Drorbn - User contributions [en]

Talk:06-240/Homework Assignment 4

2006-10-10T13:52:14Z

Mc: correct punctuation

== Divisibility by Prime Number==

Pls correct me if I were wrong.
The operation of cut away the unit digit is a distraction. If we consider the unit digit, the operation basically is a deduction of a number, and that number is divisible by 7.
The whole operation is shown as follow:

{|
|<div align="right">8641<s>5</s>
|-
|<div align="right"> 10<s>5</s>
|  105/7=21
|-
|<div align="right"> ----
|-
|<div align="right">863<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> ---</div>
|-
|<div align="right">86<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> --</div>
|-
|<div align="right">8<s>4</s></div>
|-
|<div align="right">8<s>4</s></div>
|  84/7=12
|-
|<div align="right"> -</div>
|-
|<div align="right">0</div>
|  0/7=0
|}

Since it is an operation of series subtraction by multiples of 7, therefore the number we started from is divisible by 7 iff the resulting number is divisible by 7.

Moreover, there is a relationship between the unit digit, 2, and 7. The unit digit multiple by 21(7 <math> \times </math> 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
{|border="1"
!Unit digit, <math>x</math> !! <math>x \times 3 \times 7</math>
|-
|0
|0
|-
|1
|21
|-
|2
|42
|-
|3
|63
|-
|4
|84
|-
|5
|105
|-
|6
|126
|-
|7
|147
|-
|8
|168
|-
|9
|189
|}

From the table above, I've induced the criterion for divisibility by 17 that is similar operation but the unit digit multiplies by 5 instead of 2. For divisibility by 13, the unit digit multiple by 9. Alright, I think it will be more fun if it's explained by other people. [[User:Wongpak|Wongpak]] 09:38, 5 October 2006 (EDT)

== Excellent! ==

--[[User:Drorbn|Drorbn]] 12:09, 5 October 2006 (EDT)

I found the problem solved three different ways at Jim Loy's divisibility page[http://www.jimloy.com/number/divis.htm]
He mentions the problem as stated can be found in ''The Dictionary of Curious and Interesting Numbers'' by David Wells.

I prefer his second method. Beginning from the right most digit, multiply corresponding digits by the following sequence of coefficients 1, 3, 2, 6, 4, 5. For larger numbers sequence repeats. Alternatively the sequence 1, 3, 2, -1, -3, -2 could be used.

Using the given example <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7|86415</math>

Coefficients can be determined by taking multiplying the previous coefficient by 10 and take the modulus of this new number.

For n=17, sequence begins with 1 multiply by 10 then
<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math>

So for n=17, sequence is: 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12

or alternatively: 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5

Talk:06-240/Homework Assignment 4

2006-10-10T13:50:24Z

Mc: add link descriptor

== Divisibility by Prime Number==

Pls correct me if I were wrong.
The operation of cut away the unit digit is a distraction. If we consider the unit digit, the operation basically is a deduction of a number, and that number is divisible by 7.
The whole operation is shown as follow:

{|
|<div align="right">8641<s>5</s>
|-
|<div align="right"> 10<s>5</s>
|  105/7=21
|-
|<div align="right"> ----
|-
|<div align="right">863<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> ---</div>
|-
|<div align="right">86<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> --</div>
|-
|<div align="right">8<s>4</s></div>
|-
|<div align="right">8<s>4</s></div>
|  84/7=12
|-
|<div align="right"> -</div>
|-
|<div align="right">0</div>
|  0/7=0
|}

Since it is an operation of series subtraction by multiples of 7, therefore the number we started from is divisible by 7 iff the resulting number is divisible by 7.

Moreover, there is a relationship between the unit digit, 2, and 7. The unit digit multiple by 21(7 <math> \times </math> 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
{|border="1"
!Unit digit, <math>x</math> !! <math>x \times 3 \times 7</math>
|-
|0
|0
|-
|1
|21
|-
|2
|42
|-
|3
|63
|-
|4
|84
|-
|5
|105
|-
|6
|126
|-
|7
|147
|-
|8
|168
|-
|9
|189
|}

From the table above, I've induced the criterion for divisibility by 17 that is similar operation but the unit digit multiplies by 5 instead of 2. For divisibility by 13, the unit digit multiple by 9. Alright, I think it will be more fun if it's explained by other people. [[User:Wongpak|Wongpak]] 09:38, 5 October 2006 (EDT)

== Excellent! ==

--[[User:Drorbn|Drorbn]] 12:09, 5 October 2006 (EDT)

I found the problem solved three different ways at Jim Loy's divisibility page[http://www.jimloy.com/number/divis.htm]
He mentions the problem as stated can be found in ''The Dictionary of Curious and Interesting Numbers'' by David Wells.

I prefer his second method. Beginning from the right most digit, multiply corresponding digits by the following sequence of coefficients 1 3 2 6 4 5. For larger numbers sequence repeats. Alternatively the sequence 1 3 2 -1 -3 -2 could be used.

Using the given example <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7|86415</math>

Coefficients can be determined by taking multiplying the previous coefficient by 10 and take the modulus of this new number.

