Drorbn - User contributions [en]

12-240/Classnotes for Tuesday October 09

2012-12-07T10:06:28Z

Peterlue:

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

== Theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0∙ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math> span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math> span(\beta)</math> must also contain <math> span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T10:02:52Z

Peterlue: /* Clarification on lecture notes */

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0∙ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math> span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math> span(\beta)</math> must also contain <math> span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:59:44Z

Peterlue: /* theorems */

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0∙ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:59:15Z

Peterlue: /* theorems */

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai∙ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai∙ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:57:30Z

Peterlue: /* Definition of basis */

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basis if

1/ It generates (span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:56:54Z

Peterlue:

{{12-240/Navigation}}
In this lecture, the professor concentrated on bases and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basic if

1/ It generates ( span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:55:22Z

Peterlue: /* theorems */

{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basic if

1/ It generates ( span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basis of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:54:58Z

Peterlue: /* Definition of basic */

{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.
== Definition of basis ==
β <math>\subset \!\,</math> V is a basic if

1/ It generates ( span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basic of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1009-1.jpg|Page 1
Image:12-240-1009-2.jpg|Page 2
Image:12-240-1009-3.jpg|Page 3
Image:12-240-1009-4.jpg|Page 4
Image:12-240-1009-5.jpg|Page 5
Image:12-240-1009-6.jpg|Page 6
Image:12-240-1009-7.jpg|Page 7
</gallery>

== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
<gallery>
Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Tuesday November 6

2012-12-07T09:52:50Z

Peterlue: /* Theorems */

{{12-240/Navigation}}

==Riddle==

Find A and B such that AB - BA = I

==Theorems==
1. Given U with basis <math>\alpha</math>, V with basis <math>\beta</math>, W with basis <math>\gamma,</math>
<math>[T \circ S]_\alpha^\beta = [T]_\beta^\gamma \times [S]_\alpha^\beta</math>

2. For A <math>\in M_(m \times n)</math> and B <math>\in M_(n \times p)</math> and C <math>\in M_(p \times q)</math>,
(AB)C = A(BC)

== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Nov6-1.jpeg|Page 1
Image:12-240-Nov6-2.jpeg|Page 2
Image:12-240-Nov6-3.jpeg|Page 3
</gallery>

12-240/Classnotes for Tuesday October 09

2012-12-07T09:49:09Z

Peterlue: /* theorems */

{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.
== Definition of basic ==
β <math>\subset \!\,</math> V is a basic if

1/ It generates ( span) V, span β = V

2/ It is linearly independent

== theorems ==
1/ β is a basic of V iff every element of V can be written as a linear combination of elements of β in a unique way.

proof: ( in the case β is finite)

β = {u1, u2, ..., un}

(<=) need to show that β = span(V) and β is linearly independent.

The fact that β span is the fact that every element of V can be written as a linear combination of elements of β, which is given

Assume <math>\sum \!\,</math> ai.ui = 0 ai <math>\in\!\,</math> F, ui <math>\in\!\,</math> β

<math>\sum \!\,</math> ai.ui = 0 = <math>\sum \!\,</math> 0.ui

since 0 can be written as a linear combination of elements of β in a unique way, ai=0 <math>\forall\!\,</math> i

Hence β is linearly independent

(=>) every element of V can be written as a linear combination of elements of β in a unique way.

So, suppose <math>\sum \!\,</math> ai∙ui = v = <math>\sum \!\,</math> bi∙ui

Thus <math>\sum \!\,</math> ai∙ui - <math>\sum \!\,</math> bi∙ui = 0

<math>\sum \!\,</math> (ai-bi)∙ui = 0

β is linear independent hence (ai - bi)= 0 <math>\forall\!\,</math> i

i.e ai = bi, hence the combination is unique.

== Clarification on lecture notes ==

On page 3, we find that <math>G \subseteq span(\beta)</math> then we say <math>span(G) \subseteq span(\beta)</math>. The reason is, the Theorem 1.5 in the textbook.

Theorem 1.5: The span of any subset <math>S</math> of a vector space <math>V</math> is a subspace of <math>V</math>. Moreover, any subspace of <math>V</math> that contains <math>S</math> must also contain <math>span(S)</math>

Since <math>\beta</math> is a subset of <math>V</math>, <math>span(\beta)</math> is a subspace of <math>V</math> from the first part of the Theorem 1.5. We have shown (in the lecture notes page 3) that <math>G \subseteq span(\beta)</math>. From the "Moreover" part of Theorem 1.5, since <math>span(\beta)</math> is a subspace of <math>V</math> containing <math>G</math>, <math>span(\beta)</math> must also contain <math>span(G)</math>.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
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== Lecture notes uploaded by [[User:Grace.zhu|gracez]] ==
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Image:12-240-O9-1.jpg|Page 1
Image:12-240-O9-2.jpg|Page 2
</gallery>

12-240/Classnotes for Thursday October 4

2012-12-07T09:46:52Z

Peterlue: /* Interesting inequality */

{{12-240/Navigation}}
== Reminders ==
Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

== Recap ==

Base - what were doing today

Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dots, u_n\}</math> if <math>v = a_1u_1, \dots, a_nu_n</math> for scalars from a field <math>F</math>.

Span - <math>\operatorname{span}(S)</math> is the set of all linear combinations of the set <math>S</math>.

Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

== Pre - Basis ==

'''Linear dependence'''

'''Definition''' A set S ⊂ V is called linearly dependent if you can express the zero vector as a linear combination of distinct vectors from S, excluding the non-trivial linear combination where all of the scalars are 0.

Otherwise, we call S '''linearly independent.'''

=== Examples ===

1. In '''R'''^3, take S = {u1 = (1,4,7), u2 = (2,5,8), u3 = (3,6,9)}

u1 - 2u2 + u3 = 0

S is linearly dependent.

2. In R^n, take {e_i} = {0, 0, ... 1, 0, 0, 0} where 1 is in the ith position, and (ei) is a vector with n entries.

Claim: This set is linearly independent.

Proof: Suppose (∑ ai∙ei) = 0 ({ei} is linearly dependent.)

(∑ ai∙ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

⇒ a1 = a2 = ... = an = 0!

===Comments ===

1. {u} is linearly independent.
Proof:
⇐ If u≠0, suppose au =0
By property (a∙u = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.

⇒ By definition, au = 0 for {u} only when a = 0.

2. ∅ is linearly independent.

Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V.

-> If S1 is linearly dependent, then S2 is dependent.

-> If S2 is linearly independent, then S1 is linearly independent. (Hint: contrapositive)

== Basis ==

'''Definition''': A subset β is called a basis if (1) β generates V → span(β) = V and (2) β is linearly independent.

===Examples===

1. V = {0}, β = {}

2. {ei} for F^n, this is what we call the '''standard basis'''

3. B = {(1,1),(1, -1)} is a basis for R^2

4. P_n(F) β = {x^n, x^n-1, ... , x^1, x^0}

5. P(F), β = (x^0, x^1 ... and on} ('''Infinite basis'''!)

== Interesting inequality ==

This holds is true if the field does not have characteristic 2. Can you see why?

(a,b) = (a+b)/2 ∙ (1, 1) + (a-b)/2 ∙ (1, -1)

== Lecture notes scanned by [[User:starash|starash]] ==
<gallery>
Image:12-240-1004-1.jpg|Page 1
Image:12-240-1004-2.jpg|Page 2
</gallery></math>

12-240/Classnotes for Thursday October 4

2012-12-07T09:46:14Z

Peterlue: /* Basis */

{{12-240/Navigation}}
== Reminders ==
Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

== Recap ==

Base - what were doing today

Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dots, u_n\}</math> if <math>v = a_1u_1, \dots, a_nu_n</math> for scalars from a field <math>F</math>.

Span - <math>\operatorname{span}(S)</math> is the set of all linear combinations of the set <math>S</math>.

Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

== Pre - Basis ==

'''Linear dependence'''

'''Definition''' A set S ⊂ V is called linearly dependent if you can express the zero vector as a linear combination of distinct vectors from S, excluding the non-trivial linear combination where all of the scalars are 0.

Otherwise, we call S '''linearly independent.'''

=== Examples ===

1. In '''R'''^3, take S = {u1 = (1,4,7), u2 = (2,5,8), u3 = (3,6,9)}

u1 - 2u2 + u3 = 0

S is linearly dependent.

2. In R^n, take {e_i} = {0, 0, ... 1, 0, 0, 0} where 1 is in the ith position, and (ei) is a vector with n entries.

Claim: This set is linearly independent.

Proof: Suppose (∑ ai∙ei) = 0 ({ei} is linearly dependent.)

(∑ ai∙ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

⇒ a1 = a2 = ... = an = 0!

===Comments ===

1. {u} is linearly independent.
Proof:
⇐ If u≠0, suppose au =0
By property (a∙u = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.

⇒ By definition, au = 0 for {u} only when a = 0.

2. ∅ is linearly independent.

Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V.

-> If S1 is linearly dependent, then S2 is dependent.

-> If S2 is linearly independent, then S1 is linearly independent. (Hint: contrapositive)

== Basis ==

'''Definition''': A subset β is called a basis if (1) β generates V → span(β) = V and (2) β is linearly independent.

===Examples===

1. V = {0}, β = {}

2. {ei} for F^n, this is what we call the '''standard basis'''

3. B = {(1,1),(1, -1)} is a basis for R^2

4. P_n(F) β = {x^n, x^n-1, ... , x^1, x^0}

5. P(F), β = (x^0, x^1 ... and on} ('''Infinite basis'''!)

== Interesting inequality ==

This holds is true if the field does not have characteristic 2. Can you see why?

(a,b) = (a+b)/2 * (1, 1) + (a-b)/2 * (1, -1)

== Lecture notes scanned by [[User:starash|starash]] ==
<gallery>
Image:12-240-1004-1.jpg|Page 1
Image:12-240-1004-2.jpg|Page 2
</gallery></math>

12-240/Classnotes for Thursday October 4

2012-12-07T09:44:59Z

Peterlue: /* Comments */

{{12-240/Navigation}}
== Reminders ==
Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

== Recap ==

Base - what were doing today

Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dots, u_n\}</math> if <math>v = a_1u_1, \dots, a_nu_n</math> for scalars from a field <math>F</math>.

Span - <math>\operatorname{span}(S)</math> is the set of all linear combinations of the set <math>S</math>.

Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

== Pre - Basis ==

'''Linear dependence'''

'''Definition''' A set S ⊂ V is called linearly dependent if you can express the zero vector as a linear combination of distinct vectors from S, excluding the non-trivial linear combination where all of the scalars are 0.

Otherwise, we call S '''linearly independent.'''

=== Examples ===

1. In '''R'''^3, take S = {u1 = (1,4,7), u2 = (2,5,8), u3 = (3,6,9)}

u1 - 2u2 + u3 = 0

S is linearly dependent.

2. In R^n, take {e_i} = {0, 0, ... 1, 0, 0, 0} where 1 is in the ith position, and (ei) is a vector with n entries.

Claim: This set is linearly independent.

Proof: Suppose (∑ ai∙ei) = 0 ({ei} is linearly dependent.)

(∑ ai∙ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

⇒ a1 = a2 = ... = an = 0!

===Comments ===

1. {u} is linearly independent.
Proof:
⇐ If u≠0, suppose au =0
By property (a∙u = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.

⇒ By definition, au = 0 for {u} only when a = 0.

2. ∅ is linearly independent.

Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V.

-> If S1 is linearly dependent, then S2 is dependent.

-> If S2 is linearly independent, then S1 is linearly independent. (Hint: contrapositive)

== Basis ==

'''Definition''': A subset β is called a basis if 1. β generates V → span(β) = V and 2. β is linearly independent.

===Examples===

1. V = {0}, β = {}

2. {ei} for F^n, this is what we call the '''standard basis'''

3. B = {(1,1),(1, -1)} is a basis for R^2

4. P_n(F) β = {x^n, x^n-1, ... , x^1, x^0}

5. P(F), β = (x^0, x^1 ... and on} ('''Infinite basis'''!)

== Interesting inequality ==

This holds is true if the field does not have characteristic 2. Can you see why?

(a,b) = (a+b)/2 * (1, 1) + (a-b)/2 * (1, -1)

== Lecture notes scanned by [[User:starash|starash]] ==
<gallery>
Image:12-240-1004-1.jpg|Page 1
Image:12-240-1004-2.jpg|Page 2
</gallery></math>

12-240/Classnotes for Thursday October 4

2012-12-07T09:44:25Z

Peterlue: /* Examples */

{{12-240/Navigation}}
== Reminders ==
Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

== Recap ==

Base - what were doing today

Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dots, u_n\}</math> if <math>v = a_1u_1, \dots, a_nu_n</math> for scalars from a field <math>F</math>.

