Thus the set vector w 1 vector w 2 vector w 3 is an

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Chapter 7 / Exercise 20
Numerical Analysis
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. Thus, the set { vector w 1 , vector w 2 , vector w 3 } is an orthogonal basis for S . We can now determine the projection. proj S vector x = (big vector x , vector w 1 )big bardbl vector w 1 bardbl 2 vector w 1 + (big vector x , vector w 2 )big bardbl vector w 2 bardbl 2 vector w 2 + (big vector x , vector w 3 )big bardbl vector w 3 bardbl 2 vector w 3 = 11 3 vector w 1 + 14 15 vector w 2 + 1 10 vector w 3 = 9 / 2 3 2 9 / 2 REMARK Observe that the iterative step in the Gram-Schmidt procedure is just calculating a perpendicular of a projection. In particular, the Gram-Schmidt procedure can be restated as follows: If { vector w 1 , . . . , vector w k } is a basis for an inner product space W , then let vector v 1 = vector w 1 , and for 2 i k recursively define W i 1 = Span { vector v 1 , . . . ,vector v i 1 } and vector v i = perp W i 1 vector w i . Then { vector v 1 , . . . ,vector v k } is an orthogonal basis for W . If { vector v 1 , . . . ,vector v k } is an orthonormal basis for W , then the formula for calculating a pro- jection simplifies to proj W vector v = (big vector v ,vector v 1 )big vector v 1 + · · · + (big vector v ,vector v k )big vector v k EXAMPLE 6 Let W = Span 0 1 0 , 4 5 0 3 5 be a subspace of R 3 . Find proj W 1 1 1 and perp W 1 1 1 . Solution: Observe that 0 1 0 , 4 5 0 3 5 is an orthonormal basis for W . Hence, proj W vector x = (Bigg 1 1 1 , 0 1 0 )Bigg 0 1 0 + (Bigg 1 1 1 , 4 / 5 0 3 / 5 )Bigg 4 / 5 0 3 / 5 = 4 / 25 1 3 / 25 perp W vector x = 1 1 1 4 / 25 1 3 / 25 = 21 / 25 0 28 / 25
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Chapter 7 / Exercise 20
Numerical Analysis
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60 Chapter 9 Inner Products Section 9.4 Problems 1. Consider S = Span braceleftBiggbracketleftBigg 1 0 1 0 bracketrightBigg , bracketleftBigg 2 1 1 3 bracketrightBiggbracerightBigg in

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