Diagonalization of a 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)