Menyuk ENEE605 Lectu

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Menyuk ENEE605 Lectu Eigenproblems There is perhaps no set of matrix techniques that are more important in practice! Definition: Let A = [ajk]beagien] be a given n n matrix and consider the vector equation Ax = x, then the column vector solutions x are eigenvectors (characteristic vectors) and the corresponding values of are eigenvalues (characteristic values) More generally: Let A be a linear transformation that maps a linear vector space S (usually a Hilbert space in applications) into itself , A:S S, and let x S, then the solutions to Ax = x for x are eigenvectors and the corresponding values of are eigenvalues Remark: The number of independent eigenvectors is at most n for the n n matrix. The number can be infinite in general. 8.1 Eigenproblems There is perhaps no set of matrix techniques that are more important in practice! Definition: Let A = [ajk]beagien] be a given n n matrix and consider the vector equation Ax = x, then the column vector solutions x are eigenvectors (characteristic vectors) and the corresponding values of are eigenvalues (characteristic values) More generally: Let A be a linear transformation that maps a linear vector space S (usually a Hilbert space in applications) into itself , A:S S, and let x S, then the solutions to Ax = x for x are eigenvectors and the corresponding values of are eigenvalues Remark: A key question to which we will return multiple times is: Do the eigenvectors span the linear vector space n n or more generally S ? 8.2 1 Eigenproblems 52 Example 1: Consider the simple matrix A and the 22 corresponding characteristic equation Ax = x, then in this case, we may write in components 52xx11 Ax , which becomes 22 xx22 52 x1 ()0AI x or 0 22 x2 This equation only has a solution if (A – I) is singular or det(A – I) = 0. This condition is obeyed if ( –5 – )(–2 – ) – 4=04 = 0 or 2 +7+ 7 +6=0+ 6 = 0. Solutions are 1 = and 2 = . The equation for the first eigenvalue x1 now becomes 42x11 1 0 x1 21 x21 2 where is an arbitrary constant. 8.3 Eigenproblems Example 1 (continued): Similarly, the equation for the second eigenvalue x2 becomes 12x12 2 0 x2 24x22 1 where is an arbitrary constant. More generally: The eigenvalue equation (A – I) = 0 implies aa11 12 a 1n aa a D() det(AI )21 22 2n 0 aann12 a nn This determinant is an n-th order polynomial in . We must find the roots of this polynomial to obtain the eigenvalues. This is a hard problem in general! 8.4 2 Eigenproblems Theorem 1: The eigenvalues of a square n n matrix A are the roots of the characteristic equation det(A – I) = 0, which is an n-th order polynomial. Hence, each square matrix has at least one eigenvalue and no more than n. Proof: It is a consequence of the fundamental theorem of algebra that any n-th order polynomial has n (in general complex) roots. These are the eigenvalues. The number of distinct roots is between 1 and n. Remark: Once the eigenvalues are found, the eigenvectors may be found by Gaussian elimination. Theorem 2: The space of eigenvectors corresponding to an eigenvalue is a linear manifold . This manifold must have at least dimension 1. Proof: The manifold of eigenvectors corresponds to the null space of A – I. Since this matrix is singular, we have nullity A – I 1 8.5 Eigenproblems Example 2: Find the eigenvalues of the matrix 22 3 A 21 6 12 0 The characteristic determinant det(A – I) = 0 yields the equation 32 21 45 0 (5 )(3 ) 2 0 This yields the roots (eigenvalues) The first root yields the rank 2 matrix and eigenvalue 72 3 1 AI 246and2 x 1 125 1 8.6 3 Eigenproblems Example 2: We next consider 12 3 123 AI3246 000 after row reduction 123 000 The null space has two dimensions, yielding two independent eigenvectors 23 xx 1and0 23 01 Definition: The order of an eigenvalue M is the algebraic multiplicity of . The number m of linearly independent vectors corresponding to an eigenvector is its geometric multiplicity. In general, M mThe difference M m is the defect of . A matrix that has any eigenvalues with a non-zero defect is defective. 8.7 Eigenproblems Example 2: We next consider 12 3 123 AI3246 000 after row reduction 123 000 The null space has two dimensions, yielding two independent eigenvectors 23 xx 1and0 23 01 Remark: In Example 2 , we find M m and M m . So, neither eigenvalue is defective. Remark: Defective isn’t necessarily bad! The problem of finding the parameters for critical damping of an oscillator can be framed mathematically as a search for parameters that yield a defective matrix. 