Span - <math>\operatorname{span}(S)</math> is the set of all linear combinations of the set <math>S</math>.

Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

== Pre - Basis ==

'''Linear dependence'''

'''Definition''' A set S ⊂ V is called linearly dependent if you can express the zero vector as a linear combination of distinct vectors from S, excluding the non-trivial linear combination where all of the scalars are 0.

Otherwise, we call S '''linearly independent.'''

=== Examples ===

1. In '''R'''^3, take S = {u1 = (1,4,7), u2 = (2,5,8), u3 = (3,6,9)}

u1 - 2u2 + u3 = 0

S is linearly dependent.

2. In R^n, take {e_i} = {0, 0, ... 1, 0, 0, 0} where 1 is in the ith position, and (ei) is a vector with n entries.

Claim: This set is linearly independent.

Proof: Suppose (∑ ai∙ei) = 0 ({ei} is linearly dependent.)

(∑ ai∙ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

⇒ a1 = a2 = ... = an = 0!

===Comments ===

1. {u} is linearly independent.
Proof:
⇐ If u≠0, suppose au =0
By property (a*u = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.

⇒ By definition, au = 0 for {u} only when a = 0.

2. ∅ is linearly independent.

Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V.

-> If S1 is linearly dependent, then S2 is dependent.

-> If S2 is linearly independent, then S1 is linearly independent. (Hint: contrapositive)

== Basis ==

'''Definition''': A subset β is called a basis if 1. β generates V → span(β) = V and 2. β is linearly independent.

===Examples===

1. V = {0}, β = {}

2. {ei} for F^n, this is what we call the '''standard basis'''

3. B = {(1,1),(1, -1)} is a basis for R^2

4. P_n(F) β = {x^n, x^n-1, ... , x^1, x^0}

5. P(F), β = (x^0, x^1 ... and on} ('''Infinite basis'''!)

== Interesting inequality ==

This holds is true if the field does not have characteristic 2. Can you see why?

(a,b) = (a+b)/2 * (1, 1) + (a-b)/2 * (1, -1)

== Lecture notes scanned by [[User:starash|starash]] ==
<gallery>
Image:12-240-1004-1.jpg|Page 1
Image:12-240-1004-2.jpg|Page 2
</gallery></math>

12-240/Classnotes for Thursday October 4

2012-12-07T09:43:47Z

Peterlue: /* Recap */

{{12-240/Navigation}}
== Reminders ==
Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

== Recap ==

Base - what were doing today

Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dots, u_n\}</math> if <math>v = a_1u_1, \dots, a_nu_n</math> for scalars from a field <math>F</math>.

Span - <math>\operatorname{span}(S)</math> is the set of all linear combinations of the set <math>S</math>.

Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

== Pre - Basis ==

'''Linear dependence'''

'''Definition''' A set S ⊂ V is called linearly dependent if you can express the zero vector as a linear combination of distinct vectors from S, excluding the non-trivial linear combination where all of the scalars are 0.

Otherwise, we call S '''linearly independent.'''

=== Examples ===

1. In '''R'''^3, take S = {u1 = (1,4,7), u2 = (2,5,8), u3 = (3,6,9)}

u1 - 2u2 + u3 = 0

S is linearly dependent.

2. In R^n, take {e_i} = {0, 0, ... 1, 0, 0, 0} where 1 is in the ith position, and (ei) is a vector with n entries.

Claim: This set is linearly independent.

Proof: Suppose (∑ ai*ei) = 0 ({ei} is linearly dependent.)

(∑ ai*ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

⇒ a1 = a2 = ... = an = 0!

===Comments ===

1. {u} is linearly independent.
Proof:
⇐ If u≠0, suppose au =0
By property (a*u = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.

⇒ By definition, au = 0 for {u} only when a = 0.

2. ∅ is linearly independent.

Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V.

-> If S1 is linearly dependent, then S2 is dependent.

-> If S2 is linearly independent, then S1 is linearly independent. (Hint: contrapositive)

== Basis ==

'''Definition''': A subset β is called a basis if 1. β generates V → span(β) = V and 2. β is linearly independent.

===Examples===

1. V = {0}, β = {}

2. {ei} for F^n, this is what we call the '''standard basis'''

3. B = {(1,1),(1, -1)} is a basis for R^2

4. P_n(F) β = {x^n, x^n-1, ... , x^1, x^0}

5. P(F), β = (x^0, x^1 ... and on} ('''Infinite basis'''!)

== Interesting inequality ==

This holds is true if the field does not have characteristic 2. Can you see why?

(a,b) = (a+b)/2 * (1, 1) + (a-b)/2 * (1, -1)

== Lecture notes scanned by [[User:starash|starash]] ==
<gallery>
Image:12-240-1004-1.jpg|Page 1
Image:12-240-1004-2.jpg|Page 2
</gallery></math>

12-240/Classnotes for Tuesday October 2

2012-12-07T09:43:04Z

Peterlue: /* Subspace */

{{12-240/Navigation}}
The "vitamins" slide we viewed today is [http://drorbn.net/AcademicPensieve/Classes/12-240/index.html?im=FoodsHandout.jpg here].

Today, the professor introduces more about subspace, linear combination, and related subjects.

== Subspace ==

Remind about the theorem of subspace: a non-empty subset W ⊂ V is a subspace iff is is closed under the operations of V and contains 0 of V

Proof:

First direction "=>":

if a non-empty subset W ⊂ V is a subspace , then W is a vector space over the operations of V .

=> + W is closed under the operations of V.

+ W has a unique identity of addition: <math>\forall\!\,</math> a <math>\in\!\,</math> W: 0 + a = a

Moreover, a a <math>\in\!\,</math> V. Hence 0 is also identity of addtition of V

Second direction "<=":

if a non-empty subset W ⊂ V is closed under the operations of V and contains 0 of V

we need to prove that W is a vector space over operations of V, hence, and subspace of V.

Namely, we need to show that W satisfies all axioms of a vector space, but now we just consider some axioms and leave the rest to readers.

