbf u + d \mathbf v) = c T(\mathbf u) + d T(\mathbf v).$

Then $T : V \to \mathbb K^n$ is a well-defined bijection. In fact, $T$ is a linear transformation that maps the dish $\mathbf v$ to the recipe $T(\mathbf v)$ that is required to “cook” it.

In a sense, any vector space that only requires a minimum of $n$ ingredients to cook all dishes is essentially the same as the vector space $\mathbb K^n$ . We can formalise this idea using isomorphisms, which is our next topic of discussion.

—Joel Kindiak, 24 Feb 0007H

KindiakMath

The Key Ingredients of Vector Spaces

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The Key Ingredients of Vector Spaces

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