107 lines
3.1 KiB
Markdown
107 lines
3.1 KiB
Markdown
# Lecture 12
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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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#### Proposition 3.51
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Let $C$ be an $m\times c$ 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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#### Column-Row Factorization and Rank
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#### Definition 3.52
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Let $A$ be an $m \times n$ matrix, then
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* The column rank of $A$ is the dimension of the span of the columns in $\mathbb{F}^{m,1}$.
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* The row range of $A$ is the dimension of the span of the row in $\mathbb{F}^{1,n}$.
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> Transpose: $A^t=A^\top$ refers to swapping rows and columns
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#### Theorem 3.56 (Column-Row Factorization)
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Let $A$ be an $m\times j$ matrix with column rank $c$. Then there exists an $m\times c$ matrix $C$ and $c\times x$ matrix $R$ such that $A=CR$
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Proof:
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Let $V=Span\{A_{\cdot,1},...,A_{\cdot,n}\}$, let $C_{\cdot, 1},...,C_{\cdot, c}$ be a basis of $V$. Since these forms a basis, there exists $R_{j,k}$ such that $A_{i,j}=\sum_{j=1}^c C_{i,j}R_{j,k}$, so $A_{\cdot,j}=\sum_{j=1}^c C_{\cdot,j}R_{j,k}$. This implies that $A=CR$ by construction $C$ is $m\times c$, $R$ is $c\times n$.
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Example:
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$$
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A=\begin{pmatrix}
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1&4&2\\
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2&5&8\\
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3&6&4
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\end{pmatrix}=\begin{pmatrix}
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1&4\\
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2&5\\
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3&6\\
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\end{pmatrix}\begin{pmatrix}
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1&0&-1\\
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0&1&2\\
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\end{pmatrix},rank\ A=4
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$$
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#### Definition 3.58 Rank
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The **rank of a matrix** $A$ is the column rank of $A$ denoted $rank\ A$.
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#### Theorem 3.57
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Given a matrix $A$ the column rank equals the row rank.
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Proof:
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Note that by **Theorem 3.56**, if $A$ is $m\times n$ and has column rank $c$. $A=CR$ for some $C$ is a $m\times c$ matrix, $R$ is a $c\times n$ matrices, ut the rows of $CR$ are a linear combination of the rows of $R$, and row rank of $R\leq C$. So row rank $A\leq$ column rank of $A$.
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Taking a transpose of matrix, then row rank of $A^\top$ (column rank of $A$) $\leq$ column rank of $A^\top$ (row rank $A$).
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So column rank is equal to row rank.
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### Invertibility and Isomorphisms 3D
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Invertible Linear Maps
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#### Definition 3.59
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A linear map $T\in\mathscr{L}(V,W)$ is **invertible** if there exists $S\in \mathscr{L}(W,V)$ such that $ST=I_V$ and $TS=I_W$. Such a $S$ is called an **inverse** of $T$.
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Note: $ST=I_V$ and $TS=I_W$ must **both be true** for inverse map.
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#### Lemma 3.60
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Every linear map has an unique inverse.
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Proof: Exercise and answer in the book.
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Notation: $T^{-1}$ is the inverse of $T$
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#### Theorem 3.63
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A linear map $T:V\to W$ invertible if and only if its injective and surjective.
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Proof:
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$\Rightarrow$
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$null(T)=\{0\}$ since $T(v)=0\implies (T^{-1}))(T(v))=0\implies range (T)=W$ let $w\in W$ then $T(T^{-1}(w))=w,w\in range (T)$
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$\Leftarrow$
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Find $S:W\to V$ a function such that $T(S(v))=v$ by letting $S(v)$ be the unique vector in $v$ such that $T(S(v))=v$. Goal: Show $S:W\to V$ is linear
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$$
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ST(S(w_1)+S(w_2))=S(w_1)+S(w_2)\\
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S(T(S(w_1)))+T(S(w_2))=S(w_1+w_2)
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$$
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