VS1: Consider <math>\forall\!\,</math> x,y <math>\in\!\,</math> W => a,b <math>\in\!\,</math> V

While V is a vector space

thus x + y = y + x ( and the sum <math>\in\!\,</math> W since W is closed under addition)

VS2: (x + y) + z = x + (y + z) is proven similarly

VS3: As given, 0 of V <math>\in\!\,</math> W, pick any a in W ( possible since W is not empty)

So, a <math>\in\!\,</math> V hence a + 0 = a

Thus 0 is also additive identity element of W

== Class Notes ==

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Image:12-240-Oct-2-Page-1.jpg|Page 1
Image:12-240-Oct-2-Page-2.jpg|Page 2
Image:12-240-Oct-2-Page-3.jpg|Page 3
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</gallery>

12-240/Classnotes for Tuesday September 25

2012-12-07T09:41:28Z

Peterlue: /* Theorem */

{{12-240/Navigation}}

Today's class dealt with the properties of vector spaces.

== Definition ==
Let F be a field. A vector space V over F is a set V of vectors with special element O ( of V) and two operations: (+): VxV->V, (.): FxV->V

VxV={(v,w): v,w <math>\in\!\,</math> V}

FxV={(c,v): c <math>\in\!\,</math> F, v <math>\in\!\,</math> V}

Then, (+): VxV->V is (v,w)= v+w; (.): FxV->V is (c,v)=cv

Such that

VS1 <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: x+y = y+x

VS2 <math>\forall\!\,</math> x, y, z <math>\in\!\,</math> V: x+(y+z) = (x+y)+z

VS3 <math>\forall\!\,</math> x <math>\in\!\,</math> V: 0 ( of V) +x = x

VS4 <math>\forall\!\,</math> x <math>\in\!\,</math> V, <math>\exists \!\,</math> V <math>\in\!\,</math> V: v + x= 0 ( of V)

VS5 <math>\forall\!\,</math> x <math>\in\!\,</math> V, 1 (of F) .x = x

VS6 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (ab)x = a(bx)

VS7 <math>\forall\!\,</math> a <math>\in\!\,</math> F, <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: a(x + y)= ax + ay

VS8 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (a + b)x = ax + bx

== Examples ==

==Properties ==

==Polynomials==
'''Definition''' : Pn(F) = {all polynomials of degree less than or equal to n with coefficients in F}
= {anx^n + an-1x^n-1 + ... + a1x^1 + a1x^1}

0 = 0x^n + 0x^n-1 +...+ 0x^0

addition and multiplication: as you imagine

P(f) = {all polynomials with coefficients in F}

Take F= '''Z'''/2 |F| = 2

|P(F)| = infinite

in Pn('''Z'''/2) x^3≠x^2
x^3 = 1*x^3+0x^2+0x+O = f
x^2 = 1*x^2+0x+0 = g
yet f(0)= g(0) and f(1)=g(1)

==Theorem==
1. Cancellation Laws
(a) x+z=y+z ==> x=y
(b) ax=ay,a≠0 ==> x=y
(c) x≠0 of V, ax=bx ==> a=b

2. 0 of V is unique

3. Negatives are unique (so subtraction makes sense

4.(0 of F)x = 0 of V

5. a∙0=0

6. (-a)x = -(ax) = a(-x)

7. a∙v=0 <==> a=0 or v=0

==Proof==
1. (a) x+z=y+z
Find a w s.t. z+w=0 (V.S. 4)
(x+z)+w = (y+z)+w
Use VS2
x+(z+w) = y +(z+w)
x + 0 = y + o
Use VS3 x=y

==Scanned Notes by [[User:Richardm|Richardm]]==
<gallery>
Image:12-240-0925-1vectorspaces.jpg
Image:12-240-0925-2vectorspaces.jpg
Image:12-240-0925-3vectorspaces.jpg
Image:12-240-0925-4vectorspaces.jpg
Image:12-240-0925-5vectorspaces.jpg
Image:12-240-0925-6vectorspaces.jpg
</gallery>

12-240/Classnotes for Tuesday September 25

2012-12-07T09:40:39Z

Peterlue: /* Theorem */

{{12-240/Navigation}}

Today's class dealt with the properties of vector spaces.

== Definition ==
Let F be a field. A vector space V over F is a set V of vectors with special element O ( of V) and two operations: (+): VxV->V, (.): FxV->V

VxV={(v,w): v,w <math>\in\!\,</math> V}

FxV={(c,v): c <math>\in\!\,</math> F, v <math>\in\!\,</math> V}

Then, (+): VxV->V is (v,w)= v+w; (.): FxV->V is (c,v)=cv

Such that

VS1 <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: x+y = y+x

VS2 <math>\forall\!\,</math> x, y, z <math>\in\!\,</math> V: x+(y+z) = (x+y)+z

VS3 <math>\forall\!\,</math> x <math>\in\!\,</math> V: 0 ( of V) +x = x

VS4 <math>\forall\!\,</math> x <math>\in\!\,</math> V, <math>\exists \!\,</math> V <math>\in\!\,</math> V: v + x= 0 ( of V)

VS5 <math>\forall\!\,</math> x <math>\in\!\,</math> V, 1 (of F) .x = x

VS6 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (ab)x = a(bx)

VS7 <math>\forall\!\,</math> a <math>\in\!\,</math> F, <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: a(x + y)= ax + ay

VS8 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (a + b)x = ax + bx

== Examples ==

==Properties ==

==Polynomials==
'''Definition''' : Pn(F) = {all polynomials of degree less than or equal to n with coefficients in F}
= {anx^n + an-1x^n-1 + ... + a1x^1 + a1x^1}

0 = 0x^n + 0x^n-1 +...+ 0x^0

addition and multiplication: as you imagine

P(f) = {all polynomials with coefficients in F}

Take F= '''Z'''/2 |F| = 2

|P(F)| = infinite

in Pn('''Z'''/2) x^3≠x^2
x^3 = 1*x^3+0x^2+0x+O = f
x^2 = 1*x^2+0x+0 = g
yet f(0)= g(0) and f(1)=g(1)

==Theorem==
1. Cancellation Laws
(a) x+z=y+z ==> x=y
(b) ax=ay,a≠0 ==> x=y
(c) x≠0 of V, ax=bx ==> a=b

2. 0 of V is unique

3. Negatives are unique (so subtraction makes sense

4.(0 of F)x = 0 of V

5. a*0=0

6. (-a)x= - (ax) = a(-x)

7. a*v=0 <==> a=0 or v=0

==Proof==
1. (a) x+z=y+z
Find a w s.t. z+w=0 (V.S. 4)
(x+z)+w = (y+z)+w
Use VS2
x+(z+w) = y +(z+w)
x + 0 = y + o
Use VS3 x=y

==Scanned Notes by [[User:Richardm|Richardm]]==
<gallery>
Image:12-240-0925-1vectorspaces.jpg
Image:12-240-0925-2vectorspaces.jpg
Image:12-240-0925-3vectorspaces.jpg
Image:12-240-0925-4vectorspaces.jpg
Image:12-240-0925-5vectorspaces.jpg
Image:12-240-0925-6vectorspaces.jpg
</gallery>

12-240/Classnotes for Tuesday September 25

2012-12-07T09:39:31Z

Peterlue: /* Polynomials */

{{12-240/Navigation}}

Today's class dealt with the properties of vector spaces.

