Linear Combination, Span, Generating Set.

$\newcommand{\br}{\\}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\P}{\mathbb{P}}$ $\newcommand{\F}{\mathbb{F}}$ $\newcommand{\L}{\mathcal{L}}$ $\newcommand{\spa}[1]{\text{span}(#1)}$ $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\emptyset}{\varnothing}$ $\newcommand{\otherwise}{\text{ otherwise }}$ $\newcommand{\if}{\text{ if }}$ $\newcommand{\union}{\cup}$ $\newcommand{\intercept}{\cap}$ $\newcommand{\abs}[1]{| #1 |}$ $\newcommand{\pare}[1]{\left\(#1\right\)}$ $\newcommand{\t}[1]{\text{ #1 }}$ $\newcommand{\head}{\text H}$ $\newcommand{\tail}{\text T}$ $\newcommand{\inv}[1]{{#1}^{-1}}$ $\newcommand{\nullity}[1]{\text{nullity}(#1)}$ $\newcommand{\rank}[1]{\text{rank}(#1)}$ $\newcommand{\oto}{\text{ one-to-one }}$ $\newcommand{\ot}{\text{ onto }}$ $\newcommand{\Vcw}[2]{\begin{pmatrix} #1 \br #2 \end{pmatrix}}$ $\newcommand{\Vce}[3]{\begin{pmatrix} #1 \br #2 \br #3 \end{pmatrix}}$ $\newcommand{\Vcr}[4]{\begin{pmatrix} #1 \br #2 \br #3 \br #4 \end{pmatrix}}$ $\newcommand{\Vct}[5]{\begin{pmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{pmatrix}}$ $\newcommand{\Vcy}[6]{\begin{pmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \end{pmatrix}}$ $\newcommand{\Vcu}[7]{\begin{pmatrix} #1 \br #2 \br #3 \br #4 \br #5 \br #6 \br #7 \end{pmatrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Mqr}[4]{\begin{bmatrix} #1 & #2 & #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqt}[5]{\begin{bmatrix} #1 & #2 & #3 & #4 & #5 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mrq}[4]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \end{bmatrix}}$ $\newcommand{\Mtq}[5]{\begin{bmatrix} #1 \br #2 \br #3 \br #4 \br #5 \end{bmatrix}}$ $\newcommand{\Mqw}[2]{\begin{bmatrix} #1 & #2 \end{bmatrix}}$ $\newcommand{\Mwq}[2]{\begin{bmatrix} #1 \br #2 \end{bmatrix}}$ $\newcommand{\Mww}[4]{\begin{bmatrix} #1 & #2 \br #3 & #4 \end{bmatrix}}$ $\newcommand{\Mqe}[3]{\begin{bmatrix} #1 & #2 & #3 \end{bmatrix}}$ $\newcommand{\Meq}[3]{\begin{bmatrix} #1 \br #2 \br #3 \end{bmatrix}}$ $\newcommand{\Mwe}[6]{\begin{bmatrix} #1 & #2 & #3\br #4 & #5 & #6 \end{bmatrix}}$ $\newcommand{\Mew}[6]{\begin{bmatrix} #1 & #2 \br #3 & #4 \br #5 & #6 \end{bmatrix}}$ $\newcommand{\Mee}[9]{\begin{bmatrix} #1 & #2 & #3 \br #4 & #5 & #6 \br #7 & #8 & #9 \end{bmatrix}}$
Definition: Linear Combination

$V/F$ and $S$ a non-empty subset of $V$. A vector $v \in V$ is called a linear combination of vectors from $S$ if there exists FINITE many vectors $v_1, v_2, …, v_n \in S$ and $a_1, a_2, …, a_n \in F$ s.t. $$v = a_1v_1 + a_2v_2 + … + a_nv_n$$ Furthermore, in this case, we say that $v$ is a linear combination of $v_1, v_2, …, v_n$ with coefficients $a_1, a_2, …, a_n$.

Definition: Span

$V/F, S$ is a non-empty subset of $V$. The span of $S$ or the linear span of $S$ denoted by $\spa{S}$ is the set of all possible linear combination of vectors of $S$.

We define $\spa{\emptyset} = \set{0}$.

$\abs S$ could be finite or infinite.

Theorem 1.5

$V/F$. If $S \subseteq V$, then $\spa{S} \leq V$.

Moreover, if $W \leq V, S\subseteq W$, then $\spa{S} \leq W$.

