06-240/Classnotes For Thursday October 5

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

$M_{1}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},M_{2}={\begin{pmatrix}0&1\\0&0\end{pmatrix}},M_{3}={\begin{pmatrix}0&0\\1&0\end{pmatrix}},M_{4}{\begin{pmatrix}0&0\\0&1\end{pmatrix}}$

$N_{1}={\begin{pmatrix}0&1\\1&1\end{pmatrix}},N_{2}={\begin{pmatrix}1&0\\1&1\end{pmatrix}},N_{3}={\begin{pmatrix}1&1\\0&1\end{pmatrix}},N_{4}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}$

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

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

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

$\sum a_{i}u_{i}=0\Rightarrow V_{i}a_{i}=0{\mbox{ waste not}}$

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

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. ${\mbox{If }}S_{}^{}{\mbox{ is linearly independent in }}V{\mbox{ and }}v\notin S{\mbox{ then }}S\cup \lbrace u\rbrace {\mbox{ is linearly independent.}}$

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

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

${\mbox{Assume }}v_{}^{}\in {\mbox{span }}S$ $v=\sum a_{i}u_{i}{\mbox{ where }}u_{i}\in S,a_{i}\in F$

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

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

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

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

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

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

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

$1.\beta =\emptyset {}_{}^{}{\mbox{ is a basis of }}\lbrace 0\rbrace$

$2.{}_{}^{}V{\mbox{ be }}\mathbb {R} {\mbox{ as a vector space over }}\mathbb {R}$ $\qquad {}_{}^{}\beta =\lbrace 5\rbrace {\mbox{ and }}\beta =\lbrace 1\rbrace {\mbox{ are bases.}}$

$3.{}_{}^{}{\mbox{ Let }}V{\mbox{ be }}\mathbb {C} {\mbox{ as a vector space over }}\mathbb {R} \quad \beta =\lbrace 1,i\rbrace$

$\qquad {}_{}^{}{\mbox{Check}}$

$\qquad {}_{}^{}{\mbox{1. Every complex number is a linear combination of }}\beta .$

$Z=a+bi=a\cdot 1+b\cdot i{\mbox{ with coefficients in }}\mathbb {R} {\mbox{ so }}\lbrace 1,i\rbrace {\mbox{ generates}}$

$\qquad {}_{}^{}{\mbox{2. Show }}\beta =\lbrace 1,i\rbrace {\mbox{ are linearly independent. Assume }}a\cdot 1+b\cdot i=0{\mbox{ where }}a,b\in \mathbb {R}$

${}_{}^{}\Rightarrow a+bi=0\Rightarrow a=0{\mbox{ and }}b=0$

${}_{}^{}{\mbox{4. }}V\in \mathbb {R} ^{n}=\left\lbrace {\begin{pmatrix}\vdots \end{pmatrix}}y,\qquad e_{1}={\begin{pmatrix}1\\0\\\vdots \\0\end{pmatrix}},e_{2}={\begin{pmatrix}0\\1\\\vdots \\0\end{pmatrix}},\ldots ,e_{n}={\begin{pmatrix}0\\0\\\vdots \\1\end{pmatrix}}\right\rbrace$

${}_{}^{}e_{1}\ldots e_{n}{\mbox{ are a basis of }}V$

${}_{}^{}{\mbox{They span }}{\begin{pmatrix}a_{1}\\\vdots \\a_{n}\end{pmatrix}}=\sum a_{i}e_{i}$

${}_{}^{}{\mbox{They are linearly independent. }}\sum a_{i}e_{i}=0\Rightarrow \sum a_{i}e_{i}={\begin{pmatrix}a_{1}\\\vdots \\a_{n}\end{pmatrix}}=0\Rightarrow a_{i}=0\quad \forall i$

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

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

${}_{}^{}{\mbox{1. Generate }}$

$u_{1}+u_{2}=2\Rightarrow {\frac {1}{2}}(u_{1}+u_{2})=1{\mbox{ so }}1\in {\mbox{ span }}S$

$u_{1}-u_{2}=2x\Rightarrow {\frac {1}{2}}(u_{1}-u_{2})=x{\mbox{ so }}x\in {\mbox{ span }}S$

${}_{}^{}{\mbox{ so span}}\lbrace 1,x\rbrace \subset {\mbox{ span }}\beta$

${}_{}^{}{\mbox{2. Linearly independent. Assume }}au_{1}+bu_{2}=0$

$\Rightarrow a(1+x)+b(1-x)=0\Rightarrow a+b+(a-b)x=0$

${}_{}^{}\Rightarrow a+b=0{\mbox{ and }}a-b=0$

$(a+b)+(a-b)\Rightarrow 2a=0\Rightarrow a=0$

$(a+b)-(a-b)\Rightarrow 2b=0\Rightarrow b=0$

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

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

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

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

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