== Definition ==
Let F be a field. A vector space V over F is a set V of vectors with special element O ( of V) and two operations: (+): VxV->V, (.): FxV->V

VxV={(v,w): v,w <math>\in\!\,</math> V}

FxV={(c,v): c <math>\in\!\,</math> F, v <math>\in\!\,</math> V}

Then, (+): VxV->V is (v,w)= v+w; (.): FxV->V is (c,v)=cv

Such that

VS1 <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: x+y = y+x

VS2 <math>\forall\!\,</math> x, y, z <math>\in\!\,</math> V: x+(y+z) = (x+y)+z

VS3 <math>\forall\!\,</math> x <math>\in\!\,</math> V: 0 ( of V) +x = x

VS4 <math>\forall\!\,</math> x <math>\in\!\,</math> V, <math>\exists \!\,</math> V <math>\in\!\,</math> V: v + x= 0 ( of V)

VS5 <math>\forall\!\,</math> x <math>\in\!\,</math> V, 1 (of F) .x = x

VS6 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (ab)x = a(bx)

VS7 <math>\forall\!\,</math> a <math>\in\!\,</math> F, <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: a(x + y)= ax + ay

VS8 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (a + b)x = ax + bx

== Examples ==

==Properties ==

==Polynomials==
'''Definition''' : Pn(F) = {all polynomials of degree less than or equal to n with coefficients in F}
= {anx^n + an-1x^n-1 + ... + a1x^1 + a1x^1}

0 = 0x^n + 0x^n-1 +...+ 0x^0

addition and multiplication: as you imagine

P(f) = {all polynomials with coefficients in F}

Take F= '''Z'''/2 |F| = 2

|P(F)| = infinite

in Pn('''Z'''/2) x^3≠x^2
x^3 = 1*x^3+0x^2+0x+O = f
x^2 = 1*x^2+0x+0 = g
yet f(0)= g(0) and f(1)=g(1)

==Theorem==
1. Cancellation Laws
(a) x+z=y+z ==> x=y
(b) ax+ay,a≠0 ==> x=y
(c) x≠0 of V, ax=bx ==> a=b

2. 0 of V is unique

3. Negatives are unique (so subtraction makes sense

4.(0 of F)x = 0 of V

5. a*0=0

6. (-a)x= - (ax) = a(-x)

7. a*v=0 <==> a=0 or v=0

==Proof==
1. (a) x+z=y+z
Find a w s.t. z+w=0 (V.S. 4)
(x+z)+w = (y+z)+w
Use VS2
x+(z+w) = y +(z+w)
x + 0 = y + o
Use VS3 x=y

==Scanned Notes by [[User:Richardm|Richardm]]==
<gallery>
Image:12-240-0925-1vectorspaces.jpg
Image:12-240-0925-2vectorspaces.jpg
Image:12-240-0925-3vectorspaces.jpg
Image:12-240-0925-4vectorspaces.jpg
Image:12-240-0925-5vectorspaces.jpg
Image:12-240-0925-6vectorspaces.jpg
</gallery>

12-240/Classnotes for Tuesday September 25

2012-12-07T09:38:50Z

Peterlue: /* Definition */

{{12-240/Navigation}}

Today's class dealt with the properties of vector spaces.

== Definition ==
Let F be a field. A vector space V over F is a set V of vectors with special element O ( of V) and two operations: (+): VxV->V, (.): FxV->V

VxV={(v,w): v,w <math>\in\!\,</math> V}

FxV={(c,v): c <math>\in\!\,</math> F, v <math>\in\!\,</math> V}

Then, (+): VxV->V is (v,w)= v+w; (.): FxV->V is (c,v)=cv

Such that

VS1 <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: x+y = y+x

VS2 <math>\forall\!\,</math> x, y, z <math>\in\!\,</math> V: x+(y+z) = (x+y)+z

VS3 <math>\forall\!\,</math> x <math>\in\!\,</math> V: 0 ( of V) +x = x

VS4 <math>\forall\!\,</math> x <math>\in\!\,</math> V, <math>\exists \!\,</math> V <math>\in\!\,</math> V: v + x= 0 ( of V)

VS5 <math>\forall\!\,</math> x <math>\in\!\,</math> V, 1 (of F) .x = x

VS6 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (ab)x = a(bx)

VS7 <math>\forall\!\,</math> a <math>\in\!\,</math> F, <math>\forall\!\,</math> x, y <math>\in\!\,</math> V: a(x + y)= ax + ay

VS8 <math>\forall\!\,</math> a, b <math>\in\!\,</math> F, <math>\forall\!\,</math> x <math>\in\!\,</math> V: (a + b)x = ax + bx

== Examples ==

==Properties ==

==Polynomials==
'''Definition''' : Pn(F) = {all polynomials of degree less than or equal n with coefficients in F}
= {anx^n + an-1x^n-1 + ... + a1x^1 + a1x^1}

0 = 0x^n + 0x^n-1 +...+ 0x^0

addition and multiplication: as you imagine

P(f) = {all polynomials with coefficients in F}

Take F= '''Z'''/2 |F| = 2

|P(F)| = infinite

in Pn('''Z'''/2) x^3≠x^2
x^3 = 1*x^3+0x^2+0x+O = f
x^2 = 1*x^2+0x+0 = g
yet f(0)= g(0) and f(1)=g(1)

==Theorem==
1. Cancellation Laws
(a) x+z=y+z ==> x=y
(b) ax+ay,a≠0 ==> x=y
(c) x≠0 of V, ax=bx ==> a=b

