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

$A_{j,\cdot}$ denotes the $1\times n$ matrix at the $j$ th column.
$A_{\cdot,k}$ 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

column $k$ of $CR$ is a linear combination of the columns of $C$ with coefficients given by $R_{\cdot,k}$

putting the propositions together…
row $j$ of $CR$ is a linear combination of the rows of $R$ with coefficients given by $C_{j,\cdot}$