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Math429
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Math429 L12

Lecture 12

Chapter III Linear maps

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

Matrices 3C

Proposition 3.51

Let CC be an m×cm\times c matrix and RR be a c×nc\times n matrix, then

  1. column kk of CRCR is a linear combination of the columns of CC with coefficients given by R,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,C_{j,\cdot}

Column-Row Factorization and Rank

Definition 3.52

Let AA be an m×nm \times n matrix, then

  • The column rank of AA is the dimension of the span of the columns in Fm,1\mathbb{F}^{m,1}.
  • The row range of AA is the dimension of the span of the row in F1,n\mathbb{F}^{1,n}.

Transpose: At=ATA^t=A^T refers to swapping rows and columns

Theorem 3.56 (Column-Row Factorization)

Let AA be an m×jm\times j matrix with column rank cc. Then there exists an m×cm\times c matrix CC and c×xc\times x matrix RR such that A=CRA=CR

Proof:

Let V=Span{A,1,...,A,n}V=Span\{A_{\cdot,1},...,A_{\cdot,n}\}, let C,1,...,C,cC_{\cdot, 1},...,C_{\cdot, c} be a basis of VV. Since these forms a basis, there exists Rj,kR_{j,k} such that Ai,j=j=1cCi,jRj,kA_{i,j}=\sum_{j=1}^c C_{i,j}R_{j,k}, so A,j=j=1cC,jRj,kA_{\cdot,j}=\sum_{j=1}^c C_{\cdot,j}R_{j,k}. This implies that A=CRA=CR by construction CC is m×cm\times c, RR is c×nc\times n.

Example:

A=(142258364)=(142536)(101012),rank A=4A=\begin{pmatrix} 1&4&2\\ 2&5&8\\ 3&6&4 \end{pmatrix}=\begin{pmatrix} 1&4\\ 2&5\\ 3&6\\ \end{pmatrix}\begin{pmatrix} 1&0&-1\\ 0&1&2\\ \end{pmatrix},rank\ A=4

Definition 3.58 Rank

The rank of a matrix AA is the column rank of AA denoted rank Arank\ A.

Theorem 3.57

Given a matrix AA the column rank equals the row rank.

Proof:

Note that by Theorem 3.56, if AA is m×nm\times n and has column rank cc. A=CRA=CR for some CC is a m×cm\times c matrix, RR is a c×nc\times n matrices, ut the rows of CRCR are a linear combination of the rows of RR, and row rank of RCR\leq C. So row rank AA\leq column rank of AA.

Taking a transpose of matrix, then row rank of ATA^T (column rank of AA) \leq column rank of ATA^T (row rank AA).

So column rank is equal to row rank.

Invertibility and Isomorphisms 3D

Invertible Linear Maps

Definition 3.59

A linear map TL(V,W)T\in\mathscr{L}(V,W) is invertible if there exists SL(W,V)S\in \mathscr{L}(W,V) such that ST=IVST=I_V and TS=IWTS=I_W. Such a SS is called an inverse of TT.

Note: ST=IVST=I_V and TS=IWTS=I_W must both be true for inverse map.

Lemma 3.60

Every linear map has an unique inverse.

Proof: Exercise and answer in the book.

Notation: T1T^{-1} is the inverse of TT

Theorem 3.63

A linear map T:VWT:V\to W invertible if and only if its injective and surjective.

Proof:

\Rightarrow

null(T)={0}null(T)=\{0\} since T(v)=0    (T1))(T(v))=0    range(T)=WT(v)=0\implies (T^{-1}))(T(v))=0\implies range (T)=W let wWw\in W then T(T1(w))=w,wrange(T)T(T^{-1}(w))=w,w\in range (T)

\Leftarrow

Find S:WVS:W\to V a function such that T(S(v))=vT(S(v))=v by letting S(v)S(v) be the unique vector in vv such that T(S(v))=vT(S(v))=v. Goal: Show S:WVS:W\to V is linear

ST(S(w1)+S(w2))=S(w1)+S(w2)S(T(S(w1)))+T(S(w2))=S(w1+w2)ST(S(w_1)+S(w_2))=S(w_1)+S(w_2)\\ S(T(S(w_1)))+T(S(w_2))=S(w_1+w_2)