Dear Luca,

you were right with finding the statement suspicious. Simultaneously diagonalizable matrizes would indeed commute, and it is easy to see that this is not true in general, even if one of the matrizes is assumed to be positive definite. (For example take a diagonal 2x2 Matrix with entries 1 and 2 and the 2x2 matrix with all four entries equal to 1.) So two symmetric matrizes cannot be diagonalized simultaneously in general.

The book by Fukunaga you mention indeed does something different. Namely, given a positive definite matrix X and a symmetric matrix Y, the author finds a (non-orthogonal) invertible matrix A such that AtXA and AtYA are both diagonal (so he uses the transpose rather than the inverse) and indeed AtXA is the identity matrix. So this does not actually concern diagonalization of symmetric matrizes but of bilinear forms. The result can be expressed invariantly as the fact that given two symmetric bilinear forms, one of which is positive definite, there is a basis which is orthogonal for both forms.  This is not a deep result, it can be deduced from standard linear algebra results as follows: There is an orthonormal basis for the positive definite form. The other form is represented with respect to this basis by a symmetric matrix and then the usual oprthogonal diagonalization of symmetric matrices gives you an orthonormal basis for the first form with respect to which the second form is diagonal. (The book uses this argument converted to matrix form.)

Thank you Andreas for your answer, it shed a bit of light on me.

I had already found a simple counterexample to the statement, that's why I took the courage to post here.

Please bear with me as for the technical language, linear algebra is not my field of study.

Does the difference stand all in the fact that a transpose instead of the inverse is used? I mean, if the usual inverse was used (i.e., the usual form of a similarity transformation) then the X and Y would be simultaneously diagonalised, right?

But in the end, this is not a similarity transformation, so this is not diagonalisation.

Now I have to check if in the article that is the starting point of my research, simultaneous diagonalisation is really needed (and there would be a flaw in the work) or just a common basis, orthogonal and diagonalising respectivley the two forms (and then it would simply me a misnomer error).

Dear Luca,

the difference indeed is the transpose vs. the inverse. The point is that you view invertible matrices as describing changes of basis in a vector space. Now if X is the matrix describing a linear map, then the "right" effect of a base change is mapping X to AXA-1 . If X is the (symmetric) matrix consisting of the values of a symmetric bilinear form on the elements of a basis, then the "right" effect of a base change is mapping X to AXAt. The effects are differnt, as you can see from the positive definite case. There you can choose A in such a way that AXAt is the identity matrix, reflecting the fact that any inner product admits an orthonormal basis. In contrast if you diagonalize X in the form AXA-1 you will see the eigenvalues of X (which are positive but may be different from one) on the main diagonal.

Two symmetric matrices can be simultaneously diagonalized but the resulting diagonal matrices are not given by the eigenvalues of those matrices as in ordinary diagonalization. For this reason it is not necessary that the matrices commute, only that one of them es positive definite