Hello everyone!

Actually I used the answer from another friend (as he wrote):

{vals, vects} = Eigensystem[m = {{0, 0, w}, {0, t, -w}, {w, -w, r}}, Cubics -> True];

P = Transpose@vects;

dd = Inverse@Transpose@vects.m.Transpose@vects;

By the way, it is really confused for a beginner of mathematica to get familiar with something like this:'Root[-q^2 v + s^2 v + 2 s t v + t^2 v + 4 q^2 w + (q^2 - s^2 - 2 s t - t^2 + 2 q v - 4 q w) #1 + (-2 q - v) #1^2 + #1^3 &, 1]'

If you want to see a (very large) solution in terms of radicals, try Eigensystem[mat[t, w, r],Cubics->True]

I gather that you are looking to diagonalize your matrix using a similarity transform. One approach is to use the SingularValueDecomposition function in Mathematica:

http://reference.wolfram.com/language/ref/SingularValueDecomposition.html

Analytically, since the eigenvalue problem requires solving a cubic, the expressions in the symbolic case will involve Root expressions... they can be converted to analytical expressions using ToRadicals.

However, assuming your parameters are real here is an equivalent approach. I am changing some of your notation to make things clear.

First define a function to generate your matrix in general:

mat[t_, w_, r_] := {{0, 0, w}, {0, t, -w}, {w, -w, r}}

You can find the symbolic expressions for the eigenvalues and eigenvectors using something like

Eigensystem[mat[t, w, r]]

and you can change the Root objects to algebraic expressions using, for example,

FullSimplify[
 ToRadicals[Eigensystem[mat[t, w, r]]], {t \[Element] Reals, 
  w \[Element] Reals, r \[Element] Reals}]

However, to illustrate the diagonalizatoin process let's play with numerical values. Let's first create a random example of your matrix:

testMatrix = mat[Random[], Random[], Random[]]

Now generate the eigenvalues and eigenvectors for this matrix using

eigens=Eigensystem[testMatrix]

The similarity transform is the matrix of eigenvectors

pMatrix = eigens[[2]]

and you can see this by executing

pMatrix.testMatrix.ConjugateTranspose[pMatrix] // Chop

the Chop is there to remove machine number leftovers.

This is also the same as

pMatrix.testMatrix.Inverse[pMatrix] // Chop

The same set of computations can (more generally) be done using SingularValueDecomposition.

And I see that Ivan Morozov pointed you to JordanDecomposition, which is, of course, a simpler path... ;-)

Hi, Kouhei,

Use JordanDecomposition[] function, here is a simple example:

M = {{-1, 3, -1}, {-3, 5, -1}, {-3, 3, 1}} ;
{P, A} = JordanDecomposition[M] ;
A
P.A.Inverse[P]

I.M.

He is trying to diagonalize a matrix. In his format A would be the original matrix. P would be the matrix with the eigenvectors in it and inverse P is the inverse of the P matrix. I could not even get mathematica to give me the eigenvalues of this matrix, which would also be the values along the diagonal, since P^(?1)A P=(The diagonal eigenvalues). Numbers work but with letters, mathematica gives back answers that does not make sense. If he could just get the eigenvalues, that would be all he would need. If the original matrix does not have an inverse then it cannot be diagonalized.

Thanks, I want to also know how to do this in mathematica. By hand it would take a long time, because we are working with letters and not numbers. I want to learn more about mathematica.