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Chapter 6 Eigenvalues Example In a certain town, 30 percent of the married women get divorced each year and 20 percent of the single women get married each year. There are 8000 married women and 2000 single women, and the total population remains constant. Let us investigate the long-range prospects if these percentage of marriages and divorces continue indefinitely into the future. 1 Eigenvalues and Eigenvectors Definition Let A be an n×n matrix. A scalar is said to be an eigenvalue or a characteristic value of A if there exists a nonzero vector x such that Ax x . The vector x is said to be an eigenvector or a characteristic vector belonging to . Example Let 4 2 A 1 1 and 2 x 1 The subspace N(A- I) is called the eigenspace corresponding to the eigenvalue . The polynomial p( ) det( A I ) is called the characteristic polynomial, and equation det( A I ) 0 is called the characteristic equation for the matrix A. Let A be an n×n matrix and be a scalar. The following statements are equivalent: (a) is an eigenvalue of A. (b) ( A I ) x 0 has a nontrivial solution. (c) N ( A I ) 0 (d) A I is singular. (e) det( A I ) 0 Example Find the eigenvalues and the corresponding eigenvectors of the matrix 3 2 A 3 2 2 3 1 Example Let A 1 2 1 1 3 2 Find the eigenvalues and the corresponding eigenspaces. The Product and Sum of the Eigenvalues a11 p ( ) det( A I ) a21 a1n a22 a2 n a12 an1 ann an 2 Expanding along the first column, we get n det( A I ) (a11 ) det( M 11 ) ai1 (1)i 1 det( M i1 ) i 2 1 2 n p(0) det( A) n n a i 1 i i 1 ii The sum of the diagonal elements of A is called the trace of A and is denoted by tr(A). Example If 5 18 A 1 1 Some Properties of the Eigenvalues: 1. Let A be a nonsingular matrix and let be an eigenvalue of A, then 1 is an eigenvalue of A-1. 2. Let be an eigenvalue of A and let x be an eigenvector belonging to , then m is an eigenvalue of Am and x is m an eigenvector of Am belonging to for m=1, 2, …. 3. Let f ( x) a0 x m a1 x m1 am1 x am , and let be an eigenvalue of A, then f ( ) is an eigenvalue of f ( A) . Example If the eigenvalues of matrix A are: 2, 1, -1, then find the eigenvalues for the following matrices: (a) A2 A I (b) A A1 Similar Matrices Theorem 6.1.1 Let A and B be n×n matrices. If B is similar to A, then the two matrices both have the same characteristic polynomial and consequently both have the same eigenvalues. 3 Diagonalization Theorem 6.3.1 If 1,2, , k are distinct eigenvalues of an n×n matrix A with corresponding eigenvectors x1, x2, …,xk, then x1, …, xk are linearly independent. Definition An n×n matrix A is said to be diagonalizable if there exists a nonsingular matrix X and a diagonal matrix D such that X-1AX=D We say that X diagonalizes A. Theorem 6.3.2 An n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. Remarks 1. If A is diagonalizable, then the column vectors of the diagonalizing matrix X are eigenvectors of A, and the diagonal elements of D are the corresponding eigenvalues of A. 2. The diagonalizing matrix X is not unique. Reordering the columns of a given diagonalizing matrix X or multiplying them by nonzero scalars will produce a new diagonalizing matrix. 3. If A is an n×n matrix and A has n distinct eigenvalues, then A is diagonalizable. If the eigenvalues are not distinct, then A may or may not be diagonalizable depending on whether A has n linearly independent eigenvectors. 4. If A is diagonalizable, then A can be factored into a product XDX-1. Ak XD k X 1 1k X k2 1 X kn Example Let 2 3 A 2 5 Determine whether the matrix is diagonalizable or not. Example Let 3 1 2 A 2 0 2 2 1 1 Determine whether the matrix is diagonalizable or not. Definition If an n×n matrix A has fewer than n linearly independent eigenvectors, we say that A is defective. Theorem 6.3.3 , s are s distinct eigenvalues If A is an n×n matrix and 1,2, for A, let i(1) ,i( 2) ,,i( nr ) be the basis of N (i I A) , where i 2(1) ,2( 2) ,,2( nr ) , ri r (i I A) (i 1, , s ) , then 1(1) ,1( 2) ,,1( nr ), 1 , s(1) , s( 2 ) , , s( n rs ) are linearly independent. 