2. 0 of V is unique

3. Negatives are unique (so subtraction makes sense

4.(0 of F)x = 0 of V

5. a*0=0

6. (-a)x= - (ax) = a(-x)

7. a*v=0 <==> a=0 or v=0

==Proof==
1. (a) x+z=y+z
Find a w s.t. z+w=0 (V.S. 4)
(x+z)+w = (y+z)+w
Use VS2
x+(z+w) = y +(z+w)
x + 0 = y + o
Use VS3 x=y

==Scanned Notes by [[User:Richardm|Richardm]]==
<gallery>
Image:12-240-0925-1vectorspaces.jpg
Image:12-240-0925-2vectorspaces.jpg
Image:12-240-0925-3vectorspaces.jpg
Image:12-240-0925-4vectorspaces.jpg
Image:12-240-0925-5vectorspaces.jpg
Image:12-240-0925-6vectorspaces.jpg
</gallery>

12-240/Classnotes for Tuesday September 18

2012-12-07T09:37:31Z

Peterlue: /* Complex number */

{{12-240/Navigation}}

Today's handout, "TheComplexField", can be had from {{Pensieve link|Classes/12-240|Pensieve: Classes: 12-240}}.

{{Template:12-240:Dror/Students Divider}}

In this class, the professor continued with some more theorems of field and introduced definition and theorems of complex number.

== Various properties of fields ==
'''Thrm 1''': In a field F:
1. a+b = c+b ⇒ a=c

2. b≠0, a∙b=c∙b ⇒ a=c

3. 0 is unique.

4. 1 is unique.

5. -a is unique.

6. a^-1 is unique (a≠0)

7. -(-a)=a

8. (a^-1)^-1 =a

9. a∙0=0 **Surprisingly difficult, required distributivity.

10. ∄ 0^-1, aka, ∄ b∈F s.t 0∙b=1

11. (-a)∙(-b)=a∙b

12. a∙b=0 ''iff'' a=0 or b=0

.
.
.

16. (a+b)∙(a-b)= a^2 - b^2
[Define a^2 = a∙a]
Hint: Use distributive law

'''Thrm 2''': Given a field F, there exists a map Ɩ: Z → F with the properties (∀ m,n ∈ Z):

1) Ɩ(0) =0, Ɩ(1)=1

2) Ɩ(m+n) = Ɩ(m) +Ɩ(n)

3) Ɩ(mn) = Ɩ(m)∙Ɩ(n)

Furthermore, Ɩ is unique.

'''Rough proof''':

Test somes cases:

Ɩ(2) = Ɩ(1+1) = Ɩ(1) + Ɩ(1) = 1 + 1 ≠ 2

Ɩ(3) = Ɩ(2 +1)= Ɩ(2) + Ɩ(1) = 1+ 1+ 1 ≠ 3

.
.
.

Ɩ(n) = 1 + ... + 1 (n times)

Ɩ(-3) = ?

Ɩ(-3 + 3) = Ɩ(-3) + Ɩ(3) ⇒ Ɩ(-3) = -Ɩ(3) = -(1+1+1)

''What about uniqueness?'' Simply put, we had not choice in the definition of Ɩ. All followed from the given properties.

At this point, we will be lazy and simply denote Ɩ(3) = 3_f [3 with subscript f]

∃ m≠0, m ∈ '''N''', Ɩ(m) =0

In which case, there is a smallest m>0, for which Ɩ(m)=0. '''m' is the characteristic of F.'' Denoted char(F).
Examples: char(F_2)=2, char(F_3)=3... but NOTE: char('''R''')=0

'''Thrm:''' If F is a field and char(F) >0, then char(F) is a prime number.

Proof: Suppose char(F) =m, m>0. Suppose also m is not prime: m=ts, t,s ∈ '''N'''.

Then, Ɩ(m) = 0 = Ɩ(t)∙Ɩ(s) ⇒ Ɩ(t)=0 or Ɩ(s)=0 by P12.

If Ɩ(t)=0 ⇒ t≧m ⇒ m=t, s=1 or likewise for Ɩ(s)=0, and m=s, t=1

In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎

== Complex number==

'''Abstraction, generalization, definition, examples, properties, dream, implications, realization = formalization, PROOF.'''

Consider that fact that in '''R''', ∄ x s.t. x^2 = -1

Dream: Add new number element 𝒊 to '''R''', so as to still get a field & 𝒊^2 = -1

Implications: By adding 𝒊, we must add 7𝒊, and 2+7𝒊, 3+4𝒊, (2+7𝒊)∙(3+4𝒊), (2+7𝒊)^-1, etc.

So, how do we define this field?

'''Definition'''

'''C''' = {(a,b): a,b ∈ '''R'''}
Also, 0 (of the field) = (0,0); 1( of the field) = (1,0)

Define addition: (a,b)+(c,d) = (a+c, b+d)

Define multification: (a,b)(c,d) = (ac-bd, ad+bc)

'''Theorems:'''

'''Thrm. 1.''' ('''C''', 0, 1, +, ∙) is a field.

'''Thrm. 2.''' ∃ 𝒊 ∈ '''C''' s.t. 𝒊^2 = -1

'''Thrm. 3.''' '''C''' contains '''R'''

Proof (1): Show that each of the field axioms holds for '''C'''.

Ex. F1(a): ƶ1 + ƶ2 = ƶ2 + ƶ1, where ƶ1 = (a1, b1) and ƶ2 = (a2, b2)

LHS: (a1,b1)+(a2,b2) = (a1+a2, b1+b2)

RHS: (a2,b2)+(a1,b1) = (a2+a1, b2+b1)

LHS=RHS by F1 of '''R'''

F1(b) and so on...

Proof (2):

In '''C''', consider i=(0,1)

By the definition i^2=i.i=(0.1-1.1,0.1+1.0)=(-1,0)

We also have 1(of '''c''') + (-1,0)=(1,0)+(-1,0)=(0,0)=0 (of '''c''')

Hence (-1,0) is the addictive inverse of 1, i.e, (-1,0)=-1

Thus i^2=-1. ∎

Proof 3:

Given the field '''C''' : map J: '''R''' -> '''C'''

1) J(0)=(0,0); J(1)=(1,0)

2) J(x+y)=J(x)+J(y); J(x.y)=J(x)J(y)

Define J(x)=(x,0), all will follow.

