Positive Definite Matrix

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1 1/29 Chia-Ping Chen Professor Department of Computer Science and Engineering National Sun Yat-sen University Linear Algebra

2 Positive Definite, Negative Definite, Indefinite 2/29

3 Pure Quadratic Function A pure quadratic function consists of only quadratic terms. Such a function has the following form n n f (x 1,..., x n ) = c ij x i x j i=1 j=1 where c ij s are coefficients and x i x j s are quadratic. 3/29

4 Representation by Symmetric Matrix A pure quadratic function can be represented by a symmetric matrix. For n n f (x) = c ij x i x j i=1 j=1 let a ij = 1 x 1 2 (c ij + c ji ), A = {a ij }, x =. x n then f (x) = x T Ax 4/29

5 Example: Bi-variate Case A bi-variate pure quadratic function f (x, y) = ax 2 + 2bxy + cy 2 can be represented by f (x) = x T Ax where x = [ ] x, A = y [ ] a b b c 5/29

6 A real symmetric matrix A is positive definite if x T Ax > 0, x 0 Equivalent conditions (a) x T Ax > 0, x 0. (b) A has only positive eigenvalues. (c) Leading principal sub-matrices of A have positive determinants. (d) A has only positive pivots. 6/29

7 (a) (b) If a real symmetric matrix is positive definite, then it has only positive eigenvalues. Let λ be an eigenvalue of A with an associated eigenvector x. Then Ax = λx x T Ax = λ(x T x) λ = xt Ax x T x λ > 0 7/29

8 (b) (a) If a real symmetric matrix has only positive eigenvalues, then it is positive definite. By the spectral theorem A = QΛQ T It follows that y T y {}}{{}}{ x T Ax = x T Q Λ Q T x = y T Λy = i λ i y 2 i > 0, x 0 8/29

9 (a) (c) If a real symmetric matrix is positive definite, then every leading principal sub-matrix has a positive determinant. Let A be positive definite. For x T = [ x T k 0 T ] x T Ax = [ xk T 0 ] [ ] [ ] T A k B xk B T C 0 = x T k A k x k > 0, x k 0 A k is positive definite all eigenvalues of A k are positive A k = λ k,1 λ k,k > 0 9/29

10 (c) (d) If every leading principal sub-matrix of a real symmetric matrix has a positive determinant, then it has only positive pivots. By the LDU-decomposition A = LDU where L and U are unit-triangular and D is diagonal with pivots. Partitioning the matrices, we get [ ] [ ] [ ] [ ] [ ] Ak B Lk 0 Dk 0 Uk Lk D B T = = k U k C 0 0 A k = L k D k U k A k = D k = d 1... d k d k = A k > 0, k = 1,..., n A k 1 10/29

11 (d) (a) If a real symmetric matrix has only positive pivots, then it is positive definite. When A is symmetric, the LDU-decomposition is If d ii > 0 for all i, then A = LDL T LDL T = LD 1/2 D 1/2 L T = R T R, R = D 1/2 L T It follows that x T Ax = x T R T Rx = (Rx) T (Rx) = Rx 2 > 0 11/29

12 Example A = x T Ax = 2x x x3 2 2x 1 x 2 2x 2 x 3 ( = 2 x 1 1 ) 2 2 x ( x 2 2 ) x x 3 2 > 0 λ = 2, 2 ± 2 A 1 = 2, A 2 = 3, A 3 = A = /29

13 Negative Definite Matrix A is negative definite if x T Ax < 0, x 0 Equivalent conditions (na) x T Ax < 0, x 0. (nb) A has only negative eigenvalues. (nc) Leading principal sub-matrices have alternating determinants. (nd) A has only negative pivots. 13/29

14 Local Approximation and Optimization 14/29

15 First-order Approximation The first-order approximation to a multi-variate function f (x) near x 0 is f (x 0 + δx) f (x 0 ) + i where f (x 0 ) δx i = f (x 0 ) + f (x 0 ) T δx x i f (x) = is the gradient vector of f ( ) at x. f (x) x 1. f (x) x n 15/29

16 Second-order Approximation The second-order approximation to a multi-variate function f (x) near x 0 is f (x 0 + δx) f (x 0 ) + i where f (x 0 ) x i δx i = f (x 0 ) + f (x 0 ) T δx δxt H(x 0 )δx H(x) = { 2 } f (x) x i x j is the Hessian matrix of f ( ) at x. Note that H(x) = H(x) T i j 2 f (x 0 ) x i x j δx i δx j 16/29

17 Approximation around a Stationary Point x 0 is a stationary point of f ( ) if f (x 0 ) = 0 Near a stationary point x 0, the first-order approximation of f ( ) at x 0 is f (x) f (x 0 ) and the second-order approximation of f ( ) at x 0 is f (x) f (x 0 ) (x x 0) T H(x 0 )(x x 0 ) 17/29

18 Example Approximate near x 0 = 0. f (x) = 2x 2 + 4xy + y 2 f (0) = 0, f (0) = 0, H(0) = [ ] f (x) f (0) xt H(0)x = 2x 2 + 4xy + y 2 18/29

19 Example Approximate F (x) = 7 + 2(x + y) 2 y sin y x 3 near x 0 = 0. F (0) = 7, F (0) = 0, H(0) = [ ] F (x) F (0) xt H(0)x = 7 + 2x 2 + 4xy + y 2 19/29

20 Minimum, Maximum, and Saddle Point Let x 0 be a stationary point of f (x). If H(x 0 ) is positive definite, x 0 is a local minimum of f ( ). If H(x 0 ) is negative definite, x 0 is a local maximum of f ( ). If H(x 0 ) is indefinite, x 0 is a saddle point of f ( ). For example, 0 is a saddle point of F (x). 20/29

21 Singular Value Decomposition 21/29

22 Singular Value By definition, the singular values of a real matrix A are the square roots of the non-zero eigenvalues of ( A T A ). ( A T A ) is positive semi-definite, so its eigenvalues are non-negative. By definition, the square root of a positive number is positive, so A has only positive singular values σ i > 0 22/29

23 Number of Singular Values A matrix of rank r has r singular values. Let A be m n. Then ( A T A ) is n n. A T Ax = 0 x T A T Ax = 0 Ax = 0 N ( A T A ) = N (A) dim N ( A T A ) = dim N (A) = n r So 0 is an eigenvalue of ( A T A ) with a multiplicity of n r, meaning that the sum of the multiplicities of the positive eigenvalues of ( A T A ) is n (n r) = r 23/29

24 Singular Value Decomposition The SVD of an m n matrix is A = UΣV T U is an m m orthogonal eigenvector matrix of ( AA ) T ( ) AA T = U ( ΣΣ ) T U T V is an n n orthogoal eigenvector matrix of ( A T A ) ( A T A ) = V ( Σ T Σ ) V T Σ is an m n matrix with singular values on the leading diagonal positions, and 0s elsewhere. 24/29

25 Proof 1... v r v r u r u r+1... Let V = [v ] v n where v 1,..., v r are orthonormal eigenvectors of ( A T A ) associated with non-zero eigenvalues. Let U = [u 1 ] u m where for j = 1,..., r u j = Av j σ j ( AA T ) u j = AAT Av j σ j = Aλ jv j σ j = λ j u j are eigenvectors of ( AA ) T and u r+1,..., u m are orthonormal eigenvectors of ( AA ) T of eigenvalue 0. 25/29

26 Consider U T AV. For j = 1,..., r (U T AV) ij = u T i Av j u T i Av j = u T i (σ j u j ) = σ j δ ij For j = r + 1,..., n ( A T A ) v j = 0 Av j = 0 u T i Av j = 0 Thus That is U T AV = Σ A = UΣV T 26/29

27 Example Find SVD for A = [ 1 1 ] The eigenvalues of A T A = are λ = 3, 1, 0, so the singular values of A are σ 1 = 3, σ 2 = 1 27/29

28 Orthonormal eigenvectors of ( A T A ) are v 1 = 1 1 2, v 2 = 1 1 0, v 3 = Orthonormal eigenvectors of ( AA T ) are u 1 = Av 1 = 1 [ ] 1, u σ = Av 2 = 1 [ ] 1 σ A = UΣV T = [ ] [ ] /29

29 Application to Image Processing The SVD of a matrix can be written as A = UΣV T = σ 1 u 1 v1 T + + σ r u r vr T = A A r which can be used for data compression. For a image, SVD achieves 90% compression when using 50 matrices of rank 1. Data Compression via SVD 29/29

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