Solve Eigenvalue of Complex Matrix
Date: 7/17/96 at 23:1:59 From: Shuichi Suga, Toshiba Subject: Re: Solve Eigenvalue of Complex Matrix Dear Dr.Anthony, I have heard of two methods to solve the eigenvalues of complex matrices. These are the Hause-Gibbson method and the Jacobi method. Usually the Hause-Gibbson method is the good way to solve eignenvalues and in the case of Hermite matrices, the Jacobi method is preferred. If you could, please send me information about the above two methods. Regards, Shuichi Suga
Date: 7/19/96 at 16:10:49 From: Doctor Anthony Subject: Re: Solve Eigenvalue of Complex Matrix This is quite a big topic, and I think you would be well advised to have a textbook on advanced matrices ready on-hand. I will give you some of the background you need to know about Hermitian matrices, and complex eigenvalues, and work through one or two examples, but it would take more than a brief note from me to do justice to the subject. The complex conjugate of a matrix has all its elements the complex conjugates of the original matrix. Any real element will, of course, be unchanged. So if: A = |1+i 2 -i| then A* = |1-i 2 i| | 3 1-i 2+i| | 3 1+i 2-i| If A = A* the matrix will be real If A = -A* the matrix is imaginary since all its elements must be imaginary. A matrix is symmetric if A = A(t) where A(t) is the transpose of A. Such a matrix is necessarily square and has the leading diagonal as a line of symmetry. A matrix is skew-symmetric if A = -A(t) Such a matrix is also square, but the elements of the leading diagonal must all be zero. Hermitian and skew-Hermitian Matrices ------------------------------------- A matrix A is called Hermitian if A = A(*)(t) Example | 1 1+i i| |1-i 2 4| | -i 4 3| A matrix A is skew-Hermitian if A = -A(*)(t) In this case the elements of the leading diagonal can be either zero or purely imaginary. Example | i 1 1-i| | -1 0 i | |-1-i i 2i | To find the eigenvalues and eigenvectors when there are complex eigenvalues. If A = |1 -1 -1| |1 -1 0| |1 0 -1| The characteristic equation is |1-k -1 -1| | 1 -1-k 0| | 1 0 -1-k| which gives three roots k1 = -1, k2 = i, k3 = -i Using the first eigenvalue -1 and normalizing to unit length the corresponding eigenvector is (0, 1/sqrt(2), -1/sqrt(2)) Using the second eigenvalue i we get x=(1+i)y, and y = z Now, since some of the elements of the eigenvector are complex, when normalizing to unit length we use the generalized definition of distance and require (x*)x + (y*)y + (z*)z = 1 This gives x = (1+i)/2, y = 1/2 z = 1/2 [check: (1-i)/2 * (1+i)/2 + (1/2)(1/2) + (1/2)(1/2) = 2/4 + 1/4 + 1/4 = 1] For the third eigenvalue -i, we get x = (1-i)/2, y = 1/2, z=1/2 There has not been much difference in this working from that required when the roots of the characteristic equation are all real. The main difference was using (x*)(x) + (y*)(y) + (z*)(z) = 1 instead of x^2 + y^2 + z^2 = 1 when normalizing to unit eigenvector. Hermitian matrices ------------------ A real symmetric matrix A = A(t) is just the real counterpart of a Hermitian matrix A = (A*)(tr). Since the eigenvalues of a real symmetric matrix are always real, the corresponding results apply to Hermitian matrices, namely: (a) The eigenvalues of a Hermitian matrix are real. (b) For a Hermitian matrix of order n there are n mutually orthogonal eigenvectors irrespective of whether the eigenvalues are all different or not. Example A = | 1 1+i| This is Hermitian. |1-i 2 | Its characteristic equation is |1-k 1+i| = 0 |1-i 2-k| which gives k1 = 0 and k2 = 3 (both real) The normalized eigenvectors are with k=0 [-(1+i)/sqrt(3), 1/sqrt(3)] and with k=3 [(1+i)/sqrt(6), sqrt(2/3)] The scalar product of these will give zero, indicating that the eigenvectors are orthogonal. I have indicated briefly some of the properties of Hermitian matrices, and how complex eigenvalues should be treated. However, as I mentioned at the start, this is quite a big topic and you will need to refer to appropriate textbooks to get the full story. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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