For n=17, sequence begins with 1 multiply by 10 then
<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math>

So for n=17, sequence is: 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12

or alternatively: 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5

Talk:06-240/Homework Assignment 4

2006-10-10T13:47:37Z

Mc: Different method

== Divisibility by Prime Number==

Pls correct me if I were wrong.
The operation of cut away the unit digit is a distraction. If we consider the unit digit, the operation basically is a deduction of a number, and that number is divisible by 7.
The whole operation is shown as follow:

{|
|<div align="right">8641<s>5</s>
|-
|<div align="right"> 10<s>5</s>
|  105/7=21
|-
|<div align="right"> ----
|-
|<div align="right">863<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> ---</div>
|-
|<div align="right">86<s>1</s></div>
|-
|<div align="right">2<s>1</s></div>
|  21/7=3
|-
|<div align="right"> --</div>
|-
|<div align="right">8<s>4</s></div>
|-
|<div align="right">8<s>4</s></div>
|  84/7=12
|-
|<div align="right"> -</div>
|-
|<div align="right">0</div>
|  0/7=0
|}

Since it is an operation of series subtraction by multiples of 7, therefore the number we started from is divisible by 7 iff the resulting number is divisible by 7.

Moreover, there is a relationship between the unit digit, 2, and 7. The unit digit multiple by 21(7 <math> \times </math> 3) is equal to the combination of the unit digit with its 2-time as the tenth/hundredth digit.
{|border="1"
!Unit digit, <math>x</math> !! <math>x \times 3 \times 7</math>
|-
|0
|0
|-
|1
|21
|-
|2
|42
|-
|3
|63
|-
|4
|84
|-
|5
|105
|-
|6
|126
|-
|7
|147
|-
|8
|168
|-
|9
|189
|}

From the table above, I've induced the criterion for divisibility by 17 that is similar operation but the unit digit multiplies by 5 instead of 2. For divisibility by 13, the unit digit multiple by 9. Alright, I think it will be more fun if it's explained by other people. [[User:Wongpak|Wongpak]] 09:38, 5 October 2006 (EDT)

== Excellent! ==

--[[User:Drorbn|Drorbn]] 12:09, 5 October 2006 (EDT)

I found the problem solved three different ways at [http://www.jimloy.com/number/divis.htm]
He mentions the problem as stated can be found in ''The Dictionary of Curious and Interesting Numbers''\ by David Wells.

I prefer his second method. Beginning from the right most digit, multiply corresponding digits by the following sequence of coefficients 1 3 2 6 4 5. For larger numbers sequence repeats. Alternatively the sequence 1 3 2 -1 -3 -2 could be used.

Using the given example <math>86415\Rightarrow 1(5)+3(1)+2(4)-1(6)-3(8)=-14\Rightarrow 7|86415</math>

Coefficients can be determined by taking multiplying the previous coefficient by 10 and take the modulus of this new number.

For n=17, sequence begins with 1 multiply by 10 then
<math>10\equiv 10 \mod(17),\quad 10\cdot 10\equiv 15\mod(17) \Rightarrow 15\equiv -2\mod(17)</math>

So for n=17, sequence is: 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12

or alternatively: 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5

06-240/Classnotes For Thursday October 5

2006-10-08T00:54:47Z

Mc: Add missing examples

<math>\mbox{From last class}{}_{}^{}</math>

<math>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}</math>

<math>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}</math>

<math>\mbox{The }M_i\mbox{s generate }M_{2\times 2}</math>

<math>\mbox{Fact }T\subset\mbox{ span }S\Rightarrow \mbox{ span }T\subset\mbox{ span }S </math>

<math>S\subset V\mbox{ is linearly independent }\Leftrightarrow \mbox{ whenever }u_i\in S\mbox{ are distinct}</math>

<math>\sum a_iu_i=0\Rightarrow V_ia_i=0 \mbox{ waste not}</math>
<br>

<br>
<math>\mbox{Comments}{}_{}^{}</math>
#<math>\emptyset\subset V\mbox{ is linearly independent}</math>
#<math>\lbrace u\rbrace\mbox{ is linearly independent iff }u_{}^{}\neq 0</math>
#<math>\mbox{If }S_1^{}\subset S_2\subset V</math>
##<math>\mbox{If }S_1^{}\mbox{ is linearly dependent, so is }S_2</math>
##<math>\mbox{If }S_2^{}\mbox{ is linearly dependent, so is }S_1</math>
##<math>\mbox{If }S_1^{}\mbox{ generates }V\mbox{, so does }S_2</math>
##<math>\mbox{If }S_2^{}\mbox{ does not generate }V\mbox{ neither does }S_1</math>
#<math>\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.}</math>
<br>

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

<math>\mbox{1.}\Leftarrow:\mbox{ start from second assertion and deduce first.}</math>

<math>\mbox{Assume }v_{}^{}\in \mbox{span }S</math>
<math>v=\sum a_iu_i\mbox{ where }u_i\in S, a_i\in F</math>

<math>\sum a_iu_i-1\cdot v=0\mbox{ this is a linear combination of elements in }S\cup v</math>
<math>\mbox{ in which not all coefficients are }0 \mbox{ and which add to }0_{}^{}.</math>
<math>\mbox{So }S\cup \lbrace v\rbrace\mbox{ is linearly dependent by definition}</math>
<br>
<math>\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:}</math>

<math>(*)\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</math>

<math>\mbox{If }b=0\mbox{, then }\sum a_iu_i=0\mbox{ and not }a_i\mbox{s are }0 </math>
<math>{}_{}^{}\Rightarrow S \mbox{ is linearly dependent}</math>
<math>{}_{}^{}\mbox{but initial assumption was }S\mbox{ is linearly independent.}\Rightarrow \mbox{ contradiction so }b\neq0</math>
<math>\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</math>
<br>

<br>
<math>\mbox{Definition}{}_{}^{}</math>

<math>{}_{}^{}\mbox{A basis of a vector space }V\mbox{ is a subset }\beta\subset V</math>
<math>{}_{}^{}\mbox{such that}</math>
#<math>{}_{}^{}\beta\mbox{ generates }V\mbox{ or }V=\mbox{ span }\beta</math>
#<math>{}_{}^{}\beta\mbox{ is linearly independent.}</math>
<br>

