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