06-240/Classnotes For Thursday, September 28: Difference between revisions

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===Linear Combination===
===Linear Combination===
Definition: Let (''u''<sub>i</sub>) = (''u''<sub>1</sub>, ''u''<sub>2</sub>, ..., ''u''<sub>n</sub>) be a sequence of vectors in V. A sum of the form<br>
::''a''<sub>i</sub> <math> \in </math> F, <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub> = ''a''<sub>1</sub>''u''<sub>1</sub> + ''a''<sub>2</sub>''u''<sub>2</sub>+ ... +''a''<sub>n</sub>''u''<sub>n</sub>


<math>\mbox{Definition: Let }(u_i) = (u_1,u_2,\ldots,u_n)\mbox{ be a sequence of vectors in }V</math>.
is called a "Linear Combination" of the ''u''<sub>i</sub>.

<math>\mbox{A sum of the form:}{}_{}^{}</math>

<math> a_i\in F,\sum_{i=1}^n a_i u_i = a_1u_1 + a_2u_2+\ldots+a_nu_n</math>

<math>\mbox{is called a Linear Combination of the }u_i^{ }</math>.


===Span===
===Span===
span(''u''<sub>i</sub>):= The set of all possible linear combinations of the ''u''<sub>i</sub>'s.
<math>\mbox{span}(u_i^{ }):= \lbrace\mbox{ The set of all possible linear combinations of the } u_i^{ }\rbrace</math>


<math>\mbox{If }\mathcal{S} \subset V\ \mbox{ is any subset, }</math>


<math>\mbox{span}(\mathcal{S}):= \lbrace\mbox{The set of all linear combination of vectors in }\mathcal{S}\rbrace=\left\lbrace\sum_{i=0}^n a_i u_i,\quad a_i \in F, u_i \in \mathcal{S}\right\rbrace</math>
If <math>\mathcal{S} \subseteq</math> V is any subset,
:
{| border="0" cellpadding="0" cellspacing="0"
|-
|span <math>\mathcal{S}</math>
|:= The set of all linear combination of vectors in <math>\mathcal{S}</math>
|-
|
|=<math>\left \{ \sum_{i=0}^n a_i u_i, a_i \in \mbox{F}, u_i \in \mathcal{S} \right \} \ni 0</math>
|}


even if <math>\mathcal{S}</math> is empty.
<math>\mbox{span}(\mathcal{S})\mbox{ always contains }0\mbox{ even if }\mathcal{S}=\emptyset</math>


'''Theorem'''
'''Theorem''': For any <math>\mathcal{S} \subseteq</math> V, span <math>\mathcal{S}</math> is a subspace of V.
<math>\forall\mathcal{S} \subset V\mbox{, span}(\mathcal{S})\mbox{ is a subspace of }V</math>


Proof:<br>
<math>\mbox{Proof:}{}_{}^{}</math>

1. 0 <math> \in </math> span <math>\mathcal{S}</math>.<br>
2. Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>, Let ''x'' <math> \in </math> span <math>\mathcal{S}</math>,
1. <math>0 \in\mbox{ span}(\mathcal{S})</math>.<br>
<math>\Rightarrow</math> ''x'' = <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub>, ''u''<sub>i</sub> <math> \in \mathcal{S}</math>, ''y'' = <math>\sum_{i=1}^m</math> ''b''<sub>i</sub>''v''<sub>i</sub>, ''v''<sub>i</sub> <math> \in \mathcal{S}</math>.
2. <math>\mbox{Let }x \in \mbox{ span}(\mathcal{S})\Rightarrow x =\sum_{i=1}^n a_iu_i\mbox{, }u_i\in \mathcal{S}\mbox{, }</math>

<math>\Rightarrow</math> ''x''+''y'' = <math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub> + <math>\sum_{i=1}^m</math> ''b''<sub>i</sub>''v''<sub>i</sub> = <math>\sum_{i=1}^{m+n}</math> ''c''<sub>i</sub>''w''<sub>i</sub> where ''c''<sub>i</sub>=(''a''<sub>1</sub>, ''a''<sub>2</sub>,...,''a''<sub>n</sub>, ''b''<sub>1</sub>, ''b''<sub>2</sub>,...,''b''<sub>m</sub>) and ''w''<sub>i</sub>=''c''<sub>i</sub>=(''u''<sub>1</sub>, ''u''<sub>2</sub>,...,''u''<sub>n</sub>, ''v''<sub>1</sub>, ''v''<sub>2</sub>,...,''v''<sub>m</sub>).<br>
3. ''cx''= c<math>\sum_{i=1}^n</math> ''a''<sub>i</sub>''u''<sub>i</sub>=<math>\sum_{i=1}^n</math> (''ca''<sub>i</sub>)''u''<sub>i</sub><math>\in </math> span <math>\mathcal{S}</math>.
<math>\mbox{and let }y \in \mbox{ span}(\mathcal{S})\Rightarrow y =\sum_{i=1}^m b_iv_i\mbox{, }v_i\in \mathcal{S}</math>

<math>x+y = \sum_{i=1}^n a_iu_i+ \sum_{i=1}^m b_iv_i = \sum_{i=1}^{\mbox{max}(m,n)} c_iw_i</math>

<math>\qquad\mbox{ where }c_i=(a_1+b_1,a_2+b_2,\ldots,a_{\mbox{max}(m,n)}+b_{\mbox{max}(m,n)})\mbox{ and }w_i\in\mathcal{S}</math>

