Span

[Notes from Linear Algebra Done Right | Chapter 2]

Finite-dimensional Vector Spaces

For ${\mathbb{R}}^3$:

$${\mathbb{R}}^3\mathrm{span}(1,0,0),(0,1,0),(0,0,1)$$

So:

$$\dim ({\mathbb{R}}^3)=3.$$

There are infinitely many vectors in ${\mathbb{R}}^3$, but only 3 basis vectors are needed.

Infinite-dimensional

For $\mathbb{R}[x]$, the space of all polynomials:

$$\mathbb{R}[x]\mathrm{span}1,x,x^2,x^3,\dots .$$

You cannot generate every polynomial using only finitely many of these.

$$\boxed{\dim (\mathbb{R}[x])=\infty }$$

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Linear combination

A linear combination of a list of vectors $v_1,\dots ,v_m$ is any vector formed by multiplying each vector by a scalar and adding the results.

$$\boxed{a_1{v}_1+\cdots +a_m{v}_m}$$

where

$$a_1,\dots ,a_m\in \mathbb{F}.$$

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Span

The span of a list of vectors is the set of all possible linear combinations of those vectors.

$$\boxed{\mathrm{span}(v_1,\dots ,v_m)=a_1{v}_1+\cdots +a_m{v}_m:a_1,\dots ,a_m\in \mathbb{F}}$$

Span = everything that you can reach using these vectors.

$$\boxed{\mathrm{span}(\text{empty list})=0}$$

because the only “empty” linear combination is $0$.

The span of any list of vectors is the smallest subspace containing those vectors.

$$\boxed{\mathrm{span}(v_1,\dots ,v_m)=\text{smallest subspace containing }{v}_1,\dots ,v_m}$$

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Finite-dimensional vector space

Formally, $V$ is finite-dimensional if there exists a finite list

$$v_1,\dots ,v_m$$

such that

$$\mathrm{span}(v_1,\dots ,v_m)=V.$$

Example: ${\mathbb{F}}^n$

The standard coordinate vectors

$$(1,0,\dots ,0),\dots ,(0,\dots ,0,1)$$

span ${\mathbb{F}}^n$.

Therefore:

$$\boxed{{\mathbb{F}}^n\text{ is finite-dimensional}}$$

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$\mathcal{P}(\mathbb{F})$

$\mathcal{P}(\mathbb{F})$ is the vector space of all polynomial functions

$$p:\mathbb{F}\to \mathbb{F}$$

of the form

$$p(z)=a_0+a_{1}z+\cdots +a_m{z}^m.$$

where

$$a_0,\dots ,a_m\in \mathbb{F}.$$

With the usual addition and scalar multiplication, $\mathcal{P}(\mathbb{F})$ is a vector space.

Degree : For a nonzero polynomial,

$$p(z)=a_0+a_{1}z+\cdots +a_m{z}^m$$

with

$$a_m\ne 0,$$

the degree is

$$\boxed{\mathrm{deg}p=m}.$$

The zero polynomial is assigned:

$$\boxed{\mathrm{deg}0=-\infty }.$$

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${\mathcal{P}}_m(\mathbb{F})$ : For a nonnegative integer $m$:

$$\boxed{{\mathcal{P}}_m(\mathbb{F})=p\in \mathcal{P}(\mathbb{F}):\mathrm{deg}p\le m}$$

These are all polynomials of degree at most $m$.

For example:

$${\mathcal{P}}_2(\mathbb{F})=a+bz+c{z}^2:a,b,c\in \mathbb{F}.$$

Examples: 3, 2+7z, $1-4z+8z^2$, 0 But not $z^3+1$

$$\boxed{{\mathcal{P}}_m(\mathbb{F})=\mathrm{span}(1,z,z^2,\dots ,z^m)}$$

Therefore ${\mathcal{P}}_m(\mathbb{F})$ is finite-dimensional.

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$\mathcal{P}(\mathbb{F})$ is infinite-dimensional

Take any finite list of polynomials:

$$p_1,\dots ,p_k.$$

Because there are only finitely many polynomials, there is some largest degree:

$$m=\max (\mathrm{deg}{p}_1,\dots ,\mathrm{deg}{p}_k).$$

Any linear combination

$$a_1{p}_1+\cdots +a_k{p}_k$$

has degree at most $m$.

Therefore:

$$z^{m+1}$$

cannot be in their span.

So this particular finite list does not span $\mathcal{P}(\mathbb{F})$.

And because every list is finite, no finite list can span $\mathcal{P}(\mathbb{F})$.

Therefore:

$$\boxed{\mathcal{P}(\mathbb{F})\text{ is infinite-dimensional}}$$

No finite list of polynomials can span P(F).​

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