# soln4 - MATH 245 Linear Algebra 2 Solutions to Assignment 4...

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Unformatted text preview: MATH 245 Linear Algebra 2, Solutions to Assignment 4 1: (a) Let A 1 = 1 2- 1 0 , A 2 = 3 4- 1 2 and A 3 = 3- 1- 2 4 . Apply the Gram-Schmidt Procedure to the basis U = A 1 ,A 2 ,A 3 to obtain an orthonormal basis for U = Span A 1 ,A 2 ,A 3 ⊂ M 2 × 2 . Solution: We take B 1 = A 1 = 1 2- 1 0 B 2 = A 2- h A 2 ,B 1 i | B 1 | 2 B 1 = 3 4- 1 2- 12 6 1 2- 1 0 = 1 0 1 2 B 3 = A 3- h A 3 ,B 1 i | B 1 | 2 B 1- h A 3 ,B 2 i | B 2 | 2 B 2 = 3- 1- 2 4- 3 6 1 2- 1 0- 9 6 1 0 1 2 = 1 2 6- 2- 4 8- 1 2- 1 0- 3 0 3 6 = 1 2 2- 4- 6 2 = 1- 2- 3 1 and then we normalize to get the orthonormal basis { C 1 ,C 2 ,C 3 } where C 1 = B 1 | B 1 | = 1 √ 6 1 2- 1 0 , C 2 = B 2 | B 2 | = 1 √ 6 1 0 1 2 , C 3 = B 3 | B 3 | = 1 √ 15 1- 2- 3 1 . (b) Find an orthonormal basis for P 2 with the inner product given by p,q = p (0) q (0)+ p (1) q (1)+ p (2) q (2) by applying the Gram-Schmidt Procedure to the standard basis 1 ,x,x 2 . Solution: Write p = 1, p 1 = x and p 2 = x 2 . We take q = p = 1 q 1 = p 1- h p 1 ,q i | q | 2 q = x- h x, 1 i | 1 | 2 1 = x- · 1 + 1 · 1 + 2 · 1 1 2 + 1 2 + 1 2 1 = x- 1 q 2 = p 2- h p 2 ,q i | q | 2 q- h p 2 ,q 1 i | q 1 | 2 q 1 = x 2- h x 2 , 1 i | 1 | 2 1- h x 2 ,x- 1 i | x- 1 | 2 ( x- 1) = x 2- · 1 + 1 · 1 + 4 · 1 1 2 + 1 2 + 1 2 1- (0)(- 1) + (1)(- 0) + (4)(1) (- 1) 2 + (0) 2 + (1) 2 ( x- 1) = x 2- 5 3- 2( x- 1) = x 2- 2 x + 1 3 . Note that | q | 2 = 1 2 + 1 2 + 1 2 = 3, | q 1 | 2 = (- 1) 2 + (0) 2 + (1) 2 = 2 and | q 2 | 2 = ( 1 3 ) 2 + (- 2 3 ) 2 + ( 1 3 ) = 6 9 = 2 3 , and so normalizing yields the orthonormal basis { r ,r 1 ,r 2 } with r = q | q | = 1 √ 3 , r 1 = q 1 | q 1 | = 1 √ 2 ( x- 1) , r 2 = q 2 | q 2 | = q 3 2 ( x 2- 2 x + 1 3 ) . 2: Consider P 2 as a subspace of C [- 1 , 1] with its standard inner product, which is given by f,g = Z 1- 1 fg . Applying the Gram-Schmidt Procedure to the standard basis 1 ,x,x 2 for P 2 yields the orthogonal basis p ,p 1 ,p 2 , where p = 1, p 1 = x and p 2 = x 2- 1 3 . (a) Find the polynomial f ∈ P 2 which minimizes Z 1- 1 ( f ( x )- | x | ) 2 dx . Solution: Let g ( x ) = | x | . Note that Z 1- 1 ( f ( x )-| x | ) 2 dx = f- g 2 . The unique point f ∈ P 2 nearest to g is the projection Proj P 2 ( g ), so to minimize the value of | f- g | we must take f = Proj P 2 ( g ) = h g,p i | p | 2 · p + h g,p 1 i | p 1 | 2 · p 1 + h g,p 2 i | p 2 | 2 · p 2 = R 1- 1 | x | dx R 1- 1 1 2 dx · 1 + R 1- 1 x | x | dx R 1- 1 x 2 dx · x + R 1- 1 ( x 2- 1 3 ) | x | dx R 1- 1 ( x 2- 1 3 ) 2 dx · ( x 2- 1 3 ) = 1 2 · 1 + 2 3 · x + 2 ( 1 4- 1 6 ) 2 ( 1 5- 2 9 + 1 9 ) · ( x 2- 1 3 ) = 1 2 + 1 / 12 4 / 45 ( x 2- 1 3 ) = 3 16 + 15 16 x 2 ....
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## This note was uploaded on 12/08/2011 for the course MATH 245 taught by Professor New during the Fall '09 term at Waterloo.

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soln4 - MATH 245 Linear Algebra 2 Solutions to Assignment 4...

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