<br>
<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>
<math>\mbox{Theorem}{}_{}^{}</math>

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

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

<math>{}_{}^{}\mbox{It is a combination of things we already know.}</math>
#<math>{}_{}^{}\beta\mbox{ generates}</math>
#<math>{}_{}^{}\beta\mbox{ is linearly independent}</math>

06-240/Classnotes For Thursday October 5

2006-10-07T23:34:43Z

Mc: Added Oct 5 Lecture Notes

Template:06-240/Navigation

2006-10-07T22:39:51Z

Mc: Add Oct 5 Notes link

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[06-240]]/[[Template:06-240/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|-
!#
!Week of...
!Notes and Links
|-
|align=center|1
|Sep 11
|[[06-240/About This Class|About]], [[06-240/Classnotes For Tuesday, September 12|Tue]], [[06-240/Homework Assignment 1|HW1]], [[06-240/Putnam Competition|Putnam]], [[06-240/Classnotes for Thursday, September 14|Thu]]
|-
|align=center|2
|Sep 18
|[[06-240/Classnotes For Tuesday, September 19|Tue]], [[06-240/Homework Assignment 2|HW2]], [[06-240/Classnotes For Thursday, September 21|Thu]]
|-
|align=center|3
|Sep 25
|[[06-240/Classnotes For Tuesday September 26|Tue]], [[06-240/Homework Assignment 3|HW3]], [[06-240/Class Photo|Photo]], [[06-240/Classnotes For Thursday, September 28|Thu]]
|-
|align=center|4
|Oct 2
|[[06-240/Classnotes For Tuesday October 3|Tue]], [[06-240/Homework Assignment 4|HW4]], [[06-240/Classnotes For Thursday October 5|Thu]]
|-
|align=center|5
|Oct 9
|HW5
|-
|align=center|6
|Oct 16
|
|-
|align=center|7
|Oct 23
|Term Test
|-
|align=center|8
|Oct 30
|HW6
|-
|align=center|9
|Nov 6
|HW7
|-
|align=center|10
|Nov 13
|HW8
|-
|align=center|11
|Nov 20
|HW9
|-
|align=center|12
|Nov 27
|HW10
|-
|align=center|13
|Dec 4
|
|-
|colspan=3 align=center|[[06-240/Register of Good Deeds|Register of Good Deeds]]
|-
|colspan=3 align=center|[[Image:06-240-ClassPhoto.jpg|180px]]<br>[[06-240/Class Photo|Add your name / see who's in!]]
|}
</div></div>
|}

06-240/Classnotes For Tuesday October 3

2006-10-07T22:34:02Z

Mc: Lecture Notes (MC)

<math>\mbox{Definition}{}_{}^{}</math>

<math>v\in V \mbox{ is a linear combination of elements in } S\subset V</math>
<math> \mbox{ if }\exists u_1,\ldots,u_n\in S \mbox{ and } a_1,\dots,a_n \in F \mbox{ such that } V=\sum a_i u_i</math>

<math>\mbox{Example}{}_{}^{}</math>

<math>\mbox{In }P_3(\mathbb{R})\mbox{,}</math>
<math>v_1^{}=2x^3-2x^2+12-6 \mbox{ is a linear combination of:}</math>
<math>u_1^{}=x^3-2x^2-5x-3\mbox{ and }u_2=3x^3-5x^2-4x-9</math>
<math>\mbox{but } v_2^{}=3x^3-2x^2+7x+8 \mbox{ is not.}</math>

<math>\mbox{Why?}{}_{}^{}</math>

<math>v_1^{}=2x^3-2x^2+12-6=a_1^{}u_1+a_2u_2=a_1(x^3-2x^2-5x-3)+a_2(3x^3-5x^2-4x-9)</math>
<math>v_1^{}=-4u_1+2u_2</math>

<math>\mbox{Definition}{}_{}^{}</math>

<math>\mbox{We say that a subset }S\subset V\mbox{ generates or spans }V \mbox{ if span }S=\lbrace\mbox{ all linear combinations of elements in } S\rbrace=V</math>

<math>\mbox{Examples}{}_{}^{}</math>

<math>V=M_{2\times 2}(\mathbb{R})</math>

<math>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}</math>

<math>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}</math>

<math>\mbox{Claims}{}_{}^{}</math>

#<math>\lbrace M_1^{},M_2,M_3,M_4\rbrace\mbox{ generates }V</math>
#<math>\lbrace N_1^{},N_2,N_3,N_4\rbrace\mbox{ generates }V</math>
#<math>\lbrace M_1^{},M_2,M_3\rbrace\mbox{ does not generate }V</math>
#<math>\lbrace N_1^{},N_2,N_3\rbrace\mbox{ does not generate }V</math>
<br>
<math>\mbox{Proof of 1}{}_{}^{}</math>

<math>\mbox{Given any }B=\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}\mbox{ need to find }a_1,a_2,a_3,a_4\mbox{ such that,}</math>

<math>\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}=B=a_1M_1+a_2M_2+a_3M_3+a_4M_4=\begin{pmatrix}a_1&0\\0&0\end{pmatrix}
+\begin{pmatrix}0&a_2\\0&0\end{pmatrix}
+\begin{pmatrix}0&0\\a_3&0\end{pmatrix}
+\begin{pmatrix}0&0\\0&a_4\end{pmatrix}</math>

