Drorbn - User contributions [en]

12-240/Classnotes for Tuesday October 23

2012-10-30T19:23:35Z

Dangkhoa.nguyen: /* == */

{{12-240/Navigation}}
= ===
Definition: L(V,W) is the set of all linear transformation L: V->W

u <math>\in\,\!</math> V,
0 of L(V,W) (u)=0 of W (this is a l.t.str)

If L1 and L2 <math>\in\,\!</math> L(V,W),
(L1 + L2) (u)= L1(u) +L2(u) (this is a l.t.str)

If c <math>\in\,\!</math> F and L <math>\in\,\!</math> L(V,W),
(c*L) (u)= c*L(u) (this is a l.t.str)

Theorem: L(V,W) is a vector space

Proof: "Distributivity" c(x+y)=cx+cy

In our case need to show c(L1 + L2)= cL1 + cL2

Where c <math>\in\,\!</math> F and L1 and L2 <math>\in\,\!</math> L(V,W)

(LHS) (u)

== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Tuesday October 23

2012-10-30T19:22:29Z

Dangkhoa.nguyen: /* == */

{{12-240/Navigation}}
= ===
Definition: L(V,W) is the set of all linear transformation L: V->W

u <math>\in\,\!</math> V,
0 of L(V,W) (u)=0 of W (this is a l.t.str)

If L1 and L2 <math>\in\,\!</math> L(V,W),
(L1 + L2) (u)= L1(u) +L2(u) (this is a l.t.str)

If c <math>\in\,\!</math> F and L <math>\in\,\!</math> L(V,W),
(c*L) (u)= c*L(u) (this is a l.t.str)

Theorem: L(V,W) is a vector space

Proof: "Distributivity" c(x+y)=cx+cy

In our case need to show c(L1 + L2)= cL1 + cL2

== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Tuesday October 23

2012-10-30T19:20:41Z

Dangkhoa.nguyen:

{{12-240/Navigation}}
= ===
Definition: L(V,W) is the set of all linear transformation L: V->W

u <math>\in\,\!</math> V,
0 of L(V,W) (u)=0 of W (this is a l.t.str)

If L1 and L2 <math>\in\,\!</math> L(V,W),
(L1 + L2) (u)= L1(u) +L2(u) (this is a l.t.str)

If c <math>\in\,\!</math> F and L <math>\in\,\!</math> L(V,W),
(c*L) (u)= c*L(u) (this is a l.t.str)

Theorem: L(V,W) is a vector space

== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Tuesday October 23

2012-10-30T19:20:23Z

Dangkhoa.nguyen: /* = */

{{12-240/Navigation}}
== ===
Definition: L(V,W) is the set of all linear transformation L: V->W

u <math>\in\,\!</math> V,
0 of L(V,W) (u)=0 of W (this is a l.t.str)

If L1 and L2 <math>\in\,\!</math> L(V,W),
(L1 + L2) (u)= L1(u) +L2(u) (this is a l.t.str)

If c <math>\in\,\!</math> F and L <math>\in\,\!</math> L(V,W),
(c*L) (u)= c*L(u) (this is a l.t.str)

Theorem: L(V,W) is a vector space

== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Tuesday October 23

2012-10-30T19:08:45Z

Dangkhoa.nguyen:

{{12-240/Navigation}}
== ===
Definition: L(V,W) is the set of all linear transformation L: V->W
== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-30T03:49:12Z

Dangkhoa.nguyen: /* Linear transformation */

{{12-240/Navigation}}
== Linear transformation ==
Definition: A function L: V-> W is called a linear transformation if it preserve following structures:

1) L(x + y)= L(x) + L(y)
2) L(cx)= c.L(x)
3) L(0 of V) = 0 of W

Proposition:

1) property 2 leads to property 3
2) L: V -> W is a linear transformation iff <math>\forall\,\!</math> c <math>\in\,\!</math>, <math>\forall\,\!</math> x, y <math>\in\,\!</math> V: L(cx + y)= cL(x) + L(y)
Proof:
1) take c= 0 in F and x=0 in V. Then L(cx)=cL(x) -> L(0 of F . 0 of V)=(0 of F).L(0 of V)=0 of W
2) L(cx + y)= L(cx) + L(y)= c.L(x) + L(y)

== lecture note on oct 18, uploaded by [[User:starash|starash]]==

<gallery>
Image:12-240-1018-1.jpg |page1
Image:12-240-1018-2.jpg |page2
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-30T03:38:19Z

