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Problem sheet 5: The Hessian matrix. Trond Stølen Gustavsen September, 21, 2009 Problem 1. Solution. See answers in FMEA. Problem 2. Solution. (a) We have f (x, y) = ax2 + 2bxy + cy 2 + px + qy + r. The Hessian matrix is f11 f12 2a 2b f = = . f21 f22 2b 2c The leading principal minors are 2a 2b D1 = 2a and D2 = = 4(ac − b2 ). 2b 2c So if a > 0 and ac − b2 > 0 the Hessian is positive deﬁnite, so the function is strictly convex. If a < 0 and ac − b2 > 0 we have (−1)1 D1 > 0 and (−1)2 D2 > 0, so the Hessian is negative deﬁnite. This means that the function is strictly concave. (b) We know from a theorem in Lecture 6 that a function is convex if and only if the Hessian is positive semideﬁnite. We also know that a symmetric matrix is positive semideﬁnite if all principal minors are positive or zero. The principal minors are 2a 2b , 2a and 2c. 2b 2c Thus the Hessian is positive semideﬁnite if and only if ac − b2 ≥ 0, a ≥ 0 and c ≥ 0. Thus this is a suﬃcient and a necessary condition for the function f to be convex. Similarly we get that the Hessian is negative deﬁnite if and only if ac − b2 ≥ 0 and a ≤ 0 and c ≤ 0, so this is a necessary and suﬃcient condition for the function f to be concave. Problem 3. 1 Solution. One approach is to use the result of the previous problem. We will however solve the problem directly. The function f (x, y) = −6x2 + (2a + 4)xy − y 2 + 4ay has the following partial derivatives: f1 = −12x + (2a + 4)y f2 = (2a + 4)x − 2y + 4a. The Hessian matrix becomes f11 f12 −12 2a + 4 f = = . f21 f22 2a + 4 −2 The principal minors are −12 2a + 4 −12, −2 and D2 = = 8 − 4a2 − 16a 2a + 4 −2 Since principal minors of order 1 are negative, the function is never convex. The function is concave if and only if √ √ 8 − 4a2 − 16a ≥ 0 ⇐⇒ −2 − 6 ≤ a ≤ −2 + 6. Problem 4. Consider the function f (x, y) = x4 + 16y 4 + 32xy 3 + 8x3 y + 24x2 y 2 . Find the Hessian matrix. Show that f is convex. Solution. The Hessian matrix is 48xy + 12x2 + 48y 2 96xy + 24x2 + 96y 2 f = 96xy + 24x2 + 96y 2 192xy + 48x2 + 192y 2 Completing the squares we see that 2 2 12 (x + 2y) 24 (x + 2y) f = 2 2 . 24 (x + 2y) 48 (x + 2y) 2 2 The principal minors of order one are 12 (x + 2y) ≥ 0 and 48 (x + 2y) ≥ 0. The principal minor of order two is 2 2 12 (x + 2y) 24 (x + 2y) 2 2 = 0. 24 (x + 2y) 48 (x + 2y) Thus all principal minors are non-negative, thus the function is convex (but not strictly convex). Problem 5. Solution. See answers in FMEA. Problem 6. 2 Solution. (a) π(x, y) = 13x + 8y − C(x, y) = 9x + 6y − 0.04x2 + 0.01xy − 0.01y 2 − 500. (b) π1 = 9 − 0.08x + 0.01y and π2 = 6 + 0.01x − 0.02y. The Hessian matrix is −0.08 0.01 . 0.01 −0.02 The leading principal minors are −0.08 0.01 D1 = −0.08 and D2 = = 0.0015 > 0 0.01 −0.02 Thus the function concave. Solving π1 = 9 − 0.08x + 0.01y = 0 and π2 = 6 + 0.01x − 0.02y = 0, we get that (160, 380) as the only stationary point which has to be a maximum. Problem 7. Consider the function f deﬁned on the subset S = {(x, y, z) : z > 0} of R3 by f (x, y, z) = 2xy + x2 + y 2 + z 3 . Show that S is convex. Find the stationary points of f . Find the Hessian matrix. Is f concave or convex? Does f have a global extreme point? Solution. If (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are two points in S. We must show that the line segment [(x1 , y1 , z1 ), (x2 , y2 , z2 )] = {(x, y, z)|(x, y, z) = s(x1 , y1 , z1 ) + (1 − s)(x2 , y2 , z2 ), s ∈ [0, 1]} is in S. For a point on the line segment we have z = sz1 + (1 − s)z2 . Since either s, z1 , (1 − s) and z2 are non negative, z is non negative. Since either s or (1 − s) is positive, z must be positive. This shows that the line segment is contained in S. The ﬁrst order conditions are f1 = 2x + 2y = 0 f2 = 2x + 2y = 0 f3 = 3z 2 = 0 From this we see that x = −y z=0 We conclude that there are no stationary points in S. The Hessian matrix is 2 2 0 2 2 0 . 0 0 6z Solving 2−λ 2 0 2 2−λ 0 = −λ (λ − 4) (λ − 6z) 0 0 6z − λ The eigenvalues are thus λ = 0, λ = 4 and λ = 6z. Since all eigenvalues are ≥ 0, we conclude that the Hessian is positive semideﬁnite, hence the function is convex. Since there are no stationary points in S (and since S is open), f does not have a global minimum or maximum. Problem 8. 3 Solution. The function f (x, y, z) = x4 + y 4 + z 4 + x2 + y 2 − xy + zy + z 2 has the Hessian matrix, 12x2 + 2 −1 0 −1 12y 2 + 2 1 . 0 1 12z 2 + 2 The leading principal minors are D1 = 12x2 + 2 > 0 12x2 + 2 −1 D2 = −1 12y 2 + 2 = 24x2 + 24y 2 + 144x2 y 2 + 3 > 0 12x2 + 2 −1 0 D3 = −1 12y 2 + 2 1 0 1 12z 2 + 2 = 36x2 + 48y 2 + 36z 2 + 288x2 y 2 + 288x2 z 2 + 288y 2 z 2 + 1728x2 y 2 z 2 + 4 > 0 So the Hessian is positive deﬁnite, and hence f is convex. 4