<math>=\begin{pmatrix}a_1^{}&a_2\\a_3&a_4\end{pmatrix}\Leftrightarrow
\begin{cases}b_{11}=a_1\\b_{12}=a_2\\b_{21}=a_3\\b_{22}=a_4\end{cases}</math>
<math>\mbox{A system of 4 equations with 4 unknowns}{}_{}^{}</math>
<br>
<math>\mbox{Proof of 2}{}_{}^{}</math>

<math>\begin{pmatrix}b_{11}^{}&b_{12}\\b_{21}&b_{22}\end{pmatrix}
=B=a_1N_1+a_2N_2+a_3N_3+a_4N_4=
\begin{pmatrix}0&a_1\\a_1&a_1\end{pmatrix}
+\begin{pmatrix}a_2&0\\a_2&a_2\end{pmatrix}
+\begin{pmatrix}a_3&a_3\\0&a_3\end{pmatrix}
+\begin{pmatrix}a_4&a_4\\a_4&0\end{pmatrix}</math>

<math>=\begin{pmatrix}a_2^{}+a_3+a_4&a_1+a_3+a_4\\a_1+a_2+a_4&a_1+a_2+a_3\end{pmatrix}\Leftrightarrow
\begin{cases}b_{11}=a_2+a_3+a_4\\b_{12}=a_1+a_3+a_4\\b_{21}=a_1+a_2+a_4\\b_{22}=a_1+a_2+a_3\end{cases}</math>

<math>\mbox{Trick}{}_{}^{}</math>

<math>M_1=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_1</math>
<math>M_2=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_2</math>
<math>M_3=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_3</math>
<math>M_4=\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_4</math>

<math>B=b_{11}^{}M_1+b_{12}M_3+b_{21}M_3+b_{22}M_4</math>
<math>=b_{11}\left(\frac{1}{3}\left(N_1+N_2+N_3+N_4\right)-3N_1\right)+\ldots</math>

<math>=\mbox{ a linear combination of }N_1^{},N_2,N_3,N_4</math>
<br>

<br>
<math>\mbox{Proof of 3}{}_{}^{}</math>

<math>\mbox{Indeed in }a_1^{}M_1+a_2M_2+a_3M_3=
\begin{pmatrix}a_1&a_2\\a_3&0\end{pmatrix}\mbox{ lower right corner is always } 0
</math>

<math>\mbox{for example }\begin{pmatrix}240&157\\e&\pi\end{pmatrix}\mbox{ not in span.}</math>
<br>

<br>
<math>\mbox{Proof of 4}{}_{}^{}</math>

<math>a_1^{}N_1+a_2N_2+a_3N_3=\begin{pmatrix}a_2+a_3&a_1+a_3\\a_1+a_2&a_1+a_2+a_3\end{pmatrix}</math>

<math>\begin{pmatrix}240&157\\e&\pi\end{pmatrix}\mbox{ is equal? }
\begin{cases}240=a_2+a_3\\157=a_1+a_3\\e=a_1+a_2\\\pi=a_1+a_2+a_3\end{cases}\Rightarrow\mbox{No solution}</math>
<br>

<br>
<math>\mbox{Motivation}{}_{}^{}</math>

<math>S\subset V\mbox{ is linearly dependent if it is wasteful,}</math>
<math>\mbox{i.e. if }\exists v\in V\mbox{ such that }\exists a_1^{}\ldots a_n\in F \mbox{ and }u_1^{}\ldots u_2\in S</math>
<math>\mbox{ and }\exists b_1^{}\ldots b_m\in F \mbox{ and }w_1\ldots w_m\in S</math>

<math>\mbox{so that }\sum_{i=1}^na_iu_i=v=\sum_{i=1}^mb_iw_i</math>

<math>\sum a_iu_i-\sum b_iw_i=0</math>

<math>\mbox{can be represented as }\sum c_iz_i=0</math>
<br>

<br>
<math>\mbox{Definition}{}_{}^{}</math>

<math>S\subset V\mbox{ is called linearly dependent if you can find }</math>
<math>z_1^{}\ldots z_n\in S\mbox{ different from each other and }c_1^{}\ldots c_n\in F\mbox{ so that not all of which are } 0,</math>
<math>\mbox{so that }\sum c_iz_i=0
\mbox{ otherwise, }S\mbox{ is called linearly independent}</math>
<br>

<br>
<math>\mbox{Example 1}{}_{}^{}</math>

<math>\mbox{In }\mathbb{R}, S=
\lbrace\begin{pmatrix}1&2&3\end{pmatrix},
\begin{pmatrix}4&5&6\end{pmatrix},
\begin{pmatrix}7&8&9\end{pmatrix}\rbrace\mbox{ is linearly dependent}</math>

<math>1\cdot\begin{pmatrix}1&2&3\end{pmatrix}-
2\cdot\begin{pmatrix}4&5&6\end{pmatrix}+
1\cdot\begin{pmatrix}7&8&9\end{pmatrix}=0</math>
<br>

<br>
<math>\mbox{Example 2}{}_{}^{}</math>

<math>\mathbb{R}^n, e_i=\begin{pmatrix}0\\\vdots\\1\\\vdots\\0\end{pmatrix}i^{th}\mbox{ row}</math>

<math>S=\lbrace e_1^{},\ldots,e_n\rbrace</math>

<math>\mbox{Claim }S\mbox{ is linearly independent}{}_{}^{}</math>

<math>\begin{pmatrix}0\\\vdots\\0\end{pmatrix}=0
=\sum_{i=1}^na_ie_i
=\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix}\Rightarrow
\begin{matrix}a_1=0\\a_2=0\\\vdots\\a_n=0\end{matrix}</math>

<math>\mbox{not not all }a_i^{}\mbox{ are }0\Rightarrow \mbox{ not linearly dependent.}</math>

