NoteNextra-origin/pages/Math429/Math429_L11.md

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}