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/