From now on J(x) will be writen simply x

EX: J(7)=7, J(3)=3

So, what does a+b𝒊 mean? (a, b ∈ '''R''')

a+b𝒊= Ɩ(a) + Ɩ(b)∙Ɩ(𝒊) = (a,0) + (b,0)∙(0,1) = (a,b)

Hence, (a,b) ~ a+b𝒊

Thus, we can use a+b𝒊 with less hesitation.

== Lecture 3, scanned notes upload by [[User:Starash|Starash]] and [[User:KJMorenz|KJMorenz]]==

<gallery>
Image:12-240-0918-1.jpg|Page 1
Image:12-240-0918-2.jpg|Page 2
Image:12-240-0918-3.jpg|Page 3
</gallery>
[[Image:12-240-CharacteristicOfField.jpg|400px]]
[[Image:12-240-ComplexNums.jpg|400px]]

12-240/Classnotes for Tuesday September 18

2012-12-07T09:36:30Z

Peterlue: /* Complex number */

{{12-240/Navigation}}

Today's handout, "TheComplexField", can be had from {{Pensieve link|Classes/12-240|Pensieve: Classes: 12-240}}.

{{Template:12-240:Dror/Students Divider}}

In this class, the professor continued with some more theorems of field and introduced definition and theorems of complex number.

== Various properties of fields ==
'''Thrm 1''': In a field F:
1. a+b = c+b ⇒ a=c

2. b≠0, a∙b=c∙b ⇒ a=c

3. 0 is unique.

4. 1 is unique.

5. -a is unique.

6. a^-1 is unique (a≠0)

7. -(-a)=a

8. (a^-1)^-1 =a

9. a∙0=0 **Surprisingly difficult, required distributivity.

10. ∄ 0^-1, aka, ∄ b∈F s.t 0∙b=1

11. (-a)∙(-b)=a∙b

12. a∙b=0 ''iff'' a=0 or b=0

.
.
.

16. (a+b)∙(a-b)= a^2 - b^2
[Define a^2 = a∙a]
Hint: Use distributive law

'''Thrm 2''': Given a field F, there exists a map Ɩ: Z → F with the properties (∀ m,n ∈ Z):

1) Ɩ(0) =0, Ɩ(1)=1

2) Ɩ(m+n) = Ɩ(m) +Ɩ(n)

3) Ɩ(mn) = Ɩ(m)∙Ɩ(n)

Furthermore, Ɩ is unique.

'''Rough proof''':

Test somes cases:

Ɩ(2) = Ɩ(1+1) = Ɩ(1) + Ɩ(1) = 1 + 1 ≠ 2

Ɩ(3) = Ɩ(2 +1)= Ɩ(2) + Ɩ(1) = 1+ 1+ 1 ≠ 3

.
.
.

Ɩ(n) = 1 + ... + 1 (n times)

Ɩ(-3) = ?

Ɩ(-3 + 3) = Ɩ(-3) + Ɩ(3) ⇒ Ɩ(-3) = -Ɩ(3) = -(1+1+1)

''What about uniqueness?'' Simply put, we had not choice in the definition of Ɩ. All followed from the given properties.

At this point, we will be lazy and simply denote Ɩ(3) = 3_f [3 with subscript f]

∃ m≠0, m ∈ '''N''', Ɩ(m) =0

In which case, there is a smallest m>0, for which Ɩ(m)=0. '''m' is the characteristic of F.'' Denoted char(F).
Examples: char(F_2)=2, char(F_3)=3... but NOTE: char('''R''')=0

'''Thrm:''' If F is a field and char(F) >0, then char(F) is a prime number.

Proof: Suppose char(F) =m, m>0. Suppose also m is not prime: m=ts, t,s ∈ '''N'''.

Then, Ɩ(m) = 0 = Ɩ(t)∙Ɩ(s) ⇒ Ɩ(t)=0 or Ɩ(s)=0 by P12.

If Ɩ(t)=0 ⇒ t≧m ⇒ m=t, s=1 or likewise for Ɩ(s)=0, and m=s, t=1

In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎

== Complex number==

'''Abstraction, generalization, definition, examples, properties, dream, implications, realization = formalization, PROOF.'''

Consider that fact that in '''R''', ∄ x s.t. x^2 = -1

Dream: Add new number element 𝒊 to '''R''', so as to still get a field & 𝒊^2 = -1

Implications: By adding 𝒊, we must add 7𝒊, and 2+7𝒊, 3+4𝒊, (2+7𝒊)∙(3+4𝒊), (2+7𝒊)^-1, etc.

So, how do we define this field?

'''Definition'''

'''C''' = {(a,b): a,b ∈ '''R'''}
Also, 0 (of the field) = (0,0); 1( of the field) = (1,0)

Define addition: (a,b)+(c,d) = (a+c, b+d)

Define multification: (a,b)(c,d) = (ac-bd, ad+bc)

'''Theorems:'''

'''Thrm. 1.''' ('''C''', 0, 1, +, ∙) is a field.

'''Thrm. 2.''' ∃ 𝒊 ∈ '''C''' s.t. 𝒊^2 = -1

'''Thrm. 3.''' '''C''' contains '''R'''

Proof (1): Show that each of the field axioms holds for '''C'''.

Ex. F1(a): ƶ1 + ƶ2 = ƶ2 + ƶ1, where ƶ1 = (a1, b1) and ƶ2 = (a2, b2)

LHS: (a1,b1)+(a2,b2) = (a1+a2, b1+b2)

RHS: (a2,b2)+(a1,b1) = (a2+a1, b2+b1)

LHS=RHS by F1 of '''R'''

F1(b) and so on...

Proof (2):

In '''C''', consider i=(0,1)

By the definition i^2=i.i=(0.1-1.1,0.1+1.0)=(-1,0)

We also have 1(of '''c''') + (-1,0)=(1,0)+(-1,0)=(0,0)=0 (of '''c''')

Hence (-1,0) is the addictive inverse of 1, i.e, (-1,0)=-1

Thus i^2=-1. ∎

Proof 3:

Given the field '''C''' : map J: '''R''' -> '''C'''

1) J(0)=(0,0); J(1)=(1,0)

2) J(x+y)=J(x)+J(y); J(x.y)=J(x)J(y)

Define J(x)=(x,0), all will follow.