3.<math>cx= c\sum_{i=1}^n a_iu_i=\sum_{i=1}^n(ca_i)u_i\in\mbox{ span}(\mathcal{S})</math>




''Example''
''Example''
1.
1. Let P<sub>3</sub>(<math>\Re</math>)={ax<sup>3</sup>+bx<sup>2</sup>+cx+d}<math>\subseteq</math>P(<math>\Re</math>), ''a'', ''b'', ''c'', ''d'', <math>\in \Re</math>.<BR>

''u''<sub>1</sub>=''x''<sup>3</sup>-2''x''<sup>2</sup>-5''x''-3<BR>
<math>\mbox{Let } P_3(\mathbb{R})=\lbrace ax^3+bx^2+cx+d\rbrace\subset P(\mathbb{R})\mbox{, where }a, b, c, d \in \mathbb{R}</math>.
''u''<sub>2</sub>=3''x''<sup>3</sup>-5''x''<sup>2</sup>-4''x''-9<BR>

''v''=2''x''<sup>3</sup>-2''x''<sup>2</sup>+12''x''-6<BR>
<math>\begin{matrix}u_1^{}&=&x^3-2x^2-5x-3\\
Let W=spab(''u''<sub>1</sub>, ''u''<sub>2</sub>), <BR>
u_2^{}&=&3x^3-5x^2-4x-9\\
Does ''v'' <math> \in </math> W?<BR>
v_{}^{}&=&2x^3-2x^2+12x-6\end{matrix}</math>
''v'' is in W if ''v''=''a''<sub>1</sub>''u''<sub>1</sub>+''a''<sub>1</sub>''u''<sub>2</sub><br> for some ''a''<sub>1</sub>, ''a''<sub>2</sub> <math> \in \Re </math>.

<math>\mbox{Let }W=\mbox{span}(u_1^{},u_2^{})\mbox{,}</math><br>

<br><math>\mbox{Does/Is } v \in W\mbox{ ?}</math>

<math>v\in W\mbox{ if it is a linear combination of span}(u_1^{},u_2^{})</math>

<math>v=a_1u_1 + a_2u_2 \mbox{ for some }a_1, a_2 \in \mathbb{R}</math><br>

<br><math>\mbox{If }\exists a_1,a_2\in \mathbb{R}</math>

<math>\begin{matrix}2x^3-2x^2+12x-6&=& a_1^{}(x^3-2x^2-5x-3) + a_2^{}(3x^3-5x^2-4x-9)\\
\ &=&(a_1^{}+3a_2^{})x^3 + (-2a_1^{}-5a_2^{})x^2 + (-5a_1^{}-4a_2^{})x + (-3a_1^{}-9a_2^{})\end{matrix}</math>

<math>\mbox{Need to solve}\begin{cases}
2=a_1^{}+3a_2^{}\\
-2=-2a_1^{}-5a_2^{}\\
12=-5a_1^{}-4a_2^{}\\
-6=-3a_1^{}-9a_2^{}\end{cases}</math>

<math>\mbox{Solve the four equations above and we will get }a_1^{}=-4\mbox{ and }a_2^{}=2</math>

<math>\mbox{Check if }a_1^{}=-4\mbox{ and }a_2^{}=2\mbox{ hold for all the 4 equations.}</math>


If <math>\exists</math> ''a''<sub>1</sub>, ''a''<sub>2</sub> <math>\in \Re</math>, <br>
<math>\mbox{Since it holds, } v\in W</math>
{| border="0" cellpadding="0" cellspacing="0" align="center"
|-
|2''x''<sup>3</sup>-2''x''<sup>2</sup>+12''x''-6
|= ''a''<sub>1</sub>(''x''-2''x''<sup>2</sup>-5''x''-3) + ''a''<sub>2</sub>(3''x''<sup>3</sup>-5''x''<sup>2</sup>-4''x''-9)
|
|-
|
|=(''a''<sub>1</sub>+3''a''<sub>2</sub>)''x''<sup>3</sup> + (-2''a''<sub>1</sub> -5''a''<sub>2</sub>)''x''<sup>2</sup> + (-5''a''<sub>1</sub>-4''a''<sub>2</sub>)''x'' + (-3''a''<sub>1</sub>-9''a''<sub>2</sub>)
|
|-
|&nbsp;
|
|
|-
|<div align="right"><math>\Leftrightarrow</math>2</div>
|=''a''<sub>1</sub>+3''a''<sub>2</sub>
|
|-
|<div align="right">-2</div>
|=-2''a''<sub>1</sub>-5''a''<sub>2</sub>
|
|-
|<div align="right">12</div>
|=-5''a''<sub>1</sub>-4''a''<sub>2</sub>
|
|-
|<div align="right">-6</div>
|=-3''a''<sub>1</sub>-9''a''<sub>2</sub>
|
|}
Solve the four equations above and we will get ''a''<sub>1</sub>=-4 and ''a''<sub>2</sub>=2.<br>
Check if ''a''<sub>1</sub>=-4 and ''a''<sub>2</sub>=2 hold for all the 4 equations.<br>
Since it's hold, <math>\Rightarrow</math> ''v'' <math>\in</math> W.

Revision as of 10:39, 30 September 2006

Linear Combination

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Span

Theorem

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Example 1.

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