2 Example Let 2 0 0 A 0 4 0 1 0 2 2 0 0 and B 1 4 0 3 6 2 Determine whether the two matrices are diagonalizable or not. Some Results for Real Symmetric Matrix: 1. The eigenvalues of a real symmetric matrix are all real. 2. If 1,2, , k are distinct eigenvalues of an n×n real symmetric matrix A with corresponding eigenvectors x1, x2, …,xk, then x1, …, xk are orthogonal. 3. If A is a real symmetric matrix, then there is an orthogonal matrix U that diagonalizes A, that is, U-1AU=UTAU=D, where D is diagonal. 1 2 0 Example Let A 2 2 2 0 2 3 Find an orthogonal matrix U that diagonalizes A. 2 2 2 Example Let A 2 5 4 2 4 5 Find an orthogonal matrix U that diagonalizes A. 6 Quadratic Forms Definition A quadratic equation in two variables x and y is an equation of the form (1) ax 2 2bxy cy 2 dx ey f 0 Equation (1) may be rewritten in the form a b x d y b c y (2) x Let x x y The term and x e f 0 y a b A b c x T Ax ax 2 2bxy cy 2 is called the quadratic form associated with (1). Conic Sections The graph of an equation of the form (1) is called a conic section. A conic section is said to be in standard position if its equation can be put into one of these four standard forms: (1) x 2 y 2 r 2 (2) x2 y2 1 (circle) (ellipse) y 2 x2 x2 y 2 (hyperbola) 1 or (3) 1 2 2 2 2 2 2 (4) x 2 y or y 2 x (parabola) Example Consider the conic section 3x 2 2 xy 3 y 2 8 0 This equation can be written in the form x 3 1 The matrix 1 3 3 1 x 8 y 1 3 y has eigenvalues 2 and 4 1 1 with corresponding unit eigenvectors 21 and 12 2 2 Let 1 Q 2 1 2 and set Thus 1 cos 45 2 1 sin 45 2 1 x 2 y 1 2 sin 45 cos 45 1 ' 2 x 1 y ' 2 2 0 Q AQ 0 4 T and the equation of the conic becomes ( x' )2 ( y ' )2 1 4 2 Quadratic Surfaces (1) x2 y 2 z 2 2 2 1 2 a b c (2) (ellipsoid) 2 (3) 2 (cone) 2 x y z 2 2 1 2 a b c 2 (4) (hyperboloid of one sheet) (5) x2 y 2 z 2 2 2 0 2 a b c x2 y 2 2 2z 2 a b (elliptic paraboloid) 2 2 x y z 2 2 1 2 a b c (hyperboloid of two sheets) (6) x2 y 2 2 2z 2 a b (hyperbolic paraboloid) A quadratic form including n variables is: f x1 x2 xn a11 x12 2a12 x1 x2 2a13 x1 x3 2a1n x1 xn a22 x22 2a23 x2 x3 2a2 n x2 xn ann xn2 ( x1 x2 a11 a12 xn ) a 1n a1n x1 a22 a2 n x2 T X AX a2 n ann xn a12 Theorem 6.3.5 For any quadratic form XTAX, we can find an orthogonal transformation X=CY such that YTBY is in standard form. Example Let f x1 x2 x3 x4 2 x1 x2 2 x1 x3 2 x1 x4 2 x2 x3 2 x2 x4 2 x3 x4 Find an orthogonal transformation X=CY such that YTBY is in Standard form. Example For the conic section 2 x1 x2 2 x1 x3 2 x2 x3 1 0 Find an orthogonal transformation X=CY such that YTBY is in standard form. Definition A quadratic form f(x)=xTAx is said to be definite if it takes on only one sign as x varies over all nonzero vectors in Rn. The form is positive definite if xTAx>0 for all nonzero x in Rn and negative definite if xTAx<0 for all nonzero x in Rn. A quadratic form is said to be indefinite if it takes on values that differ in sign. If f(x)=xTAx ≥0 and assumes the value 0 for some x≠0, then f(x) is said to be positive semidefinite. If f(x) ≤0 and assumes the value 0 for some x≠0, then f(x) is said to be negative semidefinite. Definition A real symmetric matrix A is said to be I. Positive definite if xTAx>0 for all nonzero x in Rn. Ⅱ. Negative definite if xTAx<0 for all nonzero x in Rn. III. Positive semidefinite if xTAx≥0 for all nonzero x in Rn. IV. Negative semidefinite if xTAx≤0 for all nonzero x in Rn. V. Indefinite if xTAx takes on values that differ in sign. Theorem 6.6.2 Let A be a real symmetric n×n matrix. Then A is positive Definite if and only if all its eigenvalues are positive. Theorem 6.6.3 Let A be a real symmetric n×n matrix. Then A is positive definite if the leading principal submatrices A1, A2, …, An of A are all positive definite. Example Determine whether the quadratic form f ( x1 , x2 , x3 ) 5 x12 6 x2 2 4 x32 4 x1 x2 4 x2 x3 is positive definite.