# Eigendecomposition of Hermitian modified by another matrix

Suppose we are interested in a matrix A $$\in \mathcal{C}^{N\times N}$$, which has elements

$$A_{ij} = a_{ij} b_{ij}$$

where the elements $$a_{ij}$$ are real and symmetric. The elements $$b_{ij} = e^{i\phi_{ij}}$$ are complex numbers of unit length. Now, while the eigendecomposition of $$a_{ij}$$ possesses convenient properties (real eigenvalues and orthogonal basis), the eigendecomposition of the full matrix A, modified by the matrix $$b_{ij}$$, is of interest.

Are there any useful statements we can make about the eigenspectrum of A? I realize that a closed-form answer is not likely in the general case; however, any restrictions of interest would be good to know as well.

• you say the matrix $b_{ij}=e^{i\phi_{ij}}$ is unitary, are you sure that is what you mean to say? (you would need orthonormal rows and columns for that...) And the technical term for this "modification" is Kronecker product, that might help a search. – Carlo Beenakker Apr 28 at 19:48
• That was unintended. Thanks for the note. – omellette Apr 28 at 20:02
• The Kronecker product (and effects on the spectrum) might be useful; however, the actual object of interest is the sub-matrix represented by the Hadamard product. There are several papers with general results on the eigenspectrum of a Hadamard product, but they generally don't consider matrices of these two classes. – omellette Apr 28 at 20:03
• ${\rm tr}\,AA^\dagger={\rm tr}\,a^2$ (where $\dagger$ denotes the Hermitian conjugate), so the sum of the absolute value squared of the eigenvalues is unchanged by the modification. – Carlo Beenakker Apr 28 at 20:14
• Thanks - this is