# lecture12 - Rn Fill in the blanks u1, u2, , uk c1u1+ c2u2 +...

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Chapter 3 Vector spaces 1 Fill in the blanks R n c 1 u 1 + c 2 u 2 +… + c k u k u 1 , u 2 , …, u k span{ u 1 , u 2 , …, u k } 1. c 1 u 1 + c 2 u 2 +… + c k u k is called a linear combination of u 1 , u 2 , …, u k . 2. The set of all linear combinations of u 1 , u 2 , …, u k is called the linear span of u 1 , u 2 , …, u k . 3. The subset span{ u 1 , u 2 , …, u k } is called a subspace of R n . 4. If there is no “ redundant ”vectorin{ u 1 , u 2 , …, u k }, we say the set is linearly independent 5. If { u 1 , u 2 , …, u k } is linearly independent and span{ u 1 , u 2 , …, u k } = V, then { u 1 , u 2 , …, u k } is called a basis for V.

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Chapter 3 Vector Spaces 2 Lecture 12 3.4 Bases 3.5 Dimensions
Lecture 12 Announcement 3 Announcement ± Workbin ² Tutorial 4 and Lab 2 solutions ² Exercise set 3 (Q1-16) solutions ² Practice Session 5 answers ± Mid-term Test next Wednesday More info during the break ± No Lab session next week

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Chapter 3 Vector spaces 4 Recall from previous lectures To Show: S = { u 1 , u 2 , …, u k } spans R n To Show: S = { u 1 , u 2 , …, u k } is lin. indep. c 1 u 1 + c 2 u 2 + ··· + c k u k = v v is any general vector in R n check whether the system is always consistent c 1 u 1 + c 2 u 2 + ··· + c k u k = 0 0 is the zero vector in R n check whether the system has non-trivial solution yes no spans does not span yes no lin.dep lin.indep Given that : S = { u 1 , u 2 , …, u k } is a subset of R n same as : span (S) = R n
Chapter 3 Vector spaces 5 Exercise (similar to Ex 3 Q18(b) ) Given u , v , w are linearly independent Are u + v , u + w , v + w linearly independent? Consider a ( u + v ) + b ( u + w ) + c ( v + w )= 0 (*) Does (*) have non-trivial coefficients for a, b, c ? Rewrite (*) : (a+b) u + (a+c) v + (b+c) w = 0 (**) (**) has only trivial coefficients for a+b, a+c, b+c a + b = 0 a + c = 0 b + c = 0 Solve: a = b = c = 0 So (*) has only trivial coefficients for a, b, c So u + v , u + w , v + w are linearly independent

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Chapter 3 Vector Spaces 6 Section 3.4 Bases Objective •Whatisa basis for a vector space? • How to show that a set is a basis? • How to find a basis for a vector space? •Whatare coordinate vectors ?
Chapter 3 Vector spaces 7 Color mixing Three primary colors: Red , Green , Blue (RGB) Different color shade combination gives “all” colors -span{ Red , Green , Blue } = Color space -{ Red , Green , Blue } is linearly independent e.g. 20% Red + 45% Green + 30% Blue The three primary colors span the color space : None of the three primary colors are redundant : An Analogy

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Chapter 3 Vector spaces 8 Example Standard basis vectors for R 3 S is a smallest possible subset of R 3 so that every vector in R 3 is a linear combination of the elements in S . e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1) S = { e 1 , e 2 , e 3 } is called a basis for R 3 -span{ e 1 , e 2 , e 3 } = R 3 -{ e 1 , e 2 , e 3 } is linearly independent building block What is a basis? e.g. (2, 3, -5) = 2 e 1 + 3 e 2 -5 e 3 No redundant vectors
Chapter 3 Vector spaces 9 Definition 3.4.2 Let S = { u 1 , u 2 , …, u k } be a subset of R n .

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## This note was uploaded on 04/15/2010 for the course MATHS MA1101R taught by Professor Vt during the Spring '10 term at National University of Singapore.

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lecture12 - Rn Fill in the blanks u1, u2, , uk c1u1+ c2u2 +...

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