8.8 4 Eigenproblems Example 3 (A defective matrix): We next consider 21 A which has the eigenvalues 122 02 T with eigenvector x1 = [1 0] and no others! To see that, we note that any T eigenvector [x1 x2] would have to satisfy 2x1 + x2 = 2x1, so that x2 = 0. Hence, in the case M = 2 and m Definition: A matrix J of the form shown below is a Jordan matrix. 1 0 J 1 0 Theorem: A Jordan matrix has m = 1. 8.9 Eigenproblems Example 4 (Real matrices with complex eigenvalues and eigenvectors): Since polynomials with real coefficients have complex roots, we expect that real matrices will have complex eigenvalues and eigenvectors. Consider 01 1 2 A for which det(AI ) ( 1) 10 1 with eigenvalues 1 = i and 2 = i. Corresponding eigenvectors are 11 xx12 and ii Note that these are a complex conjugate pair. Why? More generally: ab A has the eigenvalues aib ba with the same eigenvectors 11 xx12 and ii 8.10 5 Eigenproblems Example 5 (Boundary value problem): Consider the boundary value problem on the interval [0, 1] where we look for solutions of dx2 x 0 with the boundaryy()() conditions xx(0)( 1) 0 dt 2 These are called Dirichlet boundary conditions. The eigenvalues in this case 2 + are m = (m) , where m = 1, 2, … The corresponding eigenvectors are xm(t) = sin( mt). Changing the boundary conditions changes the eigenproblem! (a) Neumann boundary conditions: dx dx 0 dttt01 dt 2 In this case, we find m = (m) , where m = 0, 1, 2, … The corresponding eigenvectors are xm(t) = cos(mt). 8.11 Eigenproblems Example 5 (Boundary value problem): (c) Periodic boundary conditions: dx dx x(()0)(x ()1) and dttt01 dt 2 In this case, we find m = (2m) , where m = …, 2, 1, 0, 1, 2, … The corresponding eigenvectors are xm(t) = exp(i 2mt). Each of these three solutions to an eigenproblem is a complete orthogonal basis for 2(0,1). Connection to an eigenproblem is usually the key to demonstrating that a basis is complete. 8.12 6 Completeness Demonstrating that the eigenvectors of an eigenproblem are complete is generally a difficult problem… …and impppportant in applications Remark 1: For n n matrix problems, we must determine if a matrix is defective. If it is, we must add extra vectors to the span the space. Remark 2: Similar but more complicated issues arise in infinite- dimensional problems. 8.13 Completeness Why is this so important? Original problem input: Decompppgose the input into eigenvectors time, space, pixel number, sample number A complicated transformation A much simpler transformation Return to the original space Original problem output 8.14 7 Completeness A key engineering example: The Fourier transform! Input time signal Input Fourier transform frequency signal A time-dependent filter: xout()tGttxtdt ( ) in ( ) Frequency domain: xGxout() () in () Ouput Inverse Fourier transform Ouput time signal frequency signal For this approach to work, we must be able to decompose any signal. The space of eigenmodes must be complete or easily completable 8.15 Example Applications Example 1 (Markov process): (From slide 2.24) An MRI scan has detected a cancerous growth in which 30% of the cells are cancerous (state 1), 20% are abnormal (state 2), and 50% are normal (state 3). Therapy begins in January , and it is known that the transition probabilities from month to month are given by 0.8 0.1 0 P 0.1 0.7 0.1 0.1 0.2 0.9 What is the best that the treatment can do? How many months should the treatment go on? What fraction of the cells are cancerous when the treatment ends? Solution: To find the solution, we first note that: (1) all orthogonal matrices must have at least one eigenvalue whose value is equal to 1 and (2) no eigenvalues can have a real part whose values are greater than 1. 8.16 8 Example Applications Solution (continued): We first find the eigenvector that the corresponds to = 1. Our matrix becomes 0.2 0.1 0 1/ 8 PI0.1 0.3 0.1 x 2 / 8 1 0.1 0.2 0.1 5 / 8 where we look for the solution that satisfies x11 + x21 + x31 = 1. The characteristic equation for P is 48 188 24 230 100 100 10 Dividing by 1 – , we find the reduced equation 48 14 2 00.8,0.6 100 10 23 The corresponding eigenvectors may be usefully chosen: T T x2 = [1/8 0 1/8] , x3 = [1/8 1/4 1/8] 8.17 Example Applications Solution (continued): Note that the elements of x2 and x3 sum to zero. [This will be true for the eigenvectors of any stochastic matrix who have eigenvalues whose real parts are less than 1.
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