lecture13 - Lecture Lecture 13 3.5 Dimensions 3.6...

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ecture 13 Lecture 13 3.5 Dimensions 3.6 Transition Matrices Chapter 3 Vector Spaces 1
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Announcement id- rm test tomorrow ± Mid term test tomorrow ² More past year/sample test papers ab session 3 ext week ± Lab session 3 next week (Worksheets available) Lecture 13 Announcement 2
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n ubspace of n A basis for R n” is not a basis for “subspace of R n” Bases Definition R subspace of R A finite subset S of a vector space V is called a basis for V if ) nearly independent (ii) pans (i) S is linearly independent ; (ii) S spans V z z R 3 eaning y x y x V Meaning 1. A basis for V is a building block of V 2. A basis for V is a “ unit of measurement ” for Chapter 3 Vector spaces 3 vectors in V . 3. A basis for V gives a “ coordinate system ” for V .
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ection 3 5 Section 3.5 Dimensions Objective hat is the imension f a vector space? • What is the dimension of a vector space? • How to compute dimension for a vector space? • What are some equivalent conditions for a set to be a basis for a vector space? Chapter 3 Vector Spaces 4
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dimension of the vector space Number of vectors in a basis Theorem 3.5.1 Let V be a vector space which has a basis S = { u 1 , u 2 , …, u k } with k vectors . ny subset of ith ore than ectors 1. Any subset of V with more than k vectors is always linearly dependent . 2. Any subset of V with less than k vectors cannot span V . V S basis T 1 T 2 vectors ore than k vectors Chapter 3 Vector spaces 5 k vectors more than k vectors less than k vectors
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Dimension of some standard subspaces Example 3.5.4.1-3 1. The dimension of R n is n i.e. dim( R n ) = n 2. Except { 0 } and R 2 , all subspaces of R 2 are lines through the origin Span{ u } asis of the line they are of dimension 1 . 3. Except { 0 } and R 3 , all subspaces of R 3 are basis of the line either lines through the origin they are of dimension 1 , Span{ u } basis of the line Chapter 3 Vector spaces 6 or planes containing the origin, they are of dimension 2 . Span{ u, v } basis of the plane
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R 3 W x(1, 0, 0) + y(0, 1, 1) (x, y, y) Finding dimension of a subspace Example 3.5.4.4 ind a asis r and determine the imension f Find a basis for and determine the dimension of the subspace W = {(x, y, z) | y = z} of R 3 . ote: dim(W) 3 Explicit: (x, y, y) = x(1, 0, 0) + y(0, 1, 1) Note: dim(W) So W = span{(1, 0, 0), (0, 1, 1)} basis for W : {(1, 0, 0), (0, 1, 1)} im ) = linearly independent Chapter 3 Vector spaces 7 dim (W) = 2
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R 5 su + t u Solution space Dimension of solution space Example 3.5.4.5 Find a basis for and determine the dimension of 1 2 the solution space of the homogeneous system = + + 0 2 2 z x w v general t s v ⎛− 1 1 = + = + + = + + 0 2 0 0 3 2 z x w v z y x z y x w v solution = t t s z y x w 0 + = 1 0 1 0 0 0 0 1 t s u 1 u 2 linearly indep.
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lecture13 - Lecture Lecture 13 3.5 Dimensions 3.6...

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