# 06-240/Classnotes For Thursday October 5

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$\mbox{From last class}{}_{}^{}$

$M_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, M_2=\begin{pmatrix}0&1\\0&0\end{pmatrix}, M_3=\begin{pmatrix}0&0\\1&0\end{pmatrix}, M_4\begin{pmatrix}0&0\\0&1\end{pmatrix}$

$N_1=\begin{pmatrix}0&1\\1&1\end{pmatrix}, N_2=\begin{pmatrix}1&0\\1&1\end{pmatrix}, N_3=\begin{pmatrix}1&1\\0&1\end{pmatrix}, N_4\begin{pmatrix}1&1\\1&0\end{pmatrix}$

$\mbox{The }M_i\mbox{s generate }M_{2\times 2}$

$\mbox{Fact }T\subset\mbox{ span }S\Rightarrow \mbox{ span }T\subset\mbox{ span }S$

$S\subset V\mbox{ is linearly independent }\Leftrightarrow \mbox{ whenever }u_i\in S\mbox{ are distinct}$

$\sum a_iu_i=0\Rightarrow V_ia_i=0 \mbox{ waste not}$

$\mbox{Comments}{}_{}^{}$

1. $\emptyset\subset V\mbox{ is linearly independent}$
2. $\lbrace u\rbrace\mbox{ is linearly independent iff }u_{}^{}\neq 0$
3. $\mbox{If }S_1^{}\subset S_2\subset V$
1. $\mbox{If }S_1^{}\mbox{ is linearly dependent, so is }S_2$
2. $\mbox{If }S_2^{}\mbox{ is linearly dependent, so is }S_1$
3. $\mbox{If }S_1^{}\mbox{ generates }V\mbox{, so does }S_2$
4. $\mbox{If }S_2^{}\mbox{ does not generate }V\mbox{ neither does }S_1$
4. $\mbox{If }S_{}^{}\mbox{ is linearly independent in }V\mbox{ and }v\notin S\mbox{ then }S\cup\lbrace u\rbrace\mbox{ is linearly independent.}$

$\mbox{Proof}{}_{}^{}$

$\mbox{1.}\Leftarrow:\mbox{ start from second assertion and deduce first.}$

$\mbox{Assume }v_{}^{}\in \mbox{span }S$ $v=\sum a_iu_i\mbox{ where }u_i\in S, a_i\in F$

$\sum a_iu_i-1\cdot v=0\mbox{ this is a linear combination of elements in }S\cup v$ $\mbox{ in which not all coefficients are }0 \mbox{ and which add to }0_{}^{}.$ $\mbox{So }S\cup \lbrace v\rbrace\mbox{ is linearly dependent by definition}$
$\mbox{2.}:\Rightarrow\mbox{ Assume }S\cup \lbrace v\rbrace\mbox{ is linearly dependent }\Rightarrow\mbox{ a linear combination can be found, of the form:}$

$(*)\qquad\sum a_iu_i+bv=0\mbox{ where }u_i\in S\mbox{ and not all of the }a_i \mbox{ and }b \mbox{ are }0$

$\mbox{If }b=0\mbox{, then }\sum a_iu_i=0\mbox{ and not }a_i\mbox{s are }0$ ${}_{}^{}\Rightarrow S \mbox{ is linearly dependent}$ ${}_{}^{}\mbox{but initial assumption was }S\mbox{ is linearly independent.}\Rightarrow \mbox{ contradiction so }b\neq0$ $\mbox{So divide by }b\mbox{: (*) becomes }\sum\frac{a_i}{b}u_i + v = 0\Rightarrow v=-\sum\frac{a_i}{b}u_i\Rightarrow v\in \mbox{ span }S$

$\mbox{Definition}{}_{}^{}$

${}_{}^{}\mbox{A basis of a vector space }V\mbox{ is a subset }\beta\subset V$ ${}_{}^{}\mbox{such that}$

1. ${}_{}^{}\beta\mbox{ generates }V\mbox{ or }V=\mbox{ span }\beta$
2. ${}_{}^{}\beta\mbox{ is linearly independent.}$