From now on J(x) will be writen simply x

EX: J(7)=7, J(3)=3

So, what does a+b𝒊 mean? (a, b ∈ '''R''')

a+b𝒊= Ɩ(a) + Ɩ(b)+Ɩ(𝒊) = (a,0) + (b,0)∙(0,1) = (a,b)

Hence, (a,b) ~ a+b𝒊

Thus, we can use a+b𝒊 with less hesitation.

== Lecture 3, scanned notes upload by [[User:Starash|Starash]] and [[User:KJMorenz|KJMorenz]]==

<gallery>
Image:12-240-0918-1.jpg|Page 1
Image:12-240-0918-2.jpg|Page 2
Image:12-240-0918-3.jpg|Page 3
</gallery>
[[Image:12-240-CharacteristicOfField.jpg|400px]]
[[Image:12-240-ComplexNums.jpg|400px]]

12-240/Class Photo

2012-10-10T14:16:03Z

Peterlue: /* Who We Are... */

Our class on September 25, 2012:

[[Image:12-240-ClassPhoto.jpg|thumb|centre|800px|Class Photo: click to enlarge]]
{{12-240/Navigation}}

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 cellspacing=0
|-
!First Name
!Last Name
!ID wcashore
!e-mail
!Location
!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. For better line-breaking, leave a space next to the "@" in email addresses.}}
{{Photo Entry|last=Bartnicki|first=Piotr|userid=Peter|email=piotr.bartnicki@ mail.utoronto.ca|location=Left part of the last row sitting directly between two standing guys, *left* of the one in orange (from the camera's perspective) and to the right of one in a black striped shirt |comments=}}
{{Photo Entry|last=Cashore|first=Walter|userid=wcashore|email=wcashore 12 @ hotmail .com|location=third row back, in the green star wars shirt|comments=great pic guys}}
{{Photo Entry|last=Frailich|first=Rebecca|userid=Rebecca.frailich|email=rebecca. frailich@ mail. utoronto. ca|location=Last row, in between two guys standing at the back (one in red, one in black) |comments=}}
{{Photo Entry|last=Hoover|first=Ken|userid=Khoover|email=ken.hoover@ mail.utoronto.ca|location=First row, fourth from the right.|comments=}}
{{Photo Entry|last=Kennedy|first=Christopher|userid=ckennedy|email=christopherpa. kennedy@ mail. utoronto. ca|location=Third row; third from the right in white |comments=}}
{{Photo Entry|last=Klingspor|first=Josefine|userid=Josefine|email=josefine. klingspor@ mail. utoronto. ca|location=First row, second from left.|comments=}}
{{Photo Entry|last=Le|first=Quan|userid=Quanle|email=quan. le@ mail. utoronto. ca|location=Start bottom right corner, third from right. Go three steps north-west. Directly north-east from there, in blue collar shirt|comments=}}
{{Photo Entry|last=Liu|first=Zhaowei|userid=tod|email=tod. liu@ mail. utoronto .ca|location=First row, third from the right|comments=}}
{{Photo Entry|last=Lue|first=Peter|userid=Peterlue|email=peter. lue@ mail. utoronto. ca|location=On the left edge 3rd from the back in the reddish shirt|comments=}}
{{Photo Entry|last=Millson|first=Richard|userid=Richardm|email=r.millson@ mail. utoronto. ca|location=Seventh row from the front, fourth from the right, blue sweater|comments=}}
{{Photo Entry|last=McGrath|first=Celton|userid=CeltonMcGrath|email=celton. mcgrath@ mail. utoronto. ca|location=4th row front from, centre right, brown sweater|comments=}}
{{Photo Entry|last=Morenz|first=Karen|userid=KJMorenz|email=kjmorenz@ gmail.com|location=3rd-ish row from the back, centre right, purple shirt|comments=}}
{{Photo Entry|last=Pan|first=Li|userid=panli19|email=panli19@gmail.com|location=fourth row, the guy in grey fleece sweater.|comments=}}
{{Photo Entry|last=Ratz|first=Derek|userid=Derek.ratz|email=ratz.derek@gmail.com|location=2nd from the back, 2 in from the far left, yellow shirt|comments=}}
{{Photo Entry|last=Tong|first=Cheng Yu|userid=Chengyu.tong|email=chengyu. tong@ mail. utoronto. ca|location=fourth row from the front on the left side of the picture wearing green sweater and black rimmed glasses |comments=}}
{{Photo Entry|last=Vicencio-Heap|first=Felipe|userid=Heapfeli|email=felipe. vicencio. heap@ mail. utoronto. ca|location=Second row from the front, furthest to the right.|comments=}}
{{Photo Entry|last=Wamer|first=Kyle|userid=kylewamer|email=kyle. wamer @ mail. utoronto. ca|location=Second row, fifth from the left in the red shirt.|comments=}}
{{Photo Entry|last=Winnitoy|first=Leigh|userid=Leighwinnitoy|email=leigh.winnitoy@ mail. utoronto. ca|location=sixth row, near the middle of the picture|comments=}}
{{Photo Entry|last=Yang|first=Chen|userid=chen|email=neochen. yang@ mail. utoronto. ca|location=sixth row, first from the right in the black pull-over.|comments=}}
{{Photo Entry|last=Yang|first=Tianlin|userid=Tianlin.yang|email=Tianin.Yang@ mail. utoronto. ca|location=4th row, first from left in blue wind coat.|comments=}}
{{Photo Entry|last=Zhang|first=BingZhen|userid=Zetalda|email=bingzhen. zhang@ mail. utoronto. ca|location=Second last row, third from left.|comments=}}
{{Photo Entry|last=Zhao|first=TianChen|userid=Ericolony|email=zhao_ tianchen@ hotmail. com|location=fourth row, the guy in green shirt.|comments=}}
{{Photo Entry|last=Zibert|first=Vincent|userid=vincezibert|email=vincent. zibert@ mail. utoronto. ca|location=Directly beneath the white notice posted on the door on the right-hand side.|comments=}}
{{Photo Entry|last=Zoghi|first=Sina|userid=sina.zoghi|email=sina.zoghi@ utoronto .ca|location=First row, leftest left.|comments=}}
{{Photo Entry|last=Léger|first=Zacharie|userid=zach.leger8|email=zacharie. leger@ mail. utronto. ca|location= 5th row in a black T-shirt.|comments=}}
{{Photo Entry|last=Wang|first=Minqi|userid=Michael.Wang|email=wangminqi@ yahoo.cn|location=First row, fourth from the left in black oufit) |comments=}}


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