Dangkhoa.nguyen: /* Linear transformation */

{{12-240/Navigation}}
== Linear transformation ==
Definition: A function L: V-> W is called a linear transformation if it preserve following structures:

1) L(x + y)= L(x) + L(y)
2) L(cx)= c.L(x)
3) L(0 of V) = 0 of W

Proposition:

1) property 2 leads to property 3
2) L: V -> W is a linear transformation iff <math>\forall\,\!</math> c <math>\in\,\!</math>, <math>\forall\,\!</math> x, y <math>\in\,\!</math> V: L(cx + y)= cL(x) + L(y)

== lecture note on oct 18, uploaded by [[User:starash|starash]]==

<gallery>
Image:12-240-1018-1.jpg |page1
Image:12-240-1018-2.jpg |page2
</gallery>

12-240/Classnotes for Tuesday October 23

2012-10-26T02:57:43Z

Dangkhoa.nguyen:

Definition: L(V,W) is the set of all linear transformation L: V->W
== Lecture notes scanned by [[User:Zetalda|Zetalda]] ==
<gallery>
Image:12-240-Oct23-1.jpeg|Page 1
Image:12-240-Oct23-2.jpeg|Page 2
Image:12-240-Oct23-3.jpeg|Page 3
Image:12-240-Oct23-4.jpeg|Page 4
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-26T02:49:16Z

Dangkhoa.nguyen: /* Linear transformation */

{{12-240/Navigation}}
== Linear transformation ==
Definition: A function L: V-> W is called a linear transformation if it preserve following structures:

1) L(x + y)= L(x) + L(y)
2) L(cx)= c.L(x)
3) L(0 of V) = 0 of W

== lecture note on oct 18, uploaded by [[User:starash|starash]]==

<gallery>
Image:12-240-1018-1.jpg |page1
Image:12-240-1018-2.jpg |page2
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-26T02:43:54Z

Dangkhoa.nguyen:

{{12-240/Navigation}}
== Linear transformation ==

== lecture note on oct 18, uploaded by [[User:starash|starash]]==

<gallery>
Image:12-240-1018-1.jpg |page1
Image:12-240-1018-2.jpg |page2
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-18T17:05:02Z

Dangkhoa.nguyen: /* Theorems */

{{12-240/Navigation}}
===Riddle Along===
The game of 15 is played as follows. Two players alternate choosing cards numbered between 1 and 9, with repetitions forbidden, so that the game ends at most after 9 moves (or four and a half rounds). The first player to have within her/his cards a set of precisely 3 cards that add up to 15 wins.

Does this game has a winning strategy? What is it? Who wins, the first to move or the second? Why am I asking this question at this particular time?

[[Image:12-240-DeckOfCards.png|center]]

See also [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503 a video] and the [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ transcript] of that video.

{{12-240:Dror/Students Divider}}
== Theorems ==
1. If G generates, |G| <math>\ge \!\,</math> n and G contains a basis, |G|=n then G is a basis

2. If L is linearly independent, |L| <math>\le \!\,</math> n and L can be extended to be a basis. |L|=n => L is a basis.

3.W <math>\subset \!\,</math> V a subspace then W is finite dimensioned and dim W <math>\le \!\,</math> dim V

If dim W = dim V, then V = W
If dim W < dim V, then any basis of W can be extended to be a basis of V

Proof of W is finite dimensioned:

Let L be a linearly independent subset of W which is of maximal size.

Fact about '''N'''
: Every subset A of '''N''', which is:

1. Non empty

2. Bounded : <math>\exist \!\,</math> N <math>\in \!\,</math> '''N''', <math>\forall \!\,</math> a <math>\in \!\,</math> A, a <math>\le \!\,</math> N

has a maximal element: an element m <math>\in \!\,</math> A, <math>\forall\!\,</math> a <math>\in \!\,</math> A, a <math>\le \!\,</math> m ( m + 1 <math>\notin \!\,</math> A )

== class note ==

<gallery>
Image:12-240-Oct-15-Page-1.jpg |page1
Image:12-240-Oct-15-Page-2.jpg |page2
Image:12-240-Oct-15-Page-3.jpg |page3
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-18T17:00:29Z

Dangkhoa.nguyen: /* Theorems */

{{12-240/Navigation}}
===Riddle Along===
The game of 15 is played as follows. Two players alternate choosing cards numbered between 1 and 9, with repetitions forbidden, so that the game ends at most after 9 moves (or four and a half rounds). The first player to have within her/his cards a set of precisely 3 cards that add up to 15 wins.