<math>\mbox{Claim }S\subset V\mbox{ is linearly independent iff whenever }\sum a_iu_i=0
\mbox{ and distinct }u_i\in S\mbox{ then }\forall i\quad a_i=0</math>
<br>

<br>
<math>\mbox{Comments}{}_{}^{}</math>

#<math>\emptyset\subset V\mbox{ is linearly independent}</math>
#<math>\mbox{Suppose }u\in V,\quad \lbrace u\rbrace\mbox{ the singleton set is linearly independent iff }u_{}^{}\neq 0</math>

<math>\lbrace0\rbrace\mbox{ is linearly dependent. example }7\cdot0=0</math>

<math>\mbox{if }u\neq0\mbox{ assume }a\cdot u=0\mbox{, and }a\neq0\Rightarrow a_{}^{-1}au=0
\Rightarrow u=0\mbox{ contradiction results, so no such }a\mbox{ exists.}</math>
<math>\mbox{ So}{}_{}^{}\lbrace u\rbrace\mbox{is not linearly dependent, hence it is linearly independent.}</math>

Template:06-240/Navigation

2006-10-07T20:28:55Z

Mc: October 3 Lecture

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|Sep 11
|[[06-240/About This Class|About]], [[06-240/Classnotes For Tuesday, September 12|Tue]], [[06-240/Homework Assignment 1|HW1]], [[06-240/Putnam Competition|Putnam]], [[06-240/Classnotes for Thursday, September 14|Thu]]
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|Sep 18
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|Sep 25
|[[06-240/Classnotes For Tuesday September 26|Tue]], [[06-240/Homework Assignment 3|HW3]], [[06-240/Class Photo|Photo]], [[06-240/Classnotes For Thursday, September 28|Thu]]
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|align=center|4
|Oct 2
|[[06-240/Classnotes For Tuesday October 3|Tue]], [[06-240/Homework Assignment 4|HW4]]
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|Oct 9
|HW5
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|Oct 16
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|align=center|7
|Oct 23
|Term Test
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|Oct 30
|HW6
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|Nov 6
|HW7
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|Nov 13
|HW8
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|Nov 20
|HW9
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|Nov 27
|HW10
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|Dec 4
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|-
|colspan=3 align=center|[[06-240/Register of Good Deeds|Register of Good Deeds]]
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06-240/Homework Assignment 4

2006-10-05T10:29:34Z

Mc: Correct spelling of repetitions

{{06-240/Navigation}}

* '''Read''' sections 1.5 through 1.7 in our textbook. Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read.

* '''Solve''' problems 3, 8, 9, 10 and 11 on pages 41-42, but submit only your solutions of problems 8 and 9.

* '''Solve''' problems 1, 2, 4, 5, 9 and 16 on page 34, but submit only your solutions of problems 4, 5 and 9.

{|
|-
|'''Just for Fun.'''
* Take a large integer and write it in base 10. Cut away the "singles" digit, double it and subtract the result from the remaining digits. Repeat the process until the number you have left is small. Prove that the number you started from is divisible by 7 iff the resulting number is divisible by 7. Thus the example on the right shows that 86415 is divisible by 7 as 0 is divisible by 7. (I learned this trick a few days ago from [http://www.math.toronto.edu/hovinen/ Bradford Hovinen]).
* Find a similar criterion for divisibility by 17 and for all other divisibilities.
* Note that the word "divisibilities" has the largest number of repetitions of a single letter among all words in the English language (6 i's). I've known this fact for years but this is the first time that I'm finding a semi-legitimate use for that word!
|
8641<s>5</s>
10
----
863<s>1</s>
2
---
86<s>1</s>
2
--
8<s>4</s>
8
-
0
|}

06-240/Classnotes For Thursday, September 28

2006-09-30T15:41:51Z

Mc: grammar fixup

===Linear Combination===

<math>\mbox{Definition: Let }(u_i) = (u_1,u_2,\ldots,u_n)\mbox{ be a sequence of vectors in }V</math>.

<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===
<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>

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

'''Theorem'''

<math>\forall\mathcal{S} \subset V\mbox{, span}(\mathcal{S})\mbox{ is a subspace of }V</math>

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

1. <math>0 \in\mbox{ span}(\mathcal{S})</math>.<br>
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>\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''
1.

<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>.

<math>\begin{matrix}u_1^{}&=&x^3-2x^2-5x-3\\
u_2^{}&=&3x^3-5x^2-4x-9\\
v_{}^{}&=&2x^3-2x^2+12x-6\end{matrix}</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{ holds for all 4 equations.}</math>

<math>\mbox{Since it holds, } v\in W</math>

06-240/Classnotes For Thursday, September 28

2006-09-30T15:39:54Z

Mc: LaTeX cleanup

===Linear Combination===

<math>\mbox{Definition: Let }(u_i) = (u_1,u_2,\ldots,u_n)\mbox{ be a sequence of vectors in }V</math>.

<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===
<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>

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

'''Theorem'''

<math>\forall\mathcal{S} \subset V\mbox{, span}(\mathcal{S})\mbox{ is a subspace of }V</math>

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

1. <math>0 \in\mbox{ span}(\mathcal{S})</math>.<br>
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>\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''
1.

<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>.

