12-240/Classnotes for Thursday October 4: Difference between revisions

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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!
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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.
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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.




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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]] ==

Latest revision as of 05:46, 7 December 2012

Reminders

Web Fact: No link, doesn't exist!

Life Fact: Dror doesn't do email math!

Riddle: Professor and lion in a ring with , help the professor live as long as possible.

Recap

Base - what were doing today

Linear combination (lc) - We say is a linear combination of a set if for scalars from a field .

Span - is the set of all linear combinations of the set .

Generate - We say generates a vector space is .

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 starash

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