$\mbox{Examples}{}_{}^{}$

$1. \beta=\emptyset{}_{}^{}\mbox{ is a basis of }\lbrace0\rbrace$

$2. {}_{}^{}V\mbox{ be }\mathbb{R}\mbox{ as a vector space over }\mathbb{R}$ $\qquad{}_{}^{}\beta=\lbrace5\rbrace\mbox{ and }\beta=\lbrace1\rbrace\mbox{ are bases.}$

$3.{}_{}^{}\mbox{ Let }V\mbox{ be }\mathbb{C}\mbox{ as a vector space over }\mathbb{R} \quad\beta=\lbrace1,i\rbrace$

$\qquad{}_{}^{}\mbox{Check}$
$\qquad{}_{}^{}\mbox{1. Every complex number is a linear combination of }\beta.$
$Z=a+bi=a\cdot 1+b\cdot i\mbox{ with coefficients in }\mathbb{R}\mbox{ so }\lbrace1,i\rbrace\mbox{ generates}$
$\qquad{}_{}^{}\mbox{2. Show }\beta=\lbrace1,i\rbrace\mbox{ are linearly independent. Assume }a\cdot 1+b\cdot i=0\mbox{ where }a,b\in\mathbb{R}$
${}_{}^{}\Rightarrow a+bi=0\Rightarrow a=0\mbox{ and } b=0$

${}_{}^{}\mbox{4. }V\in\mathbb{R}^n= \left\lbrace\begin{pmatrix}\vdots\end{pmatrix}y,\qquad e_1=\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}, e_2=\begin{pmatrix}0\\1\\\vdots\\0\end{pmatrix},\ldots, e_n=\begin{pmatrix}0\\0\\\vdots\\1\end{pmatrix}\right\rbrace$

${}_{}^{}e_1\ldots e_n\mbox{ are a basis of }V$
${}_{}^{}\mbox{They span }\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}=\sum a_ie_i$
${}_{}^{}\mbox{They are linearly independent. }\sum a_ie_i=0\Rightarrow \sum a_ie_i= \begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}=0\Rightarrow a_i=0 \quad\forall i$

${}_{}^{}\mbox{5. In }V=P_3(\mathbb{R}),\qquad \beta=\lbrace 1,x,x^2,x^3\rbrace$

${}_{}^{}\mbox{6. In }V=P_1(\mathbb{R})=\lbrace ax+b\rbrace,\qquad \beta=\lbrace 1+x,1-x\rbrace\mbox{ is a basis}$

${}_{}^{}\mbox{1. Generate }$
$u_1+u_2=2\Rightarrow \frac{1}{2}(u_1+u_2)=1\mbox{ so }1 \in\mbox{ span }S$
$u_1-u_2=2x\Rightarrow \frac{1}{2}(u_1-u_2)=x\mbox{ so }x \in\mbox{ span }S$
${}_{}^{}\mbox{ so span}\lbrace 1,x\rbrace \subset\mbox{ span }\beta$
${}_{}^{}\mbox{2. Linearly independent. Assume }au_1+bu_2=0$
$\Rightarrow a(1+x)+b(1-x)=0\Rightarrow a+b+(a-b)x=0$
${}_{}^{}\Rightarrow a+b=0\mbox{ and }a-b=0$
$(a+b)+(a-b)\Rightarrow 2a=0\Rightarrow a=0$
$(a+b)-(a-b)\Rightarrow 2b=0\Rightarrow b=0$

$\mbox{Theorem}{}_{}^{}$

${}_{}^{}\mbox{A subset }\beta\mbox{ of a vectorspace }V \mbox{ is a basis iff every }v\in V\mbox{ can be expressed as}$ ${}_{}^{}\mbox{a linear combination of elements in }$ ${}_{}^{}\beta \mbox{ in exactly one way.}$

$\mbox{Proof}{}_{}^{}$

${}_{}^{}\mbox{It is a combination of things we already know.}$

1. ${}_{}^{}\beta\mbox{ generates}$
2. ${}_{}^{}\beta\mbox{ is linearly independent}$