MATH 5720: Unconstrained Optimization Hung Phan, UMass Lowell September 13, 2018
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1 MATH 57: Unconstrained Optimization Hung Phan, UMass Lowell September 13, 18 1 Global and Local Optima Let a function f : S R be defined on a set S R n Definition 1 (minimizers and maximizers) (i) x S is called a global minimum point (or a global minimizer) of f over S if f(x) f(x ) for all x S (ii) x S is called a global maximizer of f over S if f(x) f(x ) for all x S (iii) x S is called a local minimum point (or a local minimizer) of f over S if there exists an r > such that f(x) f(x ) for all x S B(x, r) (iv) x S is called a local minimizer of f over S if there is an r > such that f(x) f(x ) for all x S B(x, r) (v) x S is called a local maximizer of f over S if there is an r > such that f(x) f(x ) for all x S B(x, r) First order optimality condition Theorem (first order optimality condition) Let f : S R, U R n Suppose x int (S) is a local optimum point and all partial derivatives of f exist at x Then f(x ) Definition 3 (stationary point) Let f : S R, U R n Suppose that x int (S) and f is differentiable over some neighborhood of x Then x is called a stationary point if f(x ) Definition 4 (saddle point) Let f : S R, U R n Suppose that x int (S) and f is differentiable over some neighborhood of x A stationary point x is called a saddle point if x is neither a (local) minimizer nor maximizer 3 Symmetric Matrices Definition 5 Let A R n n be a symmetric matrix Then (i) A is said to be positive semidefinite, denoted by A, if x T Ax for all x R n (ii) A is said to be positive definite, denoted by A, if x T Ax > for all x R n {} (iii) A is said to be negative semidefinite, denoted by A, if x T Ax for all x R n (iv) A is said to be negative definite, denoted by A, if x T Ax < for all x R n {} (v) A is said to be indefinite if there exist x and y such that x T Ax > and y T Ay < Positive definiteness does not mean that all entries are positive 1
2 Example 6 A 1 is positive definite, while A 1 1 is not positive definite 1 Lemma 7 Let A be positive definite Then the diagonal elements of A are positive Let A be positive semidefinite Then the diagonal elements of A are nonnegative Let A be negative definite Then the diagonal elements of A are negative Let A be negative semidefinite Then the diagonal elements of A are nonpositive Lemma 8 Let A be symmetric If there exist positive and negative elements in the diagonal of A, then A is indefinite A pair (λ, x) R R n with x is called an eigenpair of the matrix A if Ax λx The value λ is called an eigenvalue and the vector x is called an eigenvector 7 4 Example 9 (finding eigenpairs) Let A Find the eigenpairs of A Solution λ is an eigenvalue Ax λx for some x (A λi)x det(a λi) We have 7 4 λ 7 λ 4 A λi λ λ So det(a λi) (7 λ)(1 λ) 4 4 λ 8λ 9 So λ 1 and λ 9 We now find an eigenvector with λ 1: consider 7 λ 4 x1 8 4 x1 λ 4 ( x1 ) 1 Next, we find an eigenvector with λ 9: consider 7 λ 4 x1 4 x1 λ 4 8 ( x1 ) 1 Theorem 1 (spectral decomposition) Let A R n n be a symmetric matrix Then there exists an orthogonal matrix U R n n (U T U UU T I) and a diagonal matrix D diag(d 1, d,, d n ) such that U T AU D In addition, the columns of U constitutes an orthonormal basis comprised of eigenvectors of A and the diagonal elements of D are the corresponding eigenvalues Theorem 11 (eigenvalue characterization) Let A R n n be a symmetric matrix Then (i) A is positive definite if and only if all its eigenvalues are positive (ii) A is positive semidefinite if and only if all its eigenvalues are nonnegative (iii) A is negative definite if and only if all its eigenvalues are negative (iv) A is positive semidefinite if and only if all its eigenvalues are nonpositive (v) A is indefinite if and only if it has at least one positive eigenvalue and at least one negative eigenvalue
3 Corollary 1 (diagonal matrices) Let D diag{d 1,, d n } be a diagonal matrix Then (i) D is positive definite if and only if d i > for all i (ii) D is positive semidefinite if and only if d i for all i (iii) D is negative definite if and only if d i < for all i (iv) D is positive semidefinite if and only if d i for all i (v) D is indefinite if and only if there exists d i > and d j < Proposition 13 Let A be a symmetric matrix Then A is positive semidefinite if and only if tr(a) and det(a) Example 14 A is positive definite; B is not positive definite Let A be an n n matrix, the determinant of the upper left k k submatrix is called the k-th principal minor and is denoted by D k (A) For example a b c A d e f a b D 1 (A) a, D (A) det, D d e 3 (A) det(a) x y z Theorem 15 (principle minors criterion) Let A be an n n symmetric matrix Then A is positive definite if and only if D i (A) > for all i 1,, n 4 Example 16 A 3 We have D 1 (A) 4, D (A) 8, and D 3 (A) 18 So A 3 4 B We have D 1 (B), D (B) D 3 (B) 1 But B is neither positive definite nor positive semidefinite positive and negative diagonal elements 1 Similary, C 1 because C In fact, B is indefinite because it has A symmetric matrix A [a ij ] n n is said to be diagonally dominant if a ii j i is said to be strictly diagonally dominant if a ii > j i a ij for every i a ij for every i A Theorem 17 Let A be a symmetric matrix If A is diagonally dominant with nonnegative diagonal entries, then A is positive semidefinite If A is strictly diagonally dominant with positive diagonal entries, then A is positive definite 3
