12-240/Classnotes for Thursday October 4
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Reminders
Web Fact: No link, doesn't exist!
Life Fact: Dror doesn't do email math!
Riddle: Professor and lion in a ring with Failed to parse (unknown function "\math"): {\displaystyle V_p = V_l<\math>, help the professor live as long as possible. == Recap == 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. Span - span(S) is the set of all linear combination of set S Generate - We say S generates a vector space V is span(S) = V == Pre - Basis == '''Linear dependance''' '''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. Otherwise, we call S '''linearly independant.''' === 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 dependant. 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 dependant.) (∑ 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 independant. 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 independant. Exercise: Prove: '''Theorem''' Suppose S1 ⊂ S2 ⊂ V. -> If S1 is linearly dependant, then S2 is dependant. -> 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>}
