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Math429
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Math429 L11

Lecture 11

Chapter III Linear maps

Assumption: U,V,WU,V,W are vector spaces (over F\mathbb{F})

Matrices 3C

Definition 3.31

Suppose TโˆˆL(V,W)T\in \mathscr{L}(V,W), v1,...,vnv_1,...,v_n a basis for VV w1,...,wmw_1,...,w_m a basis for WW. Then M(T)=M(T,(v1,...,vn),(w1,...,wm))M(T)=M(T,(v_1,...,v_n),(w_1,...,w_m)) is given by M(T)=AM(T)=An where

Tvk=A1,kw1+...+Am,kwmT_{v_k}=A_{1,k}w_1+...+A_{m,k}w_m v1v2...vnw1w2...wm(A1,1A1,2...A1,nA2,1A2,2...A2,n..................Am,1Am,2...Am,n)\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:F2โ†’F3T:\mathbb{F}^2\to \mathbb{F}^3

    T(x,y)=(x+3y,2x+5y,7x+9y)T(x,y)=(x+3y,2x+5y,7x+9y)

    M(T)=(132579)M(T)=\begin{pmatrix} 1&3\\ 2&5\\ 7&9 \end{pmatrix}

  • Let D:P3(F)โ†’P2(F)D:\mathscr{P}_3(\mathbb{F})\to \mathscr{P}_2(\mathbb{F}) be differentiation

    M(T)=(010000200003)M(T)=\begin{pmatrix} 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\\ \end{pmatrix}

Lemma 3.35

S,TโˆˆL(V,W)S,T\in \mathscr{L}(V,W), M(S+T)=M(S)+M(T)M(S+T)=M(S)+M(T)

Lemma 3.38

โˆ€ฮปโˆˆF,TโˆˆL(V,W)\forall \lambda\in \mathbb{F},T\in \mathscr{L}(V,W), M(ฮปT)=ฮปM(T)M(\lambda T)=\lambda M(T)

M:L(V,W)โ†’Fn,mM:\mathscr{L}(V,W)\to \mathbb{F}^{n,m} is a linear map

Matrix multiplication

Definition 3.41

(AB)j,k=โˆ‘r=1nAj,rBr,k(AB)_{j,k}=\sum^{n}_{r=1} A_{j,r}B_{r,k}

Theorem 3.42

TโˆˆL(U,V),SโˆˆL(V,W)T\in \mathscr{L}(U,V), S\in\mathscr{L}(V,W) then M(S,T)=M(S)M(T)M(S,T)=M(S)M(T) (dim(U)=p,dim(V)=n,dim(W)=mdim (U)=p, dim(V)=n, dim(W)=m)

Proof:

Let w1,...,vnw_1,...,v_n be a basis for VV, w1,..,wmw_1,..,w_m be a basis for WW u1,..,upu_1,..,u_p be a basis of UU.

Let A=M(S),B=M(T)A=M(S),B=M(T)

Compute M(ST)M(ST) by Definition 3.31

(ST)uk=S(T(uk))=S(โˆ‘r=1nBr,kvr)=โˆ‘r=1nBr,k(Svr)=โˆ‘r=1nBr,k(โˆ‘j=1jAj,rwj)=โˆ‘r=1n(โˆ‘j=1jAj,rBr,k)wj=โˆ‘r=1n(M(ST)j,k)wj\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} (M(ST))j,k=โˆ‘r=1nAj,rBr,k=(AB)j,k\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)M(ST)=AB=M(S)M(T)

Notation 3.44

Suppose AA is an mร—nm\times n matrix

then

  1. Aj,โ‹…A_{j,\cdot} denotes the 1ร—n1\times n matrix at the jjth column.
  2. Aโ‹…,kA_{\cdot,k} denotes the mร—1m\times 1 matrix at the kkth column.

Proposition 3.46

Suppose AA is a mร—nm\times n matrix and BB is a nร—pn\times p matrix, then

(AB)j,k=(Aj,โ‹…)โ‹…(Bโ‹…,k)(AB)_{j,k}=(A_{j,\cdot})\cdot (B_{\cdot,k})

Proof:

(AB)j,k=Aj,1B1,k+...+Aj,nBn,k(AB)_{j,k}=A_{j,1}B_{1,k}+...+A_{j,n}B_{n,k}

(Aj,โ‹…)โ‹…(Bโ‹…,k)=(Aj,โ‹…)1,1(Bโ‹…,k)1,1+...+(Aj,โ‹…)1,n(Bโ‹…,k)n,1=Aj,1B1,k+...+Aj,nBn,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 AA is an mร—nm\times n matrix and BB is an nร—pn\times p matrix, then

(A,B)โ‹…,k=A(Bโ‹…,k)(A,B)_{\cdot,k}=A(B_{\cdot,k})

Proposition 3.56

Let AA is an mร—nm\times n b=(b1...bn)b=\begin{pmatrix} b_1\\...\\b_n \end{pmatrix} a xร—1x\times 1 matrix. Then Ab=b1Aโ‹…,1+...+bnAโ‹…,nAb=b_1A_{\cdot,1}+...+b_nA_{\cdot,n}

i.e. AbAb is a linear combination of the columns of AA

Proposition 3.51

Let CC be a mร—cm\times c matrix and RR be a cร—nc\times n matrix, then

  1. column kk of CRCR is a linear combination of the columns of CC with coefficients given by Rโ‹…,kR_{\cdot,k}

    putting the propositions together...

  2. row jj of CRCR is a linear combination of the rows of RR with coefficients given by Cj,โ‹…C_{j,\cdot}