f Illit staffd Eld V Exe 41,1 E A at E X E it E XII Find eigenvalues and eigenvectors for the matrix

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1 Fri Nov 30 Symmetric matrices and the spectral theorem, 7-7 Announcements: PCA principal component analysis Prof Tom Alberts guest lecture on Tues will at least by end of lecture indicate the ideas behind genetic gemology Warm-up Exercise: IA Exe Ea XII Find eigenvalues and eigenvectors for the matrix A fi Note that the are orthogonal eigen spaces we need this eigendata in today's class staffd spanit 4, LET x z E A at E X E it E f Illit O for 9 z Eld V

2 Recall that the transpose operation swaps rows with columns, and vise verse These properties arose from the actual definition for A T, which was entry i j A T = entry j i A The i j and j i locations on a matrix are reflections across the diagonal of each other (This is the matrix version of the reflection across the line x = that we've encountered several times in this course) See how this plays out for the matrix A below, by finding the transpose three ways: Turning rows into columns; turning columns into rows; reflecting across the diagonal a 7 A = affair Def A square matrix is symmetric if and only if A T = A Exercise Which of the following matrices is symmetric, and which is not? a) 4 b) B C 0 7 O 3 yes no

3 The Spectral Theorem asserts that all n n symmetric matrices A (with real number entries) are diagonalizable, with n linearly independent real eigenvectors and associated eigenvalues Furthermore, eigenvectors with different eigenvalues are automatically orthogonal (For eigenspaces with dimension greater than one, one can use Gram Schmidt to create orthonormal bases) Thus, the eigenvector basis of n can be chosen to be orthonormal In otherwords, we may express as A P = P D A = P D P = P T e D P D p AP ptap where P is an orthogonal matrix which can also be interpreted as a change of basis matrix Let's see how this plays out in an example This will forshadow sections 7-7 You'll notice that we're using major concepts and ideas from throughout the course, which is not a bad way to be reviewing course material at this point of the semester Example Consider the curve in defined implicitly as the solution set to the equation x x = Can you identify the curve as a conic section? Can you graph it? Note the x cross-term! Does the function f, x =, x = 0, 0? Note, the gradient x x have a local maximum or local minimum at f = f x, f x = 4 x, 4 x = 0, 0 at the point 0, 0, so the origin is at least a candidate for a local max or min Exercise a Check that we can rewrite the quadratic expression as x x = x 00 x t x t Xz t z x 07 t Sx xz 0

4 Note, in general, if v, w n and if A is an n n matrix then is a v T n A n n w n I l r r matrix, ie a scalar And its value is v T A w = n n n n i = v i entry i A w = v i i = j = a i j w j = i, j = a i j v i w j So given a quadratic expression ("quadratic form") in any number of variables, x, x n rewrite the quadratic form as x T A x Ej aijxixj one can and one can choose to make A a symmetric matrix, as we did in our specific example by splitting cross terms symmetrically in the off-diagonal entries of A l g 7 x t x x x x x 70 3 quadraticform in 3 variables Z µ a Yz 7 Yz I

5 Exercise b Find the eigenvalues and eigenvectors for the matrix we're using to express our quadratic expression x x = x x warm up Solution: E = = span E = 9 = span II lil o E x y t Ex

6 Was it an accident that the two eigenvectors were orthogonal? No Here's why that will always be true as long as the eigenvalues are different, for any symmetric matrix of arbitrary size: Theorem Let A be symmetric, and let A v = v A w = w with then v w Proof: Consider the scalar v T A w On one hand v T A w = v T w = v w On the other hand v T A w = v T A T w = A v T w = vt w = So v w = v w v w So as long as we deduce that the eigenvectors v, w satsify v w = 0! Was it an accident that the eigenvalues were real s Nd recall if A hasreal entries if A Exit at bi Itis citit complex eigenvector Freaked A fi iv a bi T iv atbi eigenval Consider fu iv IT Alitit T iv tutin Catbillitivl atbilku ivjlu titi Ali ivdt a biker inflating lat bi Hill'tKilly la bi still'thttp So b o

7 Continuing example and for A = x x = x ; E = = span, E x = 9 = span This suggests creating an orthonormal eigenbasis! And we'll order the eigenvectors so that the corresponding orthogonal matrix is a rotation and not a reflection (by making the determinant of the matrix instead of ) B =, P = A P = P D A = P D P = P T D P Note that P is the change of coordinates matrix, P = P E B where as always, E = 0, 0 LgAjIfIIp For v write v = x in standard coordinates and v B = y y So the two coordinate systems are related by x = y y Wx E = P Hx B

8 Do algebra! For A symmetric and P T A P = D, the quadratic form x T A x can be diagonalized with the change of variables x = P y: x T A x = y T P T A P y = y T D y = n i = i y i In our example, x x = x x = y y y y = y y y y (because P T AP = D) = 9 y y

9 So the original curve with equation in the standard coordinate system has equation 9 y with respect to the rotated coordinate system! x x = y = Answer to a) This curve is a hyperbola! In the rotated coordinate system its equation is Answer to b) No! f, x = x y 3 y = origin is a saddle point, because in the rotated coordinate system Hand sketches: f y, y = 9 y x does not have a local min or max at 0, 0 The y

10 Maple verification: To be continued > with plots : implicitplot x x =, = 3 3, x = 3 3, grid = 00, 00 ; 3 x x 3 > plot3d x x, = 3 3, x = 3 3 ;