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content/Math429/Math429_L4.md

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+# 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.