3
$\begingroup$

Is there a relationship between the eigenvalues of individual matrices and the eigenvalues of their product?

What about the special case when one of these matrices is a diagonal (positive) matrix? I think that this topic is very difficult but, maybe, it could exist some particular case in which the answer to this question is known.

Any pointers will be very helpful.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ $det(AB) = det(A)det(B)$ and $det(M) = \lambda_1 \cdots \lambda_n$ so there is a relationship. If $AB=BA$ then they share e-vectors, with possibly different e-values... there is much to say here. $\endgroup$ – James S. Cook Jul 27 '15 at 20:06
3
$\begingroup$

If two matrices commute and are diagonalizable, then they can be simultaneously diagonalized by a common basis of eigenvectors. In this case, the eigenvalues of the product are the products of the eigenvalues of the two matrices for each common eigenvector. I think that beyond that, indeed this is a very difficult question, even if you assume one matrix is diagonal.

$\endgroup$
  • 1
    $\begingroup$ In addition to the answer given above on commuting matrices, if 0 is an eigenvalue of one of the matrices, it will also be an eigenvalue of the product, regardless of commutation. $\endgroup$ – Paul Jul 27 '15 at 20:12
1
$\begingroup$

If 0 is an eigenvalue of one of the matrices, it will also be an eigenvalue of the product, regardless of commutation. This is simple to see. Suppose A has an eigenvalue of 0. Then $\det A=0$. Since $\det AB=\det A \det B$, $\det AB =0$. So AB also has an eigenvalue equal to zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.