133 lines
3.6 KiB
Markdown
133 lines
3.6 KiB
Markdown
# Lecture 14
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## Chapter III Linear maps
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**Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)**
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### Matrices 3C
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Review
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#### Proposition 3.76
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$$
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M(Tv)=M(T)M(v)
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$$
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#### Theorem 3.78
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Let $V,W$ be finite dimensional vector space, and $T\in \mathscr{L}(V,W)$ then $dim\ range\ T=column\ rank (M(T))=rank(M(T))$
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Proof:
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$range=Span\{Tv_1,...,Tv_n\}$ compare to $Span\{M(T)_{\cdot,1},...,M(T)_{\cdot, n}\}=Span\{M(T)M(v_1),...,M(T)M(v_n)\}=Span\{M(Tv_1),...,M(Tv_n)\}$
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Since $M$ is a isomorphism, then the theorem makes sense.
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#### Change of Basis
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#### Definition 3.79, 3.80
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The identity matrix
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$$
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I=\begin{pmatrix}
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1.& 0\\
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0& '1\\
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\end{pmatrix}
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$$
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The inverse matrix of an invertible matrix $A$ denoted $A^{-1}$ is the matrix such that
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$$
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AA^{-1}=I=A^{-1}A
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$$
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Question: Let $u_1,...,u_n$ and $v_1,...,v_n$ be two bases for $V$. What is $M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V)$
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#### Proposition 3.82
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Let $u_1,...,u_n$ and $v_1,...,v_n$ be bases of $V$, then $M(I,(u_1,...,u_n),(v_1,...,v_n)),I\in \mathscr{L}(V)$ and $M(I,(v_1,...,v_n),(u_1,...,u_n)),I\in \mathscr{L}(V)$ are inverse to each other.
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Proof:
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$$
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M(I,(u_1,...,u_n),(v-1,...,v_n)),I\in \mathscr{L}(V) M(I,(v-1,...,v_n),(u_1,...,u_n))=M(I,(u_1,...,u_n),(u_1,...,u_n))
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$$
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#### Theorem 3.84 Change of Basis
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Let $u_1,...,u_n$ and $v_1,...,v_n$ be two bases for $V$ and $T\in \mathscr{L}(v), A=M(T,(u_1,...,u_n)), B=M(T,(v_1,...,v_n)), C=M(I,(u_1,...,u_n),(v_1,...,v_n))$, then $A=C^{-1}BC$
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#### Theorem 3.86
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Let $T\in \mathscr{L}(v)$ be an invertible linear map, then $M(T^{-1})=M(T)^{-1}$
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### Products and Quotients of Vector Spaces 3E
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Goals: To construct vectors spaces from other vector spaces.
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#### Definition 3.87
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Suppose $V_1,...,V_m$ vectors spaces over some field $\mathbb{F}$, then the product is given by
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$$
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V_1\times ...\times V_n=\{(v_1,v_2,...,v_n)\vert v_1\in V_1, v_2\in V_2,...,v_n\in V_n\}
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$$
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with addition given by
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$$
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(v_1,...,v_n)+(u_1,...,u_n)=(v_1+u_1,...,v_n+u_n)
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$$
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and scalar multiplication
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$$
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\lambda (v_1,...,v_n)=(\lambda v_1,...,\lambda v_n),\lambda \in \mathbb{F}
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$$
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#### Theorem 3.89
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If $v_1,...,v_n$ are vectors paces over $\mathbb{F}$ then $V_1\times ...\times V_n$ is a vector space over $\mathbb{F}$
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Example:
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$V=\mathscr{P}_2(\mathbb{R})\times \mathbb{R}^2=\{(p,v)\vert p\in \mathscr{P}_2(\mathbb{R}), v\in \mathbb{R}^2\}=\{(a_0+a_1x+a_2x,(b,c))\vert a_0,a_1,a_2,b,c\in \mathbb{R}\}$
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A basis for $V$ would be $(1,(0,0)),(x,(0,0)),(x^2,(0,0)),(0,(1,0)),(0,(0,1))$
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#### Theorem 3.92
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$$
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dim(V_1\times ...\times V_n)=dim(V_1)+...+dim(V_n)
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$$
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Sketch of proof:
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take a basis for each $V_k$, make them vectors in the product then combine the entire list of vector to be basis.
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Example:
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$\mathbb{R}^2\times \mathbb{R}^3=\{((a,b),(c,d,e))\vert a,b,c,d,e\in \R\}$
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$\mathbb{R}^2\times \mathbb{R}^3\cong \mathbb{R}^5,((a,b),(c,d,e))\mapsto(a,b,c,d,e)$
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#### Theorem 3.93
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Let $V_1,...,V_m\subseteq V$, define $\Gamma: V_1\times...\times V_m\to V_1+...+V_m$. $\Gamma(v_1,...,v_n)=v_1+...+v_n$ then $\Gamma$ is always surjective. And it is injective if and only if $V_1+...+V_m$ is a direct sum.
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Sketch of the proof:
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injective $\iff null\ T\{ (0,...,0) \} \iff$ the only way to write $0=v_1,...,v_m$ is $v_1=...=v_n=0 \iff$ then $V_1+...+V_m$ is a direct sum
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#### Theorem 3.94
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$V_1+...+V_m$ is a direct sum if and only if $dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)$
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Proof:
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Use $\Gamma$ above is an isomorphism $\iff$ $V_1+...+V_m$ is a direct sum
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Use $\Gamma$ above is an isomorphism $\implies dim(V_1+...+V_m)=dim(V_1)+...+dim(V_m)$
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