soln2 - MATH 245 Linear Algebra 2 Solutions to Assignment 2...

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MATH 245 Linear Algebra 2, Solutions to Assignment 2 1: Let U and V be vector spaces in R n . (a) Show that ( U + V ) = U V . Solution: Let x ( U + V ) . This means that x . y = 0 for all y U + V . For all u U , we also have u U + V so that x · u = 0, and so x U . Similarly, x V and hence x U V . This shows that ( U + V ) U V . Conversely, let x U V . Let w U + V , say w = u + v where u U and v V . Since x U and u U we have x . u = 0. Similarly, x . v = 0. Since x . u = x . v = 0 we have x . w = x . ( u + v ) = x . u + x . v = 0. Since x . w = 0 for every w U + V , we have x ( U + V ) . This shows that U V ( U + V ) . (b) Show that ( U V ) = U + V . Solution: By part (a) (with U and V replaced by U and V ) we have ( U V ) = ( ( U ) ( V ) ) = ( ( U + V ) ) = U + V . (c) Given that U = Col( A ) and V = Col( B ), find a matrix C such that ( U + V ) = Null( C ). Solution: When U = Col( A ) and V = Col( B ), we have U + V = Col( A,B ) so ( U + V ) = Null ( ( A,B ) t ) , and so we can take C = ( A,B ) t = ± A t B t ² . (d) Given that U = Null( A ) and V = Null( B ), find a matrix C such that ( U V ) = Col( C ). Solution: When U = Null( A ) and V = Null( B ), we have U V = Null ± A B ² so ( U V ) = Col ± A B ² t ! , and so we can take C = ± A B ² t = ( A t ,B t ) .
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2: For a vector space U in R n , we define the reflection in U to be the map Refl U : R n R n given by Refl U ( x ) = x - 2 Proj U ( x ) . (a) Let { u 1 ,u 2 , ··· ,u k } be a basis for U and let A = ( u 1 ,u 2 , ··· ,u k ) M n × k ( R ). Find the matrix of Refl U in terms of A . Solution: We have Refl U ( x ) = x - 2 Proj U ( x ) = x - 2 ( x - Proj U ( x ) ) = 2 Proj U ( x ) - x = 2 A ( A t A ) - 1 A t x - x = ( 2 A ( A t A ) - 1 A t - I ) x, so the matrix of Refl U is ± Refl U ² = 2 A ( A t A ) - 1 A t - I . (b) Show that Refl U preserves length, that is ³ ³ Refl U ( x ) ³ ³ = | x | for all x R n . Solution: We have ³ ³ Refl U ( x ) ³ ³ 2 = ³ ³ 2 A ( A t A ) - 1 A t x - x ³ ³ 2 = ( 2 A ( A t A ) - 1 A t x - x ) t ( 2 A ( A t A ) - 1 A t x - x ) = ( 2 x t A ( A t A ) - 1 A t - x t )( 2 A ( A t A ) - 1 A t x - x ) = 4 x t A ( A t A ) - 1 A t A ( A t A ) - 1 A t x - 2 x t A ( A t A ) - 1 A t x - 2 x t A ( A t A ) - 1 A t a + x t x = 4 x t A ( A t A ) - 1 A t x - 4 x t A ( A t A ) - 1 A t x + | x | 2 = | x | 2 .
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3: For two affine spaces P and Q in R n , the distance between P and Q is defined to be dist( P,Q ) = min n dist( x,y ) ± ± ± x P,y Q o . (a) Let
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soln2 - MATH 245 Linear Algebra 2 Solutions to Assignment 2...

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