4 4 Second order optimality condition Theorem 18 (necessary second order optimality condition) Let f : U R be twice continuously differentiable on the open set U, and that x is a stationary point Then (i) if x is a local minimizer of f over U, then f(x ) (ii) if x is a local maximizer of f over U, then f(x ) Theorem 19 (sufficient second order optimality condition) Let f : U R be twice continuously differentiable on the open set U, and that x is a stationary point Then (i) if f(x ), then x is a local minimizer of f over U (ii) if f(x ), then x is a local maximizer of f over U Theorem (sufficient condition for a saddle point) Let f : U R be twice continuously differentiable on the open set U, and that x is a stationary point If f(x ) is indefinite, then x is a saddle point 5 Existence of Optimum Points Theorem 1 (Weierstrass) Let f be a continuous function defined on a nonempty and compact set C Then there exist a global minimizer and a global maximizer of f over C Let f : R n R be a continuous function f is said to be coercive if lim f(x) + x Theorem Let f : R n R be continuous and coercive and let S be a nonempty closed set Then f has a global minimizer on S Theorem 3 (global optimality conditions) Let f be twice continuously differentiable over R n Suppose that f(x) for all x R n Let x be a stationary point of f Then x is a global minimizer of f 6 Examples Example 4 Let f(x) 3x 4 x x Find its stationary points and classify them Solution Consider the derivative Consider the second derivative f (x) 1x x 36 x 3 5x + 7x 3 (x 1)(x 1)(x 3) x 1, x 3 f (x) 36 1x + 84 We have f (1) and f (3) 48 > So x 3 is a local minimizer while x 1 need further investigation In fact, as f (x) does not change sign when x moves across x 1, therefore, x 1 is neither a maximizer nor a minimizer (it is an inflection point) 4
5 Example 5 Find and classify all stationary points of f(x 1, ) x Solution The stationary points are those such that the gradient equals zero 6x f(x 1, ) 1 + 6x The first equation implies 6x 1 (x 1 + ), so either x 1 or x 1 If x 1 then the second equation implies 6 4, ie, 4 If x 1, then substituting into the second equation yields So 4 and, which implies x 1 4 and x 1 respectively So there are 3 stationary points (, 4), (4, 4) and (, ) The Hessian of f is given by x1 + x f(x 1, ) 6 x 1 x 1 1 We have 4 f(, 4) 6 is positive definite because the matrix is diagonal with positive diagonal 1 elements So (, 4) is a local minimizer We have f(4, 4) 6 Because det 1 < is the product of two eigenvalues, one eigenvalue must be negative and the other must be positive So (, 4) is a saddle point We have f(, ) 6 The matrix is indefinite because it has both positive and negative 1 elements on the diagonal So (, ) is a saddle point Example 6 Find and classify all stationary points of the following functions (i) f(x 1, ) ( 1 + x 1) + ( 1) (ii) f(x 1, ) 1 + x 1 + 4x 1 (iii) f(x, y) x + y + y Quadratic Functions A quadratic function over R n is given by f(x) x T Ax + b T x + c ( x 1 ) A b x b T c 1 where A R n n is symmetric, b R n, and c R The gradient and Hessian of f have simple formulas What if A is not symmetric? Then notice that f(x) Ax + b and f(x) A x T Ax (x T Ax) T x T A T x x T ( A + A T So, we can replace A by A+AT, which is symmetric Proposition 7 Let f(x) x T Ax + b T x + c, where A R n n is symmetric Then ) x 5
6 (i) x is stationary if and only if Ax b (ii) if A, then x is a global minimizer if and only if Ax b (iii) if A, then x A 1 b is the strict global minimizer Proof (i): x is stationary f(x) Ax + b Ax b (ii): Suppose x is a global minimizer x is stationary Ax b Conversely, suppose Ax b, we will show that f(y) f(x) for all y Indeed, take any y R n, we notice that y T b b T y and x T Ay y T Ax Since A, we have (y x) T A(y x) y T Ay y T Ax x T Ay + x T Ax y T Ay y T Ax x T Ax + x T Ax y T Ay y T ( b) x T Ax + x T ( b) (y T Ay + b T y + c) (x T Ax + b T x + c) f(y) f(x) This proves x is a global minimizer (iii): similar Note that when A, the global minimizer of f is x A 1 b, thus the minimum of f is f(x ) (x ) T Ax + b T x + c ( A 1 b) T A( A 1 b) b T A 1 b + c c b T A 1 b Proposition 8 Let f(x) x T Ax + b T x + c, where A R n n is symmetric, b R n, and c R Then f is coercive if and only if A References [1] A Beck, Introduction to Nonlinear Optimization: Theory, Algorithms, and Applications with MATLAB, MOS-SIAM Series on Optimization (14) 6
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