HW3 - Due 02/06. Each answer must be mathematically justified. Don t forget your name. 1 2, A = 2 2

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1 HW3 - Due 02/06 Each answer must be mathematically justified Don t forget your name Problem 1 Find a 2 2 matrix B such that B 3 = A, where A = 2 2 If A was diagonal, it would be easy: we would just take the cubic roots of the diagonal elements But A is not diagonal However A is diagonalizable because it is symmetric, so let us first diagonalize A The eigenvalues solve (1 λ)( 2 λ) 4 = λ 2 + λ 6 = (λ 2)(λ + 3) Thus 2 and 3 are eigenvalues We now find eigenvectors: [ 2 2Id = is eigenvector; ] Id = is eigenvector We deduce that A = P DP 1, P = 2 0, D = 0 3 P 1 = Now we set E the diagonal matrix whose elements are the cubic roots of the elements of D; and we define [ 3 ] 2 0 B = P EP 1, E = Then B 3 = P EP 1 P EP 1 P EP 1 = P E 3 P 1 = P DP 1 = A Therefore B responds to the problem Problem 2 Is there a 3 3 symmetric matrix A with distinct eigenvalues and eigenvectors, 3 4 5, 6 7 8? 9 Symmetric matrices admit an orthogonal basis of eigenvectors Because here these eigenvalues are distinct, such a matrix can exist if and only if the above vectors are orthogonal But = = Therefore there are no such matrices 1

2 2 Problem 3 Test whether these quadratic forms are positive definite or negative definite: (a) q 1 (x 1, x 2, x 3 ) = 10x x x x 1 x 2 + 2x 2 x 3 + 2x 3 x 1, (b) q 2 (x 1, x 2, x 3 ) = x x 2 2 4x x 1 x 2 2x 2 x 3 + 4x 1 x 3 (a) q 1 is represented by the symmetric matrix Q 3 = We use the minors test: 10 > 0, > 0 and = > Thus q 1 is definite positive (b) q 2 is represented by the symmetric matrix Q 2 = We use the minors test: 1 > 0, 1 2 = 1 > 0 and = = = 17 Thus q 2 is indefinite Problem 4 Write the quadratic form q(x 1, x 2, x 3 ) = x x x x 1 x 2 + 2x 1 x 3 + 6x 2 x 3 as a sum of squares Deduce that q is positive definite We use the algorithm seen in class: q(x 1, x 2, x 3 ) = (x 1 + x 2 + x 3 ) 2 + x x x 2 x 3 = (x 1 + x 2 + x 3 ) 2 + (x 2 + 2x 3 ) 2 + x 2 3 It is clear that q is positive semidefinite it is a sum of squares To check that it is definite positive, we must observe that these squares do not vanish simultaneously unless x 1 = x 2 = x 3 = 0 If that happens, then x 1 + x 2 + x 3 = 0 x 2 + 2x 3 = 0 x 3 = 0 This is a triangular system and (0, 0, 0) is the unique solution Hence q is definite positive

3 Problem 5 Find the equation of the tangent plane of the graphs of the function (a) f(x, y) = y 3 2y+x at (x, y) = (1, 1); (b) g(x, y, z) = xz+2y 2 z 2 at (x, y, z) = ( 1, 1, 0) (a) We start by computing x f(1, 1) and y f(1, 1): x f(x, y) = 1, x f(1, 1) = 1, y f(x, y) = 3y 2 2, y f(1, 1) = 1, f(1, 1) = 0 Therefore the tangent plane at (1, 1) has equation x 2 + y = z (b) The same strategy yields x g(x, y, z) = z, x g( 1, 1, 0) = 0, y g(x, y, z) = 4yz 2, y g( 1, 1, 0) = 0, z g(x, y, z) = x + 4y 2 z, z g( 1, 1, 0) = 1 In addition, g( 1, 1, 0) = 0 thus the equation of the tangent plane at ( 1, 1, 0) is t = z Problem 6 Find all local minimum and maximum of the following functions: (a) f(x, y) = x 2 + 2x + 2y + 2y 2 + 2xy + 1; (b) g(x, y) = 4x 2 + 4xy + 2y 2 3; (c) h(x, y) = 3x + 3xy + y 3 (a) We compute the directional derivatives: x f(x, y) = 2x y = 2(1 + x + y), p y f = 2 + 4y + 2x = 2(1 + 2y + x) We now look for critical points: we have to solve the system { { { 1 + x + y = x + y = 0 x = y + x = 0 y = 0 y = 0 We now compute the Hessian of f at ( 1, 0): 2 1 f (x, y) = We observe that the Hessian is definite positive: 2 > 0 and > 0 Therefore ( 1, 0) corresponds to a local minimum (b) We compute the directional derivatives: x g(x, y) = 8x + 4y = 4(2x + y), p y g = 4y + 4x = 4(x + y) We now look for critical points: we have to solve the system { { 2x + y = 0 x = 0 x + y = 0 y = 0 We now compute the Hessian of g at (0, 0): g (x, y) =

4 4 We observe that the Hessian is definite positive: 8 > 0 and > 0 Therefore (0, 0) corresponds to a local minimum (c) We first compute the derivatives and the Hessian: 0 3 x h(x, y) = 3 + 3y, y h(x, y) = 3x + 3y 2, h (x, y) = 3 6y The derivatives vanish simultaneously at ( 1, 1) The Hessian at this points is: 0 3 h ( 1, 1) = 3 6 It is indefinite because its determinant is 9 < 0 Thus h has a saddle points at ( 1, 1) and no local minimum or maximum Problem 7 Let f(x, y) = x 2 + 2xy + y 3 + x 3 (a) Show that f has a saddle point at (0, 0) (b) Find some numbers a and b such that the function g(t) = f(at, bt) has a local minimum at t = 0 (c) Find some numbers c and d such that the function h(t) = f(ct, dt) has a local maximum at t = 0 (a) We compute the first derivatives and the Hessian of f: 2 + 6x 2 x f(x, y) = 2x + 2y + 3x 2, y f(x, y) = 2x + 3y 2, f (x, y) = 0 6y Both partial derivatives vanish at (0, 0); hence (0, 0) is a critical point In addition the Hessian at (0, 0) is 2 2 f (0, 0) = 2 0 This matrix is indefinite: 2 > 0 and = 2 < 0 Therefore f has a saddle point at (0, 0) (b) According to Taylor s formula, near (0, 0) f(x, y) f(0, 0)+ t f (0, 0)(x, y)+ Hence, t (x, y)f (0, 0)(x, y)+ x 2 +2xy+ = (x+y) 2 y 2 + g(t) = f(at, bt) (at + bt) b 2 t 2 + For g to have a minimum at 0, we should try to have g look like a positive quadratic form: the term b 2 t 2 above would then vanish Therefore set a = 1, b = 0 to get g(t) = f(t, 0) = t 2 + t 3 This function clearly has a local minimum at 0 because g (0) = 0 and g (0) = 2 > 0 Therefore a = 1 and b = 0 respond to the question (c) We apply the same argument as above and get: h(t) = f(ct, dt) (ct + dt) 2 d 2 t 2 +

5 This time, we would like to have a local maximum Thus the term (ct + dt) 2 above should vanish To this end we pick c = 1 and d = 1 We get then h(t) = f(t, t) = t 2 2t 2 + t 3 t 3 = t 2 This function clearly has a local maximum at 0, hence c = 1 and d = 1 respond to the problem 5