<math>\begin{matrix}u_1^{}&=&x^3-2x^2-5x-3\\
u_2^{}&=&3x^3-5x^2-4x-9\\
v_{}^{}&=&2x^3-2x^2+12x-6\end{matrix}</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>

<math>\mbox{Since it holds, } v\in W</math>

06-240/Class Photo

2006-09-29T11:59:29Z

Mc: Added Mick

{{06-240/Navigation}}

Our class on September 28, 2006:

[[Image:06-240-ClassPhoto.jpg|thumb|center|500px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are===

{| align=center border=1
|-
!First Name
!Last Name
!UserID
!Email
!In Photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Carberry|first=Mick|userid=MC|email=Mick.Carberry@utoronto.ca|location=long haired, bearded old guy in back|comments= }}
|}

Talk:06-240/Homework Assignment 3

2006-09-29T11:47:44Z

Mc: Question 1?

Question 1 is a series of true/false questions. Do you just want the answer to 1f)? MC

06-240/Classnotes For Thursday, September 21

2006-09-27T10:37:58Z

Mc: Italic correction

A force has a direction and a magnitude.

==<center><u>'''Force Vectors'''</u></center>==
#<math>\mbox{There is a special force vector called 0.}</math>
#<math>\mbox{They can be added.}</math>
#<math>\mbox{They can be multiplied by any scalar.}</math>

====''Properties''====

<math>\mbox{(convention: }x,y,z\mbox{ }\mbox{ are vectors; }a,b,c\mbox{ }\mbox{ are scalars)}</math>
#<math> x+y=y+x \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+0=x \ </math>
#<math> \forall x\; \exists\ y \ \mbox{ s.t. }x+y=0</math>
#<math> 1\cdot x=x \ </math>
#<math> a(bx)=(ab)x \ </math>
#<math> a(x+y)=ax+ay \ </math>
#<math> (a+b)x=ax+bx \ </math>

=====Definition=====

Let F be a field "of scalars". A vector space over F is a set V, of "vectors", along with two operations

: <math> +: V \times V \to V </math>
: <math> \cdot: F \times V \to V \mbox{, so that:}</math>
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
#<math> \exists\ 0 \in V s.t.\; \forall x \in V\ x+0=x </math>
#<math> \forall x \in V\; \exists\ y \in V\ s.t. \ x+y=0</math>
#<math> 1\cdot x=x\ </math>
#<math> a(bx)=(ab)x\ </math>
#<math> a(x+y)=ax+ay\ </math>
#<math> \forall x \in V\ ,\forall a,b \in F\ (a+b)x=ax+bx </math>
-----
9. <math> x \mapsto \vert x\vert \in \mathbb{R} \ \vert x+y\vert \le \vert x\vert+\vert y\vert </math>
====''Examples''====
'''Ex.1.'''
<math> F^n= \lbrace(a_1,a_2,a_3,\ldots,a_{n-1},a_n):\forall i\ a_i \in F \rbrace </math> <br/>
<math> n \in \mathbb{Z}\ , n \ge 0 </math> <br/>
<math> x=(a_1,\ldots,a_2)\ y=(b_1,\ldots, b_2)\ </math> <br/>
<math> x+y:=(a_1+b_1,a_2+b_2,\ldots,a_n+b_n)\ </math> <br/>
<math> 0_{F^n}=(0,\ldots,0) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,\ldots,aa_n) </math> <br/>
<math> \mbox{In } \mathbb{Q}^3 \ \left( \frac{3}{2},-2,7\right)+\left( \frac{-3}{2}, \frac{1}{3},240\right)=\left(0, \frac{-5}{3},247\right) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
'''Ex.2.'''
<math> V=M_{m\times n}(F)=\left\lbrace\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots &
& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \right\rbrace </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
<math>\mbox{Addition by adding entry by entry:}</math>

<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>

<math>\mbox{Multiplication by multiplying scalar c to all entries by M.}</math>

<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>

<math>\mbox{Zero matrix has all entries = 0:}</math>

<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
& \vdots \\ 0 & \cdots & 0\end{pmatrix} </math> <br/>
'''Ex.3.'''
<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
'''Ex.4.'''
<math>\mbox{F is a vector space over itself.}</math> <br/>
'''Ex.5.'''
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
'''Ex.6.'''
<math>\mbox{Let S be a set. Let}</math> <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>
<math> (f+g)(t)=f(t)+g(t)\ for\ any\ t\in S </math> <br/>
<math> (af)(t)=a\cdot f(t)\ </math>

06-240/Classnotes For Thursday, September 21

2006-09-27T10:31:22Z

Mc: More formatting

A force has a direction & a magnitude.

==<center><u>'''Force Vectors'''</u></center>==
#<math>\mbox{There is a special force vector called 0.}</math>
#<math>\mbox{They can be added.}</math>
#<math>\mbox{They can be multiplied by any scalar.}</math>

====''Properties''====

<math>\mbox{(convention: }x,y,z \mbox{ are vectors; }a,b,c \mbox{ are scalars)}</math>
#<math> x+y=y+x \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+0=x \ </math>
#<math> \forall x\; \exists\ y \ \mbox{ s.t. }x+y=0</math>
#<math> 1\cdot x=x \ </math>
#<math> a(bx)=(ab)x \ </math>
#<math> a(x+y)=ax+ay \ </math>
#<math> (a+b)x=ax+bx \ </math>

=====Definition=====

Let F be a field "of scalars". A vector space over F is a set V, of "vectors", along with two operations

