# soln6 - MATH 245 Linear Algebra 2, Solutions to Assignment...

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Unformatted text preview: MATH 245 Linear Algebra 2, Solutions to Assignment 6 1: (a) For the quadratic curve 7 x 2 + 8 xy + y 2 + 5 = 0, find the coordinates of each vertex, find the equation of each asymptote, and sketch the curve. Solution: Let K ( x,y ) = 7 x 2 + 8 xy + y 2 . Note that K ( x,y ) = ( x y ) A x y where A = 7 4 4 1 . The characteristic polynomial of A is A- λI = 7- λ 4 4 1- λ = λ 2- 8 λ- 9 = ( λ- 9)( λ + 1) so the eigenvalues are λ 1 = 9 and λ 2 =- 1. For λ = 9 we have A- λI =- 2 4 4- 8 ∼ 1- 2 so we can choose the unit eigenvector u 1 = 1 √ 5 2 1 for λ 1 . The other eigenspace will be orthogonal since A is symmetric, so we can choose the unit eigenvector u 2 = 1 √ 5- 1 2 for λ 2 . Let P = ( u 1 ,u 2 ) = 1 √ 5 2- 1 1 2 and let D = 9- 1 . Then we have P * AP = D . Write x y = P s t or equivalently s t = P * x y . Then K ( x,y ) = ( x y ) A x y = ( s t ) P * AP s t = ( s t ) D s t = 9 s 2- t 2 and so K ( u,v ) =- 5 ⇐⇒ 9 s 2- t 2 =- 5 ⇐⇒ t 2 5- s 2 5 / 9 = 1 . This is the hyperbola in the st-plane with vertices at (0 , ± √ 5) and asymptotes t = ± 3 s . We calculate the points ( x,y ) corresponding to ( s,t ) = (0 , ± √ 5) (the vertices) and ( s,t ) = ( √ 5 , ± 3 √ 5) (points on the asymptotes): P ± √ 5 = ± 2- 1 1 2 1 = ±- 1 2 P √ 5 ± 3 √ 5 = 2- 1 1 2 1 ± 3 =- 1 7 , 5- 5 . Thus the curve K ( x,y ) =- 5 is a hyperbola with vertices at ( x,y ) = (- 1 , 2) and (1 ,- 2) and asymptotes y =- x and y =- 7 x . t y s x (b) For the real quadratic form K ( x,y,z ) = 3 x 2 + ay 2 + bz 2- 6 xy +2 xz- 4 yz , sketch the set of points ( a,b ) for which K is positive-definite. Solution: We have K ( x,y,z ) = ( x y z ) A x y z where A = 3- 3 1- 3 a- 2 1- 2 b . Note that det A 1 × 1 = det(3) = 3, det A 2 × 2 = det 3- 3- 3 a = 3 a- 9 and det A 3 × 3 = det A = 3 ab + 6 + 6- 12- 9 b- a = 3 ab- 9 b- a . For K to be positive-definite we need det A 2 × 2 > 0, that is 3 a- 9 > 0 so a > 3, and we need det A 3 > 0, that is 3 ab- 9 b- a > 0 or equivalently 3 b ( a- 3) > a . For a > 3 this gives b > a 3( a- 3) = 1 3 + 1 a- 3 . Thus the required set of points ( a,b ) lies above and to the right of the hyperbola y = 1 3 + 1 a- 3 , a > 3. b a 2: Let U and V be non-trivial subspaces of R n with U ∩ V = { } . Recall that angle( U,V ) = min angle( u,v ) 0 6 = u ∈ U, 6 = v ∈ V . (a) Show that angle( U,V ) = cos- 1 ( σ ) where σ is the largest singular value of the linear map P : U → V given by P ( x ) = Proj V ( x ). Solution: Recall from Assignment 2 that for fixed 0 6 = u ∈ R n we have min 6 = v ∈ V angle( u,v ) = cos- 1 Proj V u | u | . Thus we have angle( U,V ) = min 6 = u ∈ U min 6 = v ∈ V angle( u,v ) = min 6 = u ∈ U cos- 1 Proj V u | u | = cos- 1 max 6 = u ∈ U Proj V u | u | = cos- 1 max u ∈ U, | u | =1 Proj V ( u ) = cos- 1 max u ∈ U, | u | =1 P ( u ) = cos- 1 σ where σ is the largest singular value of P . (b) Let u 1 = 1 1- 1 1 , u 2 = 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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soln6 - MATH 245 Linear Algebra 2, Solutions to Assignment...

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