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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 GramSchmidt 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 GramSchmidt 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 GramSchmidt 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.
 Fall '09
 NEW
 Linear Algebra, Algebra

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