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# 5.1 Linear span

As before, let $$V$$ denote a vector space over $$\mathbb{F}$$. Given vectors $$v_1,v_2,\ldots,v_m\in V$$, a vector $$v\in V$$ is a linear combination of $$(v_1,\ldots,v_m)$$ if there exist scalars $$a_1,\ldots,a_m\in\mathbb{F}$$ such that

$v = a_1 v_1 + a_2 v_2 + \cdots + a_m v_m.$

Definition 5.1.1.  The linear span (or simply span) of $$(v_1,\ldots,v_m)$$ is defined as

$\Span(v_1,\ldots,v_m) := \{ a_1 v_1 + \cdots + a_m v_m \mid a_1,\ldots,a_m \in \mathbb{F} \}.$

Lemma 5.1.2. Let $$V$$ be a vector space and $$v_1,v_2,\ldots,v_m\in V$$. Then

1. $$v_j\in \Span(v_1,v_2,\ldots,v_m)$$.
2. $$\Span(v_1,v_2,\ldots,v_m)$$ is a subspace of $$V$$.
3. If $$U\subset V$$ is a subspace such that $$v_1,v_2,\ldots v_m\in U$$, then $$\Span(v_1,v_2,\ldots,v_m)\subset U$$.

Proof.  Property~1 is obvious. For Property~2, note that $$0\in\Span(v_1,v_2,\ldots,v_m)$$ and that $$\Span(v_1,v_2,\ldots,v_m)$$ is closed under addition and scalar multiplication.  For Property~3, note that a subspace $$U$$ of a vector space $$V$$ is closed under addition and scalar multiplication. Hence, if $$v_1,\ldots,v_m\in U$$, then any linear combination $$a_1v_1+\cdots +a_m v_m$$ must also be an element of $$U$$.

Lemma 5.1.2 implies that $$\Span(v_1,v_2,\ldots,v_m)$$ is the smallest subspace of $$V$$ containing each of $$v_1,v_2,\ldots,v_m$$.

Definition 5.1.3. If $$\Span(v_1,\ldots,v_m)=V$$, then we say that $$(v_1,\ldots,v_m)$$ spans $$V$$ and we call $$V$$ finite-dimensional. A vector space that is not finite-dimensional is called infinite-dimensional.

Example 5.1.4.  The vectors $$e_1=(1,0,\ldots,0)$$, $$e_2=(0,1,0,\ldots,0), \ldots, e_n=(0,\ldots,0,1)$$ span $$\mathbb{F}^n$$. Hence $$\mathbb{F}^n$$ is finite-dimensional.

Example 5.1.5.  The vectors $$v_1=(1,1,0)$$ and $$v_2=(1,-1,0)$$ span a subspace of $$\mathbb{R}^3$$. More precisely, if we write the vectors in $$\mathbb{R}^3$$ as 3-tuples of the form $$(x,y,z)$$, then $$\Span(v_1,v_2)$$ is the $$xy$$-plane in $$\mathbb{R}^3$$.

Example 5.1.6. Recall that if $$p(z)=a_mz^m + a_{m-1} z^{m-1} + \cdots + a_1z + a_0\in \mathbb{F}[z]$$ is a polynomial with coefficients in $$\mathbb{F}$$ such that $$a_m\neq 0$$, then we say that $$p(z)$$ has degree $$m$$. By convention, the degree of the zero polynomial $$p(z)=0$$ is $$-\infty$$. We denote the degree of $$p(z)$$ by $$\deg(p(z))$$. Define

$$\mathbb{F}_m[z] =$$ set  of all polynomials in $$\mathbb{F}[z]$$ of degree at most m.

Then $$\mathbb{F}_m[z]\subset \mathbb{F}[z]$$ is a subspace since $$\mathbb{F}_m[z]$$ contains the zero polynomial and is closed under addition and scalar multiplication. In fact, $$\mathbb{F}_m[z]$$ is a finite-dimensional subspace of $$\mathbb{F}[z]$$ since

$\mathbb{F}_m[z] = \Span(1,z,z^2,\ldots,z^m).$

At the same time, though, note that $$\mathbb{F}[z]$$ itself is infinite-dimensional. To see this, assume the contrary, namely that

$\mathbb{F}[z] = \Span(p_1(z),\ldots,p_k(z))$

for a finite set of $$k$$ polynomials $$p_1(z),\ldots,p_k(z)$$. Let $$m=\max(\deg p_1(z),\ldots,\deg p_k(z))$$. Then $$z^{m+1}\in\mathbb{F}[z]$$, but $$z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))$$.