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content/Math429/Math429_L7.md

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+# Lecture 7
+
+## Chapter II Finite Dimensional Subspaces
+
+### Dimension 2C
+
+Intuition: $\mathbb{R}^2$ is two dimensional. $\mathbb{R}^n$ is $n$ dimensional.
+
+#### Definition 2.35
+
+The **dimension** of a finite dimensional vector space denoted $dim(V)$ is the length of any basis of $V$.
+
+Potential issue:
+
+* Why does it not matter which basis I take...
+
+#### Theorem 2.34
+
+Any two basis of a finite dimensional vector spaces have the same length.
+
+Proof: 
+
+Let $V$ be a finite dimensional vector space, and let $B_1,B_2$ (list of vectors) be two basis of $V$. $B_1$ is linearly independent and $B_2$ spans $V$, so by (Theorem 2.22) the length of $B_1$ is less than or equal to $B_2$, By symmetry the length of $B_2$ is less than or equal to the length of $B_1$ so length of $B_1$ = length of $B_2$.
+
+Examples:
+
+$dim\{\mathbb{F}^2\}=2$ because $(0,1),(1,0)$ forms a basis
+
+$dim\{\mathscr{P}_m\}=m+1$ because $z^0,...,z^m$ forms a basis
+
+$dim_{\mathbb{C}}\{\mathbb{C}\}=1$ as a $\mathbb{C}$ vector space, because $1$ forms a basis
+
+$dim_{\mathbb{R}}\{\mathbb{C}\}=2$ as a $\mathbb{R}$ vector space, because $1,i$ forms a basis
+
+#### Proposition 2.37
+
+If a vector space is finite dimensional, then every linearly independent list of length $dim\{V\}$ is a basis.
+
+#### Proposition 2.42
+
+If a vector space is finite dimensional, then every spanning list of length $dim\{V\}$ is a basis for $V$.
+
+Sketch of Proof:
+
+If it's not a basis, extend reduce to a basis, but then that contradicts with **Theorem 2.34**
+
+#### Proposition 2.39
+
+If $U$ is a subspace of a finite dimensional vector space $V$ and $dim\{V\}=dim\{U\}$ then $U=V$
+
+Proof:
+
+Suppose $u_1,...,u_n$ is basis for $U$, then it is linearly independent in $V$. but $dim\{V\}=dim\{U\}$, by **Proposition 2.37**, $u_1,...,u_n$ is a basis of $V$.
+
+So $U=V$.
+
+#### Theorem 2.43   
+
+Let $V_1$ and $V_2$ be subspaces of a finite dimensional vector space $V$, then $dim\{V_1+V_2\}=dim\{V_1\}+dim\{V_2\}-dim\{V_1\bigcap V_2\}$
+
+Proof:
+
+Let $u_1,...,u_m$ be a basis for $V_1\bigcap V_2$,   
+then extend by $v_1,...,v_k,u_1,...,u_m$ to a basis of $V_1$,  
+then extend to $u_1,...,u_m,w_1,..,w_l$ a basis of $V_2$.
+
+Then I claim $v_1,...,v_k,u_1,...,u_m,w_1,...,w_l$ is a basis of $V_1+V_2$
+
+Note: given the above statement, we have $dim\{V_1+V_2\}=k+m+l=(k+m)+(m+l)-m=dim\{V_1\}+dim\{V_2\}-dim\{V_1\bigcap V_2\}$
+
+So showing $v_1,...,v_k,u_1,...,u_m,w_1,...,w_l$ is a basis suffices.
+
+Since $V_1,V_2\subseteq Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}$, $V_1+V_2\subseteq Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}$.
+
+Since $V_1+V_2$ is the smallest subspace contains both $V_1,V_2$, $v_i,u_k,w_j\in V_1+V_2$, $V_1+V_2= Span\{v_1,...,v_k,u_1,...,u_m,w_1,...,w_l\}$
+
+So the list above spans $V_1+V_2$.
+
+Suppose $a_1 v_1+...+a_k v_k=-b_1 u_1-...-b_m u_m-c_1 w_1-...- c_e w_l\in V_1\bigcap V_2$...