Does this game has a winning strategy? What is it? Who wins, the first to move or the second? Why am I asking this question at this particular time?

[[Image:12-240-DeckOfCards.png|center]]

See also [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503 a video] and the [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ transcript] of that video.

{{12-240:Dror/Students Divider}}
== Theorems ==
1. If G generates, |G| <math>\ge \!\,</math> n and G contains a basis, |G|=n then G is a basis

2. If L is linearly independent, |L| <math>\le \!\,</math> n and L can be extended to be a basis. |L|=n => L is a basis.

3.W <math>\subset \!\,</math> V a subspace then W is finite dimensioned and dim W <math>\le \!\,</math> dim V

If dim W = dim V, then V = W
If dim W < dim V, then any basis of W can be extended to be a basis of V

Proof of W is finite dimensioned:

Let L be a linearly independent subset of W which is of maximal size.

Fact about '''N'''
: Every subset A of '''N''', which is:

1. Non empty

2. Bounded : <math>\exist \!\,</math> N <math>\in \!\,</math> '''N''', <math>\forall \!\,</math> a <math>\in \!\,</math> A, a <math>\le \!\,</math> N

has a maximal element: an element m <math>\in \!\,</math> A, <math>\forall\!\,</math> a <math>\in \!\,</math> A, a <math>\le \!\,</math> m ( m + 1 <math>\not in\!\,</math> A

== class note ==

<gallery>
Image:12-240-Oct-15-Page-1.jpg |page1
Image:12-240-Oct-15-Page-2.jpg |page2
Image:12-240-Oct-15-Page-3.jpg |page3
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-18T16:58:12Z

Dangkhoa.nguyen: /* Theorems */

{{12-240/Navigation}}
===Riddle Along===
The game of 15 is played as follows. Two players alternate choosing cards numbered between 1 and 9, with repetitions forbidden, so that the game ends at most after 9 moves (or four and a half rounds). The first player to have within her/his cards a set of precisely 3 cards that add up to 15 wins.

Does this game has a winning strategy? What is it? Who wins, the first to move or the second? Why am I asking this question at this particular time?

[[Image:12-240-DeckOfCards.png|center]]

See also [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503 a video] and the [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ transcript] of that video.

{{12-240:Dror/Students Divider}}
== Theorems ==
1. If G generates, |G| <math>\ge \!\,</math> n and G contains a basis, |G|=n then G is a basis

2. If L is linearly independent, |L| <math>\le \!\,</math> n and L can be extended to be a basis. |L|=n => L is a basis.

3.W <math>\subset \!\,</math> V a subspace then W is finite dimensioned and dim W <math>\le \!\,</math> dim V

If dim W = dim V, then V = W
If dim W < dim V, then any basis of W can be extended to be a basis of V

Proof of W is finite dimensioned:

Let L be a linearly independent subset of W which is of maximal size.

Fact about '''N'''
: Every subset A of '''N''', which is:

1. Non empty

2. Bounded : <math>\exist \!\,</math> N <math>\in \!\,</math> '''N''', <math>\forall \!\,</math> a <math>\in \!\,</math> A, a <math>\le \!\,</math>

== class note ==

<gallery>
Image:12-240-Oct-15-Page-1.jpg |page1
Image:12-240-Oct-15-Page-2.jpg |page2
Image:12-240-Oct-15-Page-3.jpg |page3
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-18T16:54:16Z

Dangkhoa.nguyen: /* Theorems */

{{12-240/Navigation}}
===Riddle Along===
The game of 15 is played as follows. Two players alternate choosing cards numbered between 1 and 9, with repetitions forbidden, so that the game ends at most after 9 moves (or four and a half rounds). The first player to have within her/his cards a set of precisely 3 cards that add up to 15 wins.

Does this game has a winning strategy? What is it? Who wins, the first to move or the second? Why am I asking this question at this particular time?

[[Image:12-240-DeckOfCards.png|center]]

See also [https://media.library.utoronto.ca/play.php?DJ6CPFxByy2J&id=8503 a video] and the [https://cmc.math.ca/home/videos/game-of-15-and-isomorphisms/ transcript] of that video.