: <math> +: V \times V \to V </math>
: <math> \cdot: F \times V \to V \mbox{, so that:}</math>
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
#<math> \exists\ 0 \in V s.t.\; \forall x \in V\ x+0=x </math>
#<math> \forall x \in V\; \exists\ y \in V\ s.t. \ x+y=0</math>
#<math> 1\cdot x=x\ </math>
#<math> a(bx)=(ab)x\ </math>
#<math> a(x+y)=ax+ay\ </math>
#<math> \forall x \in V\ ,\forall a,b \in F\ (a+b)x=ax+bx </math>
-----
9. <math> x \mapsto \vert x\vert \in \mathbb{R} \ \vert x+y\vert \le \vert x\vert+\vert y\vert </math>
====''Examples''====
'''Ex.1.'''
<math> F^n= \lbrace(a_1,a_2,a_3,\ldots,a_{n-1},a_n):\forall i\ a_i \in F \rbrace </math> <br/>
<math> n \in \mathbb{Z}\ , n \ge 0 </math> <br/>
<math> x=(a_1,\ldots,a_2)\ y=(b_1,\ldots, b_2)\ </math> <br/>
<math> x+y:=(a_1+b_1,a_2+b_2,\ldots,a_n+b_n)\ </math> <br/>
<math> 0_{F^n}=(0,\ldots,0) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,\ldots,aa_n) </math> <br/>
<math> \mbox{In } \mathbb{Q}^3 \ \left( \frac{3}{2},-2,7\right)+\left( \frac{-3}{2}, \frac{1}{3},240\right)=\left(0, \frac{-5}{3},247\right) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
'''Ex.2.'''
<math> V=M_{m\times n}(F)=\left\lbrace\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots &
& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \right\rbrace </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
<math>\mbox{Addition by adding entry by entry:}</math>

<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>

<math>\mbox{Multiplication by multiplying scalar c to all entries by M.}</math>

<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>

<math>\mbox{Zero matrix has all entries = 0:}</math>

<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
& \vdots \\ 0 & \cdots & 0\end{pmatrix} </math> <br/>
'''Ex.3.'''
<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
'''Ex.4.'''
<math>\mbox{F is a vector space over itself.}</math> <br/>
'''Ex.5.'''
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
'''Ex.6.'''
<math>\mbox{Let S be a set. Let}</math> <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>
<math> (f+g)(t)=f(t)+g(t)\ for\ any\ t\in S </math> <br/>
<math> (af)(t)=a\cdot f(t)\ </math>

06-240/Classnotes For Thursday, September 21

2006-09-27T09:55:38Z

Mc: Fix up LaTeX

A force has a direction & a magnitude.

==<center><u>'''Force Vectors'''</u></center>==
#There is a special force vector called 0.
#They can be added.
#They can be multiplied by any scalar.

====''Properties''==== (convention: x,y,z-vectors; a,b,c-scalars)
# <math> x+y=y+x \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+0=x \ </math>
#<math> \forall x\; \exists\ y \ s.t.\ x+y=0 \ </math>
#<math> 1\cdot x=x \ </math>
#<math> a(bx)=(ab)x \ </math>
#<math> a(x+y)=ax+ay \ </math>
#<math> (a+b)x=ax+bx \ </math>

=====Definition===== Let F be a field "of scalars". A vector space over F is a set V (of "vectors") along with two operations:
: <math> +: V \times V \to V </math>
: <math> \cdot: F \times V \to V </math>, so that
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
#<math> \exists\ 0 \in V s.t.\; \forall x \in V\ x+0=x </math>
#<math> \forall x \in V\; \exists\ y \in V\ s.t. \ x+y=0</math>
#<math> 1\cdot x=x\ </math>
#<math> a(bx)=(ab)x\ </math>
#<math> a(x+y)=ax+ay\ </math>
#<math> \forall x \in V\ ,\forall a,b \in F\ (a+b)x=ax+bx </math>
-----
9. <math> x \mapsto \vert x\vert \in \mathbb{R} \ \vert x+y\vert \le \vert x\vert+\vert y\vert </math>
====''Examples''====
'''Ex.1.'''
<math> F^n= \lbrace(a_1,a_2,a_3,\ldots,a_{n-1},a_n):\forall i\ a_i \in F \rbrace </math> <br/>
<math> n \in \mathbb{Z}\ , n \ge 0 </math> <br/>
<math> x=(a_1,\ldots,a_2)\ y=(b_1,\ldots, b_2)\ </math> <br/>
<math> x+y:=(a_1+b_1,a_2+b_2,\ldots,a_n+b_n)\ </math> <br/>
<math> 0_{F^n}=(0,\ldots,0) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,\ldots,aa_n) </math> <br/>
<math> In \ \mathbb{Q}^3 \ \left( \frac{3}{2},-2,7\right)+\left( \frac{-3}{2}, \frac{1}{3},240\right)=\left(0, \frac{-5}{3},247\right) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
'''Ex.2.'''
<math> V=M_{m\times n}(F)=\left\lbrace\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots &
& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \right\rbrace </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
Addition by adding entry by entry:

<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>

Multiplication by multiplying scalar c to all entries by M.

<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>

Zero matrix has all entries = 0:

<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
& \vdots \\ 0 & \cdots & 0\end{pmatrix} </math> <br/>
'''Ex.3.'''
<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
'''Ex.4.'''
F is a vector space over itself. <br/>
'''Ex.5.'''
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
'''Ex.6.'''
Let S be a set. Let <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>
<math> (f+g)(t)=f(t)+g(t)\ for\ any\ t\in S </math> <br/>
<math> (af)(t)=a\cdot f(t)\ </math>

06-240/Classnotes For Thursday, September 21

2006-09-26T13:42:38Z

Mc: Some updates

A force has a direction & a magnitude.

==<center><u>'''Force Vectors'''</u></center>==
#There is a special force vector called 0.
#They can be added.
#They can be multiplied by any scalar.