Proof [expand]

case 1: if $S$ is $\emptyset, y$ definition $\spa{s} = {0} \leq V$.

case 2: $S$ is non-empty $(S \neq \emptyset)$ $\exists V \in S$

(W1).$0 = 0\cdot v \in \spa{S}$

(W2).$x, y \in \spa{S}$

$x \in \spa{S} \implies \exists v_1, v_2, …, v_m \in S$ and $a_1, a_2, …, a_n \in F$ s.t. $x = a_1v_1 + a_2v_2 + … + a_nv_n$

$y \in \spa{S} \implies \exists w_1, w_2, …, w_n \in S$ and $b_1, b_2, …, b_n \in F$ s.t. $y = b_1w_1 + b_2w_2 + … + b_nw_n$

$x + y = a_1v_1 + a_2v_2 + … + a_mv_m + b_1w_1 + b_2w_2 + … + b_nw_n$

$\set{v_1, v_2, …,v_n, w_1, w_2, …, w_n}$ at most $m+n$ elements

$x+ y \in \spa{S}$

(W3).$x\in \spa{S}$ and $\lambda \in F$ $\lambda x \in \spa{S}$ $x = a_1v_1 + a_2v_2 + … + a_nv_n$ $\lambda x = \lambda a_1v_1 + \lambda a_2v_2 + … + \lambda a_nv_n$ $\lambda x \in \spa{S}$

Q.E.D.

Theorem

$V/F$. If $S \subseteq V$, then $\spa{S} \leq V$ and $\spa{S}$ is the smallest subspace containing $S$.

Proof [expand]

To prove $\spa{S}$ is the smallest subspace, we need to prove that $$\forall W \leq V, W \supseteq S \implies \spa{S} \leq W$$

Since $x\in \spa{S}, \exists x_1, x_2,…, x_n \in S$ and $a_1, a_2, …, a_n \in F$ s.t.

$$x = a_1x_2 + a_2x_2 + … + a_nx_n$$

$S\subseteq W$ $\therefore x_i \in S \implies x_i \in W$

$x_i \in S \implies x_i \in W \implies a_ix_i \in W$

i.e. $a_1x_1 + a_2x_2 + … + a_nx_n \in W$

$\implies x\in W$

$\therefore \spa{S} \leq W$

Definition: Generating Set

Let $V$ be a vector space over a field $F$. a subset $S \subseteq V$ is called a generating set if $\spa{S} = V$.

In this case we say $S$ generates $V$.

Example: $\R^2/\R$

$S= \set{(1,0), (0,1)}$, then $S$ generates $\R^2$.

Proof [expand]

$a,b \in \R,$ then $(a,b) \in \R^2.$

$(a, b) = a(1,0) + b(0,1).$

$\R^2 \subseteq \spa{S}$.

Also, $\spa{S} \subseteq \R^2$, since $S \subseteq R^2$(Theorem 1.5)

$\spa{S} = \R^2$, i.e. $S$ generates $\R^2$.

Example

Prove that $T=\set{(1, -1), (2, 3)}$ is a generating set.

Proof [expand]

$a, b \in \R^2$.

$(a,b) = x(1, -1) + y (2,3) = (x+2y, -x+ 3y)$

$$ \begin{align} x+2y = a \br -x+3y = b \end{align} $$

solve these equations and we get:

$(a,b) = \frac{3a-2b}5 (1, -1) + \frac{a+b}5(2,3)$

$(a,b) \in \spa{T}$

$\R^2 \subseteq \spa{T}$

$T \subseteq \R^2$, then $\spa{T} \subseteq \R^2$(Theorem 1.5)

$\therefore \spa{T} = \R^2 $ i.e. $T$ generates $\R^2 $.

Example: $\P\_3(x)/\R$

$S= \set{1, x-x^2, x^2, x^3+x^2}$

Proof [expand]

$$ \begin{align*} f(x) &= ax^3 + bx^2 + cx + d \br &= z_1 \cdot 1 + z_2 \cdot (x-x^2) + z_3 \cdot x^2 + z_4 (x^3+ x) ,z_1, z_2, z_3, z_4 \in R \br &= z_1+z_2x-z_2x^2 + z_3x^2+z_4x^3+z_4x^2 \br &= z_4x^3 + (z_4+z_3-z_2) x^2 +z_2x+z_1 \end{align*} $$


$$ z_1 = d, z_2 = c, z_3 = b -a +c, z_4 = a $$

Lemma

$V / F$,

$$\spa{S}=V, S \subseteq T, T \subseteq V \implies \spa{T}=V$$

Proof [expand]

$S \subseteq T \implies \spa{S} \subseteq \spa{T}$

$\because \spa{S} = V, \spa{T} \leq V$

$\therefore \spa{S} \leq V \subseteq \spa{T} \subseteq V$

$\spa{T} = V$ i.e. $T$ is also a generating set of $V$.