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