====''Properties''==== (convention: x,y,z-vectors; a,b,c-scalars)
# <math> x+y=y+x \ </math>
#<math> x+(y+z)=(x+y)+z \ </math>
#<math> x+0=x \ </math>
#<math> \forall x\; \exists\ y \ s.t.\ x+y=0 \ </math>
#<math> 1.x=x \ </math>
#<math> a(bx)=(ab)x \ </math>
#<math> a(x+y)=ax+ay \ </math>
#<math> (a+b)x=ax+bx \ </math>

=====Definition===== Let F be a field "of scalars". A vector space over F is a set V (of "vectors") along with two operations:
: <math> +: V \times V \to V </math>
: <math> \cdot: F \times V \to V </math>, so that
#<math> \forall x,y \in V\ x+y=y+x </math>
#<math> \forall x,y \in V\ x+(y+z)=(x+y)+z </math>
#<math> \exists\ 0 \in V s.t.\; \forall x \in V\ x+0=x </math>
#<math> \forall x \in V\; \exists\ y \in V\ s.t.\ x+y=0</math>
#<math> 1.x=x\ </math>
#<math> a(bx)=(ab)x\ </math>
#<math> a(x+y)=ax+ay\ </math>
#<math> \forall x \in V\ ,\forall a,b \in F\ (a+b)x=ax+bx </math>
-----
9. <math> x \mapsto |x| \in \mathbb{R} \ \ |x+y| \le |x|+|y| </math>
====''Examples''====
'''Ex.1.'''
<math> F^n= \lbrace(a_1,a_2,a_3,...,a_{n-1},a_n):\forall i\ a_i \in F \rbrace </math> <br/>
<math> n \in \mathbb{Z}\ , n \ge 0 </math> <br/>
<math> x=(a_1,...,a_2)\ y=(b_1,...,b_2)\ </math> <br/>
<math> x+y:=(a_1+b_1,a_2+b_2,...,a_n+b_n)\ </math> <br/>
<math> 0_{F^n}=(0,...,0) </math> <br/>
<math> a\in F\ ax=(aa_1,aa_2,...,aa_n) </math> <br/>
<math> In \ \mathbb{Q}^3 \ ( \frac{3}{2},-2,7)+( \frac{-3}{2}, \frac{1}{3},240)=(0, \frac{-5}{3},247) </math> <br/>
<math> 7\left( \frac{1}{5},\frac{1}{7},\frac{1}{9}\right)=\left( \frac{7}{5},1,\frac{7}{9}\right) </math> <br/>
'''Ex.2.'''
<math> V=M_{m\times n}(F)=\lbrace\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots &
& \vdots \\ a_{m1} & \cdots & a_{mn}\end{pmatrix}: a_{ij} \in F \rbrace </math> <br/>
<math> M_{3\times 2}( \mathbb{R})\ni \begin{pmatrix} 7 & -7 \\ \pi & \mathit{e} \\ -5 & 2 \end{pmatrix} </math> <br/>
Addition by adding entry by entry:

<math> M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}+\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \end{pmatrix}</math> <br/>

Multiplication by multiplying scalar c to all entries by M.

<math> c\cdot M_{2\times 2}\ \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=\begin{pmatrix} c\cdot a_{11} & c\cdot a_{12} \\ c\cdot a_{21} & c\cdot a_{22} \end{pmatrix}</math> <br/> <br/>

Zero matrix has all entries = 0:

<math> 0_{M_{m\times n}}=\begin{pmatrix} 0 & \cdots & 0 \\ \vdots &
& \vdots \\ 0 & \cdots & 0\end{pmatrix} </math> <br/>
'''Ex.3.'''
<math> \mathbb{C}</math> form a vector space over <math> \mathbb{R}</math>. <br/>
'''Ex.4.'''
F is a vector space over itself. <br/>
'''Ex.5.'''
<math> \mathbb{R}</math> is a vector space over <math> \mathbb{Q}</math>. <br/>
'''Ex.6.'''
Let S be a set. Let <br/>
<math> \mathcal{F}(S,\mathbb{R})=\big\{f:S\to \mathbb{R} \big\} </math> <br/>
<math> f,g \in \mathcal{F}(S,\mathbb{R}) </math> <br/>
<math> (f+g)(t)=f(t)+g(t)\ for\ any\ t\in S </math> <br/>
<math> (af)(t)=a\cdot f(t)\ </math>

Talk:06-240/Homework Assignment 2

2006-09-24T12:43:58Z

Mc: Answer Wongpak

===p12: Q1===
(f) An <math>m \times n</math> matrix has ''m'' columns and ''n'' rows. (True or False)
According to the answer at the back, it's FALSE. Can anyone please explain why? Thank you. [[User:Wongpak|Wongpak]] 22:01, 23 September 2006 (EDT)

By convention it is ''m'' rows and ''n'' columns. MC

Talk:06-240/About This Class

2006-09-19T14:09:26Z

Mc: Text in bookstore

This is about the textbook, does anyone know where to purchase it? I've looked at the U of T bookstore, but the closest I could find was Linear Algebra (blah blah) 5th Edition.

The U of T bookstore did have copies of the textbook. I picked up mine last Tuesday after class. The Linear Algebra texts are not together downstairs in the bookstore. The used bookstore across the street may have it. MC

Talk:06-240/Homework Assignment 1

2006-09-15T17:48:33Z

Mc:

What information should be included on the homework assignments besides the answers to the assignment?
Is student name, Math 240, Homework Assignment 1 and date sufficient?
MC

Talk:06-240/Homework Assignment 1

2006-09-15T17:47:53Z

Mc: HW submission details

What information should be included on the homework assignments besides the answers to the assignment?
Is student name, Math 240, Homework Assignment 1 and date sufficient?