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content/Math429/Math429_L14.md

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