12-240/Classnotes for Thursday October 4 - Revision history

Peterlue: /* Interesting inequality */

2012-12-07T09:46:52Z

Interesting inequality

← Older revision		Revision as of 05:46, 7 December 2012
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	This holds is true if the field does not have characteristic 2. Can you see why?		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)		(a,b) = (a+b)/2 ∙ (1, 1) + (a-b)/2 ∙ (1, -1)

	== Lecture notes scanned by [[User:starash\|starash]] ==		== Lecture notes scanned by [[User:starash\|starash]] ==

Peterlue: /* Basis */

2012-12-07T09:46:14Z

Basis

← Older revision		Revision as of 05:46, 7 December 2012
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	== Basis ==		== Basis ==

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

Peterlue: /* Comments */

2012-12-07T09:44:59Z

Comments

← Older revision		Revision as of 05:44, 7 December 2012
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	Proof:		Proof:
	⇐ If u≠0, suppose au =0		⇐ If u≠0, suppose au =0
	By property (~~a*u~~ = 0, a = 0 or u = 0), a = 0. Thus, {u} is linearly independent.		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.		⇒ By definition, au = 0 for {u} only when a = 0.

Peterlue: /* Examples */

2012-12-07T09:44:25Z

Examples

← Older revision		Revision as of 05:44, 7 December 2012
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	Claim: This set is linearly independent.		Claim: This set is linearly independent.

	Proof: Suppose (∑ ~~ai*ei~~) = 0 ({ei} is linearly dependent.)		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		(∑ ai∙ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0

	⇒ a1 = a2 = ... = an = 0!		⇒ a1 = a2 = ... = an = 0!

Peterlue: /* Recap */

2012-12-07T09:43:47Z

Recap

← Older revision		Revision as of 05:43, 7 December 2012
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	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>.		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 ~~combination~~ of set <math>S</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>.		Generate - We say <math>S</math> generates a vector space <math>V</math> is <math>\operatorname{span}(S) = V</math>.

Quanle: /* Pre - Basis */

2012-10-21T19:17:28Z

Pre - Basis

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	== Pre - Basis ==		== Pre - Basis ==

	'''Linear ~~dependance~~'''		'''Linear dependence'''

	'''Definition''' A set S ⊂ V is called linearly ~~dependant~~ 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.		'''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 ~~independant~~.'''		Otherwise, we call S '''linearly independent.'''

	=== Examples ===		=== Examples ===
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	u1 - 2u2 + u3 = 0		u1 - 2u2 + u3 = 0

	S is linearly ~~dependant~~.		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.		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.
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	Claim: This set is linearly independent.		Claim: This set is linearly independent.

	Proof: Suppose (∑ ai*ei) = 0 ({ei} is linearly ~~dependant~~.)		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		(∑ ai*ei) = 0 ⇔ a1(1, 0, ..., 0) + ... + an(0, ..., 0, 1) = 0 ⇔ (a1, 0, ... , 0) + ... + (0, ... , an) = 0
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	===Comments ===		===Comments ===

	1. {u} is linearly ~~independant~~.		1. {u} is linearly independent.
	Proof:		Proof:
	⇐ If u≠0, suppose au =0		⇐ If u≠0, suppose au =0
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	⇒ By definition, au = 0 for {u} only when a = 0.		⇒ By definition, au = 0 for {u} only when a = 0.

	2. ∅ is linearly ~~independant~~.		2. ∅ is linearly independent.


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

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

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

Pratyush: /* Recap */ fixing errors

2012-10-11T12:42:19Z

Recap: fixing errors

← Older revision		Revision as of 08:42, 11 October 2012
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	Base - what were doing today		Base - what were doing today

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

Pratyush: /* Recap */ changing math to math mode

2012-10-11T12:41:23Z

Recap: changing math to math mode

← Older revision		Revision as of 08:41, 11 October 2012
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	Base - what were doing today		Base - what were doing today

	Linear combination (lc) - We say v is a linear combination of a set S = {u1 ~~...~~ un} if v = ~~a1u1~~ ~~...~~ ~~anun~~ for scalars from a field F.		Linear combination (lc) - We say <math>v</math> is a linear combination of a set <math>S = \{u_1, \dotsc, u_n\}</math> if <math>v = a_1u_1, \dotsc, a_nu_n</math> for scalars from a field <math>F</math>.

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

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

	== Pre - Basis ==		== Pre - Basis ==

Pratyush: /* Reminders */ Putting math stuff in math mode

2012-10-11T12:36:00Z

Reminders: Putting math stuff in math mode

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	Life Fact: Dror doesn't do email math!		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.		Riddle: Professor and lion in a ring with <math>V_p = V_l</math>, help the professor live as long as possible.

	== Recap ==		== Recap ==

Pratyush: /* Reminders */

2012-10-11T12:34:49Z

Reminders

← Older revision		Revision as of 08:34, 11 October 2012
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	Life Fact: Dror doesn't do email math!		Life Fact: Dror doesn't do email math!

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

	== Recap ==		== Recap ==
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	Image:12-240-1004-1.jpg\|Page 1		Image:12-240-1004-1.jpg\|Page 1
	Image:12-240-1004-2.jpg\|Page 2		Image:12-240-1004-2.jpg\|Page 2
	</gallery>		</gallery></math>