{{12-240:Dror/Students Divider}}
== Theorems ==
1. If G generates, |G| <math>\ge \!\,</math> n and G contains a basis, |G|=n then G is a basis

2. If L is linearly independent, |L| <math>\le \!\,</math> n and L can be extended to be a basis. |L|=n => L is a basis.

3.W <math>\subset \!\,</math> V a subspace then W is finite dimensioned and dim W <math>\le \!\,</math> dim V

If dim W = dim V, then V = W
If dim W < dim V, then any basis of W can be extended to be a basis of V

Proof of W is finite dimensioned:

== class note ==

<gallery>
Image:12-240-Oct-15-Page-1.jpg |page1
Image:12-240-Oct-15-Page-2.jpg |page2
Image:12-240-Oct-15-Page-3.jpg |page3
</gallery>

12-240/Classnotes for Thursday October 18

2012-10-18T16:48:02Z

Dangkhoa.nguyen: /* Theorems */

12-240/Classnotes for Thursday October 18

2012-10-18T16:47:25Z

Dangkhoa.nguyen: /* Theorems */

12-240/Classnotes for Thursday October 18

2012-10-18T16:45:38Z

Dangkhoa.nguyen:

12-240/Classnotes for Thursday October 18

2012-10-18T16:44:11Z

Dangkhoa.nguyen: /* class note */

File:12-240-Oct-15-Page-3.jpg

2012-10-18T16:43:49Z

Dangkhoa.nguyen:

12-240/Classnotes for Thursday October 18

2012-10-18T16:42:56Z

Dangkhoa.nguyen: /* class note */

File:12-240-Oct-15-Page-2.jpg

2012-10-18T16:42:36Z

Dangkhoa.nguyen:

12-240/Classnotes for Thursday October 18

2012-10-18T16:38:51Z

Dangkhoa.nguyen: /* class note */

12-240/Classnotes for Thursday October 18

2012-10-18T16:38:39Z

Dangkhoa.nguyen: /* class note */

12-240/Classnotes for Thursday October 18

2012-10-18T16:37:49Z

Dangkhoa.nguyen: /* class note */

12-240/Classnotes for Thursday October 18

2012-10-18T16:37:34Z

Dangkhoa.nguyen:

File:12-240-Oct-15-Page-1.jpg

2012-10-18T16:37:12Z

Dangkhoa.nguyen:

12-240/Classnotes for Thursday October 11

2012-10-12T21:38:25Z

Dangkhoa.nguyen: /* Proofs */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
1) β2 generate and β1 is linearly independent

From replacement theorem

|β2|<math>\ge\!\,</math> |β1| , ( role reversal), |β1|<math>\ge\!\,</math> |β2|

Then |β2|= |β1|

3) a) (|G| <math>\ge\!\,</math> n)

by dim V = n, exist basic β of V with n elements, Take L = β in the replacement lemma, |G| = n1

|L| <math>\le\!\,</math> n1= |G|

Hence n <math>\le\!\,</math> |G|

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T21:37:56Z

Dangkhoa.nguyen: /* Proofs */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
1) β2 generate and β1 is linearly independent

From replacement theorem

|β2|<math>\ge\!\,</math> |β1| , ( role reversal), |β1|<math>\ge\!\,</math> |β2|

Then |β2|= |β1|

3) a) (|G| <math>\ge\!\,</math> n)

#1: by dim V = n, exist basic β of V with n elements, Take L = β in the replacement lemma, |G| = n1

|L| <math>\le\!\,</math> n1= |G|

Hence n <math>\le\!\,</math> |G|

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T21:37:36Z

Dangkhoa.nguyen: /* Proofs */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
1) β2 generate and β1 is linearly independent

From replacement theorem

|β2|<math>\ge\!\,</math> |β1| , ( role reversal), |β1|<math>\ge\!\,</math> |β2|

Then |β2|= |β1|

3) a) (|G| <math>\ge\!\,</math> n)

#1: by dim V = n, exist basic β of V with n elements, Take L = β in the replacement lemma, |G| = n1

|L| <math>\le\!\,</math> n1= |G|

Hence n <math>\le\!\,</math> |G|

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T21:30:47Z

Dangkhoa.nguyen: /* Proofs */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
1) β2 generate and β1 is linearly independent

From replacement theorem

|β2|<math>\ge\!\,</math> |β1| , ( role reversal), |β1|<math>\ge\!\,</math> |β2|

Then |β2|= |β1|

3) a) (|G| <math>\ge\!\,</math> n)

#1: by dim V = n, exist basic β of V with n elements,

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T21:26:01Z

Dangkhoa.nguyen: /* Proofs */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
1) β2 generate and β1 is linearly independent

From replacement theorem

|β2|<math>\ge\!\,</math> |β1| , ( role reversal), |β1|<math>\ge\!\,</math> |β2|

Then |β2|= |β1|

3)

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
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</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T21:15:42Z

Dangkhoa.nguyen:

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.
== Proofs ==
== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T21:13:25Z

Dangkhoa.nguyen: /* 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.

== 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
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</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T21:12:59Z

Dangkhoa.nguyen: /* 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>\for all\!\</math> i

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

== 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
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</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T21:10:34Z

Dangkhoa.nguyen:

{{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

== 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
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</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T21:09:30Z

Dangkhoa.nguyen: /* 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

=> <math>\sum \!\,</math> ai.ui - <math>\sum \!\,</math> bi.ui = 0 => <math>\sum \!\,</math> (ai-bi).ui = 0

== 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
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</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T20:59:30Z

Dangkhoa.nguyen: /* 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.

== 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
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Image:12-240-1009-7.jpg|Page 7
</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T20:57:20Z

Dangkhoa.nguyen: /* 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

== 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>

12-240/Classnotes for Tuesday October 09

2012-10-12T20:56:49Z

Dangkhoa.nguyen: /* 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>\all\!\,</math> i

== 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
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Image:12-240-1009-7.jpg|Page 7
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T20:43:35Z

Dangkhoa.nguyen: /* corollaries */

{{12-240/Navigation}}

In this lecture, the professor concentrate on corollaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== corollaries ==
1/ If V has a finite basic β1, then any other basic β2 of V is also finite and |β1|=|β2|

2/ "dim V" makes sense

dim V = |β| if V has a finite basic β

Otherwise, dim V = <math>\infty \!\,</math>

ex: dim P(F)= <math>\infty \!\,</math>

3/ Assume dim V = n < <math>\infty \!\,</math> then,

a) If G generate V then |G|<math>\ge \!\,</math> n & some set of G is a basic of V. ( If |G|= n, itself is a basic)

b) If L is linearly independent then |L|<math>\le \!\,</math> n, if |L|=n then L is a basic, if |L|< n then L can be extended to become a basic.

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T20:38:19Z

Dangkhoa.nguyen: /* theorems */

12-240/Classnotes for Thursday October 11

2012-10-12T20:33:47Z

Dangkhoa.nguyen: /* corollaries */

12-240/Classnotes for Tuesday October 09

2012-10-12T20:28:22Z

Dangkhoa.nguyen: /* Definition of basic */

{{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 ==

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

12-240/Classnotes for Tuesday October 09

2012-10-12T20:28:02Z

Dangkhoa.nguyen: /* Definition of basic */

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

1/ It generates ( span) V, span β = V

2/ It is linearly independent

== theorems ==

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

12-240/Classnotes for Thursday October 11

2012-10-12T20:23:16Z

Dangkhoa.nguyen:

12-240/Classnotes for Tuesday October 09

2012-10-12T20:22:09Z

Dangkhoa.nguyen:

{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.
== Definition of basic ==
== theorems ==

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

12-240/Classnotes for Thursday October 11

2012-10-12T20:21:11Z

Dangkhoa.nguyen:

{{12-240/Navigation}}

In this lecture, the porfessor concentrate on collaries of basic and dimension.
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Thursday October 11

2012-10-12T20:20:32Z

Dangkhoa.nguyen: /* Annoucements */

{{12-240/Navigation}}

In this lecture, the porfessor concentrate on collaries of basics
== Annoucements ==
TA Office Hours (Still pending!) @ 215 Huron St., 10th floor.

Peter - 11am - 1pm

Brandon 1pm - 3pm

Topic: Replacement Theorem

== Lecture notes scanned by [[User:Oguzhancan|Oguzhancan]] ==
<gallery>
Image:12-240-1011-1.jpg|Page 1
Image:12-240-1011-2.jpg|Page 2
Image:12-240-1011-3.jpg|Page 3
</gallery>

12-240/Classnotes for Tuesday October 09

2012-10-12T20:19:16Z

Dangkhoa.nguyen: /* Lecture notes scanned by Oguzhancan */

{{12-240/Navigation}}
In this lecture, the professor concentrate on basics and related theorems.

== 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
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Image:12-240-1009-7.jpg|Page 7
</gallery>

12-240/Classnotes for Tuesday October 2

2012-10-04T18:22:44Z

Dangkhoa.nguyen:

{{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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