Two matrices A and B are called simultaneously diagonalizable by congruence if there exists a non-singular matrix U such that
U^{T}AU and U^{T}BU are diagonal matrices. A known sufficient condition for the simultaneous diagonalization of A and B is that A is symmetric positive semi-definite and B is symmetric. Please note that this notion is different from the (probably better known) simultaneous diagonalization in the sense that U^{-1}AU and U^{-1}BU are diagonal, which becomes
U^{T}AU and U^{T}BU are diagonal with U orthogonal when A and B are symmetric. The necessary and sufficient condition for the latter simultaneous diagonalization is that A and B commute. The sufficient condition above for the simultaneous diagonalization by congruence shows that for this simultaneous diagonalization A and B does not need to commute.
I hope my description above is not too confusing and it is clear what is the difference between the two simultaneous diagonalizations. The answer to my question may be well known from some textbook in Linear Algebra. I would be grateful if you could send me a reference, book or paper, where the necessary and sufficient condition is given.
I just found the following on
http://math.stackexchange.com/questions/88022/congruence-and-diagonalizations
If A,B are real symmetric and A is nonsingular, then they are simultaneously congruent to diagonal matrices if and only if C=A{−1}B is diagonalizable by similarity transform (Horn and Johnson, Matrix Analysis, Theorem 4.5.15.).
I don't have the book with me now, I will check tomorrow when I am back to my office.
A matrix is diagonalizable by similarity if and only if it is normal, so the necessary and sufficient condition would be that C=A{−1}B is normal, so this answers my question in most of the cases.
Hence, it should follow that if A is symmetric and invertible and B is symmetric, then A^{-1}B is normal. I can't prove this.
Sorry, I made a mistake. If A matrix is diagonalizable by similarity, then it does not need to be normal. A matrix is diagonalizable by an orthogonal matrix if and only if it is normal, but in the definition of similarity the matrix does not need to be orthogonal. So C=A^{−1}B does not need to be normal. Is there a necessary and sufficient condition for the diagonalization by similarity?
By Horn and Johnson, Matrix Analysis, Theorem 1.3.7 a necessary and sufficient condition for the diagonalization by similarity of an n\times n matrix C is to exist a set of n linearly independent vectors, each of which is an eigenvector of C. Probably there is no result more explicit than this.
Dear Sandor,
"We can diagonalize two symmetric matrices Σ₁ and Σ₂ simultaneously by a linear transformation."
I suggest you to take a look on p.31 of the following book:
Keinosuke Fukunaga, Introduction to Statistical Pattern Recognition, Second Edition, ACADEMIC PRESS, Boston.
Best regards,
Wiwat
Dear Wiwat,
Thank you for the reference. I am not sure what type of diagonalization could be this, because according to
Horn and Johnson, Matrix Analysis, Theorem 4.5.15
there are extra conditions which needs to be put both for the diagonalization by similarity and the diagonalization by congruence.
I am interested in diagonalization by congruence. I suspect one of the matrices Σ₁ and Σ₂ has to be positive semidefinite. This is the sufficient condition given in my first posting.
Best regards,
Sandor
Wiwat,if SIGMA1=Identity,the standard basis works.A different symmetric matrix SIGMA2 will need a different basis.So it cannot be simultaneous diagonalization by a common basis.
Gathering the above results it follows:
Assuming that A is nonsingular, the necessary and sufficient condition for the simultaneous diagonalization of A and B by congruence is that there exist a set of n linearly independent vectors, each of which is an eigenvector of C=A^{−1}B.
There are some interesting method for nationalization at www.oddmaths.info .
To answer your second question, a matrix is diagonalizable if and only if the algebraic multiplicities of its eigenvalues are equal to their respective geometric multiplicities. The algebraic multiplicity is the order with which the eigenvalue appears as a root of the characteristic polynomial, while the geometric multiplicity of the eigenvalue k is the dimension of the nullspace of A-kI. See
https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors
for further discussion
Thanks for all the neighbors upstairs for adding the interesting discussions. I accidentally meet this problem in quantum mechanics. My problem is: what is the necessary and sufficient condition for the commutation of two matrices A and B. If A and B can be simultaneously diagonalized by C, then they commutates. This leads to A and B (order n) should be normal and have n different eigenvalues (exist a set of n linearly independent vectors ) or have common eigenvectors if they are special Hermitte matrices. But this is not all the case for commutation, for example, if B=f(A) or A=g(B), they clearly commute with each other. I do not know the relations between them and more nervous results [1].
Yes, I am getting off the question. So back to the track and I need some new answers, thanks a lot!
[1]. SE DE L M. The Necessary and Sufficient Condition for a Set of Matrices
to Commute and Related Results[J]. Proceedings of the World Congress
on Engineering 2009 Vol II, 2009(2): 1-10.
Dear Lin Zhang,
This seems a very interesting paper. Thank you for recommending it.
Best regards,
Sandor
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