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content/Math429/Math429_L11.md

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+# Lecture 11
+
+## Chapter III Linear maps
+
+**Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)**
+
+### Matrices 3C
+
+#### Definition 3.31
+
+Suppose $T\in \mathscr{L}(V,W)$, $v_1,...,v_n$ a basis for $V$ $w_1,...,w_m$ a basis for $W$. Then $M(T)=M(T,(v_1,...,v_n),(w_1,...,w_m))$ is given by $M(T)=A$n where
+
+$$
+T_{v_k}=A_{1,k}w_1+...+A_{m,k}w_m
+$$
+
+$$
+\begin{matrix}
+    & & v_1& & v_2&&...&v_n&
+\end{matrix}\\
+\begin{matrix}
+w_1\\w_2\\.\\.\\.\\w_m
+\end{matrix}
+\begin{pmatrix}
+A_{1,1} & A_{1,2} &...& A_{1,n}\\
+A_{2,1} & A_{2,2} &...& A_{2,n}\\
+. & . &...&.\\
+. & . &...&.\\
+. & . &...&.\\
+A_{m,1} & A_{m,2} &...& A_{m,n}\\
+\end{pmatrix}
+$$
+
+Example:
+
+* $T:\mathbb{F}^2\to \mathbb{F}^3$
+  
+    $T(x,y)=(x+3y,2x+5y,7x+9y)$
+
+    $M(T)=\begin{pmatrix}
+        1&3\\
+        2&5\\
+        7&9
+    \end{pmatrix}$
+
+* Let $D:\mathscr{P}_3(\mathbb{F})\to \mathscr{P}_2(\mathbb{F})$ be differentiation
+
+    $M(T)=\begin{pmatrix}
+        0&1&0&0\\
+               0&0&2&0\\
+        0&0&0&3\\
+    \end{pmatrix}$
+
+#### Lemma 3.35
+
+$S,T\in \mathscr{L}(V,W)$, $M(S+T)=M(S)+M(T)$
+
+#### Lemma 3.38
+
+$\forall \lambda\in \mathbb{F},T\in \mathscr{L}(V,W)$, $M(\lambda T)=\lambda M(T)$
+
+$M:\mathscr{L}(V,W)\to \mathbb{F}^{n,m}$ is a linear map
+
+#### Matrix multiplication
+
+#### Definition 3.41
+
+$$
+(AB)_{j,k}=\sum^{n}_{r=1} A_{j,r}B_{r,k}
+$$
+
+#### Theorem 3.42
+
+$T\in \mathscr{L}(U,V), S\in\mathscr{L}(V,W)$ then $M(S,T)=M(S)M(T)$ ($dim (U)=p, dim(V)=n, dim(W)=m$)
+
+Proof:
+
+Let $w_1,...,v_n$ be a basis for $V$, $w_1,..,w_m$ be a basis for $W$ $u_1,..,u_p$ be a basis of $U$.
+
+Let $A=M(S),B=M(T)$
+
+Compute $M(ST)$ by **Definition 3.31**
+
+$$
+\begin{aligned}
+(ST)u_k&=S(T(u_k))\\
+&=S(\sum^n_{r=1}B_{r,k}v_r)\\
+&=\sum^n_{r=1} B_{r,k}(S_{v_r})\\
+&=\sum^n_{r=1} B_{r,k}(\sum^j_{j=1}A_{j,r} w_j)\\
+&=\sum^n_{r=1} (\sum^j_{j=1}A_{j,r}B_{r,k})w_j\\
+&=\sum^n_{r=1} (M(ST)_{j,k})w_j\\
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+(M(ST))_{j,k}&=\sum^n_{r=1}A_{j,r}B_{r,k}\\
+&=(AB)_{j,k}
+\end{aligned}
+$$
+
+$$
+M(ST)=AB=M(S)M(T)
+$$
+
+#### Notation 3.44
+
+Suppose $A$ is an $m\times n$ matrix
+
+then 
+
+1. $A_{j,\cdot}$ denotes the $1\times n$ matrix at the $j$th column.
+2. $A_{\cdot,k}$ denotes the $m\times 1$ matrix at the $k$th column.
+
+#### Proposition 3.46
+
+Suppose $A$ is a $m\times n$ matrix and $B$ is a $n\times p$ matrix, then 
+
+$$
+(AB)_{j,k}=(A_{j,\cdot})\cdot (B_{\cdot,k})
+$$
+
+Proof:
+
+$(AB)_{j,k}=A_{j,1}B_{1,k}+...+A_{j,n}B_{n,k}$
+
+$(A_{j,\cdot})\cdot (B_{\cdot,k})=(A_{j,\cdot})_{1,1}(B_{\cdot,k})_{1,1}+...+(A_{j,\cdot})_{1,n}(B_{\cdot,k})_{n,1}=A_{j,1}B_{1,k}+...+A_{j,n}B_{n,k}$
+
+#### Proposition 3.48
+
+Suppose $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, then
+
+$$
+(A,B)_{\cdot,k}=A(B_{\cdot,k})
+$$
+
+#### Proposition 3.56
+
+Let $A$ is an $m\times n$ $b=\begin{pmatrix}
+    b_1\\...\\b_n
+\end{pmatrix}$ a $x\times 1$ matrix. Then $Ab=b_1A_{\cdot,1}+...+b_nA_{\cdot,n}$
+
+i.e. $Ab$ is a linear combination of the columns of $A$
+
+#### Proposition 3.51
+
+Let $C$ be a $m\times c$ matrix and $R$ be a $c\times n$ matrix, then 
+
+1. column $k$ of $CR$ is a linear combination of the columns of $C$ with coefficients given by $R_{\cdot,k}$
+
+    *putting the propositions together...*
+
+2. row $j$ of $CR$ is a linear combination of the rows of $R$ with coefficients given by $C_{j,\cdot}$