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content/Math429/Math429_L10.md

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+# Lecture 10
+
+## Chapter III Linear maps
+
+**Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)**
+
+### Vector Space of Linear Maps 3A
+
+Review
+
+#### Theorem 3.21 (The Fundamental Theorem of Linear Maps, Rank-nullity Theorem)
+
+Suppose $V$ is finite dimensional, and $T\in \mathscr{L}(V,W)$, then $range(T)$ is finite dimensional ($W$ don't need to be finite dimensional). and 
+
+$$
+dim(V)=dim(null (T))+dim(range(T))
+$$
+
+Proof:
+
+Let $u_1,...,u_m$ be a basis for $null(T)$, then we extend to a basis of $V$ given by $u_1,...,u_m,v_1,...,v_m$, we have $dim(V)=m+n$. Claim that $Tv_1,...,Tv_n$ forms a basis for $range (T)$. Need to show 
+
+* Linearly independent. (in Homework 3)
+* These span $range(T)$.  
+  
+  Let $w\in range(T)$ the there exists $v\in V$ such that $Tv=W$, $u_1,...,u_m,v_1,...,v_m$ are basis so $\exists a_1,...,a_m,b_1,...,b_n$ such that $v=a_1u_1+...+a_mu_m+b_1v_1+...+b_n v_n$. $Tv=a_1Tu_1+...+a_mTu_m+b_1Tu_1+...+b_nTv_n$. 
+
+  Since $u_k\in null(T)$, So $Tv_1,...,Tv_n$ spans range $T$ and so form a basis. Thus $range(T)$ is finite dimensional and $dim(range(T))=n$. So $dim(V)=dim(null (T))+dim(range(T))$
+
+#### Theorem 3.22
+
+Suppose $V,W$ are finite dimensional with $dim(V)>dim(W)$, then there are no injective maps from $V$ to $W$.
+
+#### Theorem 3.24
+
+Suppose $V,W$ are finite dimensional with $dim(V)<dim(W)$, then there are no surjective maps from $V$ to $W$.
+
+Ideas of Proof: relies on **Theorem 3.21** $dim(null(T))>0$
+
+### Linear Maps and Linear Systems 3EX-1
+
+Suppose we have a homogeneous linear system * with $m$ equation and $n$ variables.
+
+$$
+A_{11} x_1+ ... + A_{1n} x_n=0\\
+...\\
+A_{m1} x_1+ ... + A_{mn} x_n=0
+$$
+
+which is equivalent to 
+
+$$
+A\begin{bmatrix}
+    x_1\\...\\x_n
+\end{bmatrix}=\vec{0}
+$$
+
+also equivalent to
+
+$$
+T(v)=0,\textup{ for some }T
+$$
+
+$$
+T(x_1,...,x_n)=(A_{11} x_1+ ... + A_{1n},...,A_{m1} x_1+ ... + A_{mn} x_n),T\in \mathscr{L}(\mathbb{R}^n,\mathbb{R}^m)
+$$
+
+Solution to * is $null(T)$.
+
+#### Proposition 3.26
+
+A homogeneous linear system with more variables than equations has non-zero solutions.
+
+Proof:
+
+Using $T$ as above, note that since $n>m$, use **Theorem 3.22**, implies that $T$ cannot be injective. So, $null (T)$ contains a non-zero vector.
+
+#### Proposition 3.28
+
+An in-homogenous system with more equations than variables has no solutions for some choices of constants. ($A\vec{x}=\vec{b}$ for some $\vec{b}$ this has no solution)
+
+### Matrices 3A
+
+#### Definition 3.29
+
+For $m,n>0$ and $m\times n$ matrix $A$ is a rectangular array with elements of the $\mathbb{F}$ given by
+
+$$
+A=\begin{pmatrix}
+    A_{1,1}& ...&A_{1,n}\\
+    ... & & ...\\
+    A_{n,1}&...&A_{m,n}\\
+\end{pmatrix}
+$$
+
+### Operations on matrices
+
+Addition:
+
+$$
+A+B=\begin{pmatrix}
+    A_{1,1}+B_{1,1}& ...&A_{1,n}+B_{1,n}\\
+    ... & & ...\\
+    A_{n,1}+A_{n,1}&...&A_{m,n}+B_{m,n}\\
+\end{pmatrix}
+$$
+
+**for $A+B$, $A,B$ need to be the same size**
+
+Scalar multiplication: 
+
+$$
+\lambda A=\begin{pmatrix}
+    \lambda A_{1,1}& ...& \lambda A_{1,n}\\
+    ... & & ...\\
+    \lambda A_{n,1}&...& \lambda A_{m,n}\\
+\end{pmatrix}
+$$
+
+#### Definition 3.39
+
+$\mathbb{F}^{m,n}$ is the set of $m$ by $n$ matrices.
+
+#### Theorem 3.40
+
+$\mathbb{F}^{m,n}$ is a vector space (over $\mathbb{F}$) with $dim(\mathbb{F}^{m,n})=m\times n$
+
+### Matrix multiplication 3EX-2
+
+Let $A$ be a $m\times n$ matrix and $B$ be an $n\times s$ matrix
+
+$$
+(A,B)_{i,j}= \sum^n_{r=1} A_{i,r}\cdot B_{r,j}
+$$
+
+Claim:
+
+This formula comes from multiplication of linear maps.
+
+#### Definition 3.44
+
+Linear maps to matrices, let $V$, $W$, $Tv_i$ written in terms of $w_i$.
+
+$$
+M(T)=\begin{pmatrix}
+    Tv_1\vert Tv_2\vert ...\vert Tv_n
+\end{pmatrix}
+$$