Math 3191 Applied Linear Algebra

Transcription

1 Math 3191 Applied Linear Algebra Lecture 5: Quadratic Forms Stephen Billups University of Colorado at Denver Math 3191Applied Linear Algebra p.1/16

2 Diagonalization of Symmetric Matrices Recall: A symmetric matrix A is one satisfying A = A T. NOTE: A must be square to be symmetric. Some interesting (and useful) facts: 1. If A is symmetric, then it has a complete set of orthogonal eigenvectors. The dimension of each eigenspace is equal to the (algebraic) multiplicity of the corresponding eigenvalue. (And we can find an orthogonal basis for each eigenspace). Eigenvectors corresponding to different eigenvalues are not just linearly independent, but also orthogonal. (Theorem 1). If A is symmetric, then it is diagonalizable (actually orthogonally diagonizable). That is A = P DP 1 = P DP T, where P is an orthogonal matrix, and D is diagonal. Math 3191Applied Linear Algebra p./16

3 Theorem 1 If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal. Proof Let v 1 and v be eigenvectors corresponding to distinct eigenvalues λ 1 and λ. Observe that v T Av 1 = v1 T A T v = v1 T Av = v T (λ 1 v 1 ) = v1 T (λ v ) = λ 1 (v 1 v ) = λ (v 1 v ) = v 1 v = 0. Math 3191Applied Linear Algebra p.3/16

4 Quadratic Forms A quadratic form is a function Q : IR n IR that can be written in the form Q(x) = x T Ax, with A symmetric. Examples: Q(x) = x T Ax, with A = We could rewrite this in the form Q(x) = 3x 1 4x 1 x + 7x. Math 3191Applied Linear Algebra p.4/16

5 Going the other way Question: Is any function of the form Q(x) = ax 1 + bx 1 x + cx a quadratic form? 3 a b/ Answer: Yes. Set A = 4 5. b/ c Question: What about Q(x) = 5x 1 + 3x + x 3 x 1 x + 8x x 3? 3 5 1/ 0 Answer: Set A = 6 4 1/ Math 3191Applied Linear Algebra p.5/16

6 What if A isn t symmetric? If A isn t symmetric, the function Q(x) = x T Ax is still a quadratic form: Define  = A+AT then «x T Âx = x T A + A T x = 1 x T Ax + x T A T x = 1 x T Ax + x T Ax = x T Ax. Because of this, it is safe to assume that A is symmetric when we examine quadratic forms. Math 3191Applied Linear Algebra p.6/16

7 Change of Variable in a Quadratic Form If x is a variable in IR n, then a change of variable is an equation of the form x = P y or y = P 1 x, where P is an invertible matrix and y is a new variable in IR n. Key Idea: If we choose P carefully, we can convert the quadratic form Q(x) = x T Ax into a simpler quadratic form ˆQ(y) = y T Dy, where D is diagonal! In order to make this work, we compute the orthogonal decomposition of A = P DP T, and then use the change of variables x = P y. Then x T Ax = (P y) T AP y = y T P T AP y = y T Dy. Math 3191Applied Linear Algebra p.7/16

8 Example Make a change of variable to transform the quadratic form Q(x) = x 1 8x 1x 5x into a quadratic form with no cross-product terms. Solution: Step 1: Write Q in its matrix form: Q(x) = x T Ax, with A = Step : Calculate the orthogonal diagonalization of A: Char. equation: (1 λ)( 5 λ) 16 = 0 = λ = 3, λ = 7. 3 Corresponding eigenvectors: λ = 3 : 4 5, λ = 7 : Normalized eigenvectors (scaled to get unit vectors): / λ = 3 : 4 1/ 5, λ = 7 : 4 1/ 5 5 / 5. 5 Math 3191Applied Linear Algebra p.8/16

9 Orthogonal diagonalization: A = P DP T, with P = 4 / 5 1/ 5 1/ 5 / 5 3 5, D = Step 3: Perform change of variable x = P y: x 1 8x 1 x 5x = x T Ax = y T Dy = 3y 1 7y. Math 3191Applied Linear Algebra p.9/16

10 Principal Axes Theorem Let A be an n n symmetric matrix. Then there is an orthogonal change of variable x = P y, that transforms the quadratic form x T Ax into a quadratic form y T Dy with no cross-product term. The columns of P in the theorem are called the principal axes of the quadratic form. The vector y is the coordinate vector of x relative to the orthonormal basis of IR n given by these principal axes. Math 3191Applied Linear Algebra p.10/16

11 Geometric View of Prinical Axes If A is an invertible symmetric matrix, and c is a constant, then then set of points satisfying x T Ax = c is one of an ellipse (or circle), a hyperbola, two intersecting lines, a single point, or no points at all. Math 3191Applied Linear Algebra p.11/16

12 If A is diagonal, then the graph of this set of points is in standard position. b x ellipse: a x 1 x 1 x + a b = 1 a > 0, b > 0 x b hyperbola: a x 1 x 1 x a b = 1 a > 0, b > 0 FIGURE An ellipse and a hyperbola in standard position. Chapter 7 Lay, Linear Algebra and Its Applications, Second Edition Update Copyright c 000 by Addison Wesley Longman. All rights reserved. A7..0 Math 3191Applied Linear Algebra p.1/16

13 If A is not diagonal, then the graph is rotated out of standard position. Finding the prinicpal axes (which are the eigenvectors) amounts to finding a new coordinate system for which the graph will be in standard position. y x y x 1 (a) 5x 4x 1 x + 5x = 48 1 x y 1 1 x 1 (b) x 8x 1 x 5x = 16 1 y 1 FIGURE 3 An ellipse and a hyperbola not in standard position. Math 3191Applied Linear Algebra p.13/16

14 Graphs of Quadratic Forms z z x 1 x x 1 x (a) z = 3x + 7x 1 (b) z = 3x 1 z z x 1 x 1 x x x (c) z = 3x 1 7x (d) z = 3x 1 7x FIGURE 4 Graphs of quadratic forms. Chapter 7 Lay, Linear Algebra and Its Applications, Second Edition Update Copyright c 000 by Addison Wesley Longman. All rights reserved. A7..04 Math 3191Applied Linear Algebra p.14/16

15 Classifying Quadratic Forms A quadratic form Q is 1. Positive definite if Q(x) > 0 for all x 0.. Negative definite if Q(x) < 0 for all x Indefinite if Q(x) assumes both positive an negative values for different choices of x. Graphically, the graph z = Q(x) is convex up if Q is positive definite, concave down if Q is negative definite, A saddle if Q is indefinite. Math 3191Applied Linear Algebra p.15/16

16 Quadratic Forms and Eigenvalues Theorem 5: Let A be an n n symmetrix matrix. Then a quadratic form x T Ax is: 1. positive definite if and only if the eigenvalues of A are all positive.. negative definite if and only if the eigenvalues are all negative. 3. indefinite if A has both positive and negative eigenvalues. Math 3191Applied Linear Algebra p.16/16