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}