Diagonalization of a Matrix

Date: 12/10/98 at 13:24:59
From: Michael
Subject: Diagonalization of a 3x3 real symmetric matrix

Do you know a quick algorithm for diagonalizing a 3x3 real symmetric 
matrix, also obtaining the transformation matrix?

Given A, a real symmetric 3x3 matrix. Find:
 
   P, a nonsingular 3x3 matrix
   D, a diagonal 3x3 matrix

such that:

   A = P D P^(-1)

Date: 12/10/98 at 16:58:01
From: Doctor Rob
Subject: Re: Diagonalization of a 3x3 real symmetric matrix

The diagonal entries of D are the eigenvalues of A. The columns of P 
are the eigenvectors of A. I proceed as follows:

Find the characteristic polynomial of A: det(x*I-A) = f(x). Find the
three roots of f(x) = 0. Those are the eigenvalues a1, a2, and a3.
Find a vector v1 such that (A-a1*I)*v1' = 0. It is a vector in the 
right nullspace of A - a1*I. Do the same to find v2 in the right 
nullspace of A - a2*I, and v3 in the right nullspace of A - a3*I. Then 
v1, v2, and v3 are eigenvectors of A, and the columns of P are 
v1', v2', and v3', in that order. P^(-1) has to be computed from P in 
the usual way.

This may not be quick or simple, but it is effective.

Example:

       ( 3  4 -1)
   A = ( 4  3  1)
       (-1  1  1)

   f(x) = x^3 - 7*x^2 - 3*x + 21
        = (x-7)*(x-sqrt[3])*(x+sqrt[3])

so the eigenvalues are 7, sqrt[3], and -sqrt[3].

                 (-4  4 -1)
   (A-7*I)*v1' = ( 4 -4  1)*v1' = 0
                 (-1  1 -6)
   
   v1 = (1, 1, 0)

                       (3-sqrt[3]     4        -1    )
   (A-sqrt[3]*I)*v2' = (    3     3-sqrt[3]     1    )*v2' = 0 
                       (   -1         1     1-sqrt[3])

   v2 = (1-sqrt[3], sqrt[3]-1, 2)

                       (3+sqrt[3]     4        -1    )
   (A+sqrt[3]*I)*v3' = (    3     3+sqrt[3]     1    )*v3' = 0
                       (   -1         1     1+sqrt[3])

   v3 = (1+sqrt[3], -1-sqrt[3], 2)

       (1  1-sqrt[3]  1+sqrt[3])
   P = (1 -1+sqrt[3] -1-sqrt[3]) 
       (0      2          2    )

            (    1/2         1/2            0      )
   P^(-1) = (-sqrt[3]/12  sqrt[3]/12 (3+sqrt[3])/12).
            ( sqrt[3]/12 -sqrt[3]/12 (3-sqrt[3])/12)

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/