NoteNextra-origin/pages/Math429/Math429_L4.md

Lecture 4

Office hour after lecture: Cupules I 109

Chapter II Finite Dimensional Subspaces

Span and Linear Independence 2A

Definition 2.2

Linear combination

Given a list (a finite list), of \mathbb{F} vectors \vec{v_1},...,\vec{v_m}. A linear combination of \vec{v_1},...,\vec{v_m} is a vector \vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m},a_i\in \mathbb{F} (Adding vectors with different weights)

Definition 2.4

Span

The set of all linear combinations of \vec{v_1},...,\vec{v_m} is called the span of \{\vec{v_1},...,\vec{v_m}\}

Span \{\vec{v_1},...,\vec{v_m}\}=\{\vec{v}\in V, \vec{v}=a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m}\textup{ for some }a_i\in \mathbb{F}\}

Note: When there is a nonzero vector in \{\vec{v_1},...,\vec{v_m}\}, the span is a infinite set.

Example:

Consider V=\mathbb{R}^3, find the span of the vector \{(1,2,3),(1,1,1)\},

The span is \{a_1\cdot (1,2,3),a_2\cdot (1,1,1):a_1,a_2\in \mathbb{R}\}=\{(a_1+a_2,2a_1+a_2,3a_1+a_2):a_1,a_2\in \mathbb{R}\}

(-1,0,1)\in Span((1,2,3),(1,1,1))

(1,0,1)\cancel{\in} Span((1,2,3),(1,1,1))

Theorem 2.6

The span of a list of vectors in V is the smallest subspace of V containing this list.

Proof:

1. Span is a subspace

   Span\{\vec{v_1},...,\vec{v_m}\}=\{a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m}\textup{ for some }a_i\in \mathbb{F}\}

   • The zero vecor is inside the span by letting all the a_i=0
   • Closure under addition: a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m}+b_1\vec{v_1}+b_2\vec{v_2}+...+b_m\vec{v_m}=(a_1+b_1)\vec{v_1}+(a_2+b_2)\vec{v_2}+...+(a_m+b_m)\vec{v_m}\in Span\{\vec{v_1},...,\vec{v_m}\}
   • Closure under multiplication: c(a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m})=(ca_1)\vec{v_1}+(ca_2)\vec{v_2}+...+(ca_m)\vec{v_m}\in Span\{\vec{v_1},...,\vec{v_m}\}

2. Span is the smallest subspace containing the given list.

   For each i\in\{1,...,m\}, \vec{v_i}=0\vec{v_1}+...+0\vec{v_{i-1}}+\vec{v_i}+0\vec{v_{i+1}}+...+0\vec{v_m}\in Span\{\vec{v_1},...,\vec{v_m}\}

   If W is a subspace of V containing Span\{\vec{v_1},...,\vec{v_m}\}, then W is closed under addition and scalar multiplication.

   Thus for any a_1,...,a_m\in \mathbb{F},a_1\vec{v_1}+a_2\vec{v_2}+...+a_m\vec{v_m}\in W. So Span\{\vec{v_1},...,\vec{v_m}\}\subset W

Definition 2.ex.1

Spanning set

If a vector space V=Span\{\vec{v_1},...,\vec{v_m}\}, then we say \{\vec{v_1},...,\vec{v_m}\} spans V, which is the spanning set of V.

A vector space is called finite dimensional if it spanned by a finite list.

Example:

\mathbb{F}^n is finite dimensional

\mathbb{R}=Span\{(1,0,0),(0,1,0),(0,0,1)\}

(a,b,c)=a(1,0,0)+b(0,1,0)+c(0,0,1)

Definition

Polynomial

A polynomial is a function p:\mathbb{F}\to \mathbb{F} such that p(Z)=\sum_{i=0}^{m} a_i z^i,a_i\in \mathbb{F}

Let \mathbb{P}(\mathbb{F}) be the set of polynomials over \mathbb{F}, then \mathbb{P}(\mathbb{F}) has the structure of a vector space.

If we consider the degree of polynomials, then f=a_1f_1+...+a_mf_m, with degree f\leq max\{deg(f_1,...,f_m)\}

\mathbb{P}(\mathbb{F}) is a infinite dimensional vector space.

Let \mathbb{P}_m(\mathbb{F}) be the set of polynomials of degree at mote m, then \mathbb{P}_m(\mathbb{F}) is a finite dimensional vectro space.

\mathbb{P}_m(\mathbb{F})=Span\{1,z,z^2,...z^m\}

Linear independence

How to find a "good" spaning set for a finite dimensional vector space.

Example:

V=\mathbb{R^2}

\mathbb{R^2}=Span\{(1,0),(0,1)\}

\mathbb{R^2}=Span\{(1,0),(0,1),(0,0),(1,1)\}

\mathbb{R^2}=Span\{(1,2),(3,1),(4,25)\}

Definition 2.15

A list of vector \vec{v_1},...,\vec{v_m} in V is called linearly independent if the only choice for a_1,...,a_m\in \mathbb{F} such that a_1\vec{v_1}+...+a_m\vec{v_m}=\vec{0} is a_1=...=a_m=0

If not, then there must \exists\vec{v_i} that can be expressed by other vectors in the set.