Tutorial 7 Solutions

Tutorial 7 Solutions - span W thus they form a basis for W...

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Tutorial 7, SOLUTIONS MATH 1104 1-Last name: First name: 2-Last name: First name: 3-Last name: First name: 4-Last name: First name: 5-Last name: First name: ——————————————————————————- 1: Let B = u 1 = 1 2 - 1 1 , u 2 = - 2 1 1 1 . a: Show that B is an orthogonal set. b: Let W = span { B } . Show that B is a basis for W . c: Let y = 2 - 1 2 1 , find the coordinate vector of y relative to B . d: Find the orthogonal projection of y onto W . e: Decompose y into two vectors, one in W the other one in W . f: Find the distance from y to W . Solution: a: u 1 .u 2 = . so u 1 u 2 . b: Since u 1 and u 2 are orthogonal then they are independent. They also
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Unformatted text preview: span W thus they form a basis for W . c: y = c 1 u 1 + c 2 u 2 , where c 1 = y.u 1 u 1 .u 1 =-1 7 and c 2 = y.u 2 u 2 .u 2 =-2 7 . Then -1 7-2 7 is the coordinate vector of y relative to B . d: b y = y.u 1 u 1 .u 1 u 1 + y.u 2 u 2 .u 2 u 2 = 3 7-4 7-1 7-1 7 . y = b y + z , where b y W and z W , and z = y-b y = 11 7-3 7 15 7 8 7 . f: It is k y-b y k = k z k = 1 7 11 2 + 3 2 + 15 2 + 8 2 . g: Solve the system u 1 .x = 0 , u 2 .x = 0 , where , x = [ x 1 x 2 x 3 x 4 ] T R 4 . 1...
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This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton.

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