Difference between revisions of "06-240/Classnotes For Thursday October 5"
From Drorbn
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<math>\mbox{Examples}{}_{}^{}</math> | <math>\mbox{Examples}{}_{}^{}</math> | ||
+ | <math>1. \beta=\emptyset{}_{}^{}\mbox{ is a basis of }\lbrace0\rbrace</math> | ||
+ | |||
+ | <math>2. {}_{}^{}V\mbox{ be }\mathbb{R}\mbox{ as a vector space over }\mathbb{R}</math> | ||
+ | <math>\qquad{}_{}^{}\beta=\lbrace5\rbrace\mbox{ and }\beta=\lbrace1\rbrace\mbox{ are bases.}</math> | ||
+ | |||
+ | <math>3.{}_{}^{}\mbox{ Let }V\mbox{ be }\mathbb{C}\mbox{ as a vector space over }\mathbb{R} \quad\beta=\lbrace1,i\rbrace</math> | ||
+ | |||
+ | :<math>\qquad{}_{}^{}\mbox{Check}</math> | ||
+ | |||
+ | :<math>\qquad{}_{}^{}\mbox{1. Every complex number is a linear combination of }\beta.</math> | ||
+ | ::<math>Z=a+bi=a\cdot 1+b\cdot i\mbox{ with coefficients in }\mathbb{R}\mbox{ so }\lbrace1,i\rbrace\mbox{ generates}</math> | ||
+ | |||
+ | :<math>\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}</math> | ||
+ | ::<math>{}_{}^{}\Rightarrow a+bi=0\Rightarrow a=0\mbox{ and } b=0</math> | ||
+ | |||
+ | <math>{}_{}^{}\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</math> | ||
+ | |||
+ | :<math>{}_{}^{}e_1\ldots e_n\mbox{ are a basis of }V</math> | ||
+ | ::<math>{}_{}^{}\mbox{They span }\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}=\sum a_ie_i</math> | ||
+ | ::<math>{}_{}^{}\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</math> | ||
+ | |||
+ | <math>{}_{}^{}\mbox{5. In }V=P_3(\mathbb{R}),\qquad \beta=\lbrace 1,x,x^2,x^3\rbrace</math> | ||
+ | |||
+ | <math>{}_{}^{}\mbox{6. In }V=P_1(\mathbb{R})=\lbrace ax+b\rbrace,\qquad \beta=\lbrace 1+x,1-x\rbrace\mbox{ is a basis}</math> | ||
+ | :<math>{}_{}^{}\mbox{1. Generate }</math> | ||
+ | ::<math>u_1+u_2=2\Rightarrow \frac{1}{2}(u_1+u_2)=1\mbox{ so }1 \in\mbox{ span }S</math> | ||
+ | ::<math>u_1-u_2=2x\Rightarrow \frac{1}{2}(u_1-u_2)=x\mbox{ so }x \in\mbox{ span }S</math> | ||
+ | ::<math>{}_{}^{}\mbox{ so span}\lbrace 1,x\rbrace \subset\mbox{ span }\beta</math> | ||
+ | :<math>{}_{}^{}\mbox{2. Linearly independent. Assume }au_1+bu_2=0</math> | ||
+ | ::<math>\Rightarrow a(1+x)+b(1-x)=0\Rightarrow a+b+(a-b)x=0</math> | ||
+ | ::<math>{}_{}^{}\Rightarrow a+b=0\mbox{ and }a-b=0</math> | ||
+ | ::<math>(a+b)+(a-b)\Rightarrow 2a=0\Rightarrow a=0</math> | ||
+ | ::<math>(a+b)-(a-b)\Rightarrow 2b=0\Rightarrow b=0</math> | ||
<br> | <br> | ||
Revision as of 20:54, 7 October 2006