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# 5.3: Bases

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A basis of a finite-dimensional vector space is a spanning list that is also linearly independent. We will see that all bases for finite-dimensional vector spaces have the same length. This length will then be called the dimension of our vector space.

Definition 5.3.1. A list of vectors $$(v_1,\ldots,v_m)$$ is a basis for the finite-dimensional vector space $$V$$ if $$(v_1,\ldots,v_m)$$ is linearly independent and $$V = \Span(v_1,\ldots,v_m)$$.

If $$(v_1,\ldots,v_m)$$ forms a basis of $$V$$, then, by Lemma 5.2.6, every vector $$v\in V$$ can be uniquely written as a linear combination of $$(v_1,\ldots,v_m)$$.

Example 5.3.2. $$(e_1,\ldots,e_n)$$ is a basis of $$\mathbb{F}^n$$. There are, of course, other bases. For example, $$((1,2),(1,1))$$ is a basis of $$\mathbb{F}^2$$. Note that the list $$((1,1))$$ is also linearly independent, but it does not span $$\mathbb{F}^2$$ and hence is not a basis.

Example 5.3.3. $$(1,z,z^2,\ldots,z^m)$$ is a basis of $$\mathbb{F}_m[z]$$.

Theorem 5.3.4. If $$V=\Span(v_1,\ldots,v_m)$$, then either $$(v_1,\ldots,v_m)$$ is a basis of $$V$$ or some $$v_i$$ can be removed to obtain a basis of $$V$$.

Proof. Suppose $$V=\Span(v_1,\ldots,v_m)$$. We start with the list $$\mathcal{S}=(v_1,\ldots,v_m)$$ and iteratively run through all vectors $$v_k$$ for $$k=1,2,\ldots,m$$ to determine whether to keep or remove them from $$\mathcal{S}$$:

Step 1. If $$v_1=0$$, then remove $$v_1$$ from $$\mathcal{S}$$. Otherwise, leave $$\mathcal{S}$$ unchanged.

Step $$k$$. If $$v_k\in \Span(v_1,\ldots,v_{k-1})$$, then remove $$v_k$$ from $$\mathcal{S}$$. Otherwise, leave $$\mathcal{S}$$ unchanged.

The final list $$\mathcal{S}$$ still spans $$V$$ since, at each step, a vector was only discarded if it was already in the span of the previous vectors. The process also ensures that no vector is in the span of the previous vectors. Hence, by the Linear Dependence Lemma 5.2.7, the final list $$\mathcal{S}$$ is linearly independent. It follows that $$\mathcal{S}$$ is a basis of $$V$$.

Example 5.3.5. To see how Basis Reduction Theorem 5.3.4 works, consider the list of vectors

$\mathcal{S} = ((1, -1, 0), (2, -2, 0), (-1, 0 , 1), (0, -1, 1), (0, 1, 0)).$

This list does not form a basis for $$\mathbb{R}^{3}$$ as it is not linearly independent. However, it is clear that $$\mathbb{R}^{3} = \Span(\mathcal{S})$$ since any arbitrary vector $$v = (x, y, z) \in \mathbb{R}^{3}$$ can be written as the following linear combination over $$\mathcal{S}$$:

$v = (x + z) (1, -1, 0) + 0 (2, -2, 0) + (z) (-1, 0 , 1) + 0 (0, -1, 1) + (x + y + z) (0, 1, 0).$

In fact, since the coefficients of $$(2, -2, 0)$$ and $$(0, -1, 1)$$ in this linear combination are both zero, it suggests that they add nothing to the span of the subset

$\mathcal{B} = ((1, -1, 0), (-1, 0 , 1), (0, 1, 0))$

of $$\mathcal{S}$$. Moreover, one can show that $$\mathcal{B}$$ is a basis for $$\mathbb{R}^{3}$$, and it is exactly the basis produced by applying the process from the proof of Theorem 5.3.4 (as you should be able to verify).

Corollary 5.3.6. Every finite-dimensional vector space has a basis.

Proof. By definition, a finite-dimensional vector space has a spanning list. By the Basis Reduction Theorem 5.3.4, any spanning list can be reduced to a basis.

Theorem 5.3.7. Every linearly independent list of vectors in a finite-dimensional vector space $$V$$ can be extended to a basis of $$V$$.

Proof. Suppose $$V$$ is finite-dimensional and that $$(v_1,\ldots,v_m)$$ is linearly independent.

Since $$V$$ is finite-dimensional, there exists a list $$(w_1,\ldots,w_n)$$ of vectors that spans $$V$$. We wish to adjoin some of the $$w_k$$ to $$(v_1,\ldots,v_m)$$ in order to create a basis of $$V$$.

Step 1. If $$w_1\in \Span(v_1,\ldots,v_m)$$, then let $$\mathcal{S}=(v_1,\ldots,v_m)$$. Otherwise, $$\mathcal{S}=(v_1,\ldots,v_m,w_1)$$.

Step $$k$$. If $$w_k\in\Span(\mathcal{S})$$, then leave $$\mathcal{S}$$ unchanged. Otherwise, adjoin $$w_k$$ to $$\mathcal{S}$$.

After each step, the list $$\mathcal{S}$$ is still linearly independent since we only adjoined $$w_k$$ if $$w_k$$ was not in the span of the previous vectors. After $$n$$ steps, $$w_k\in \Span(\mathcal{S})$$ for all $$k=1,2,\ldots,n$$. Since $$(w_1,\ldots,w_n)$$ was a spanning list, $$\mathcal{S}$$ spans $$V$$ so that $$\mathcal{S}$$ is indeed a basis of $$V$$.

Example 5.3.8. Take the two vectors $$v_1=(1,1,0,0)$$ and $$v_2=(1,0,1,0)$$ in $$\mathbb{R}^4$$. One may easily check that these two vectors are linearly independent, but they do not form a basis of $$\mathbb{R}^4$$. We know that $$(e_1,e_2,e_3,e_4)$$ spans $$\mathbb{R}^4$$. (In fact, it is even a basis.) Following the algorithm outlined in the proof of the Basis Extension Theorem, we see that $$e_1\not\in \Span(v_1,v_2)$$. Hence, we adjoin $$e_1$$ to obtain $$\mathcal{S}=(v_1,v_2,e_1)$$. Note that now

$e_2=(0,1,0,0) = 1v_1+0v_2+(-1)e_1$

so that $$e_2\in\Span(v_1,v_2,e_1)$$, and so we leave $$\mathcal{S}$$ unchanged. Similarly,

$e_3=(0,0,1,0) = 0v_1+1v_2+(-1)e_1,$

and hence $$e_3\in\Span(v_1,v_2,e_1)$$, which means that we again leave $$\mathcal{S}$$ unchanged. Finally, $$e_4\not\in \Span(v_1,v_2,e_1)$$, and so we adjoin it to obtain a basis $$(v_1,v_2,e_1,e_4)$$ of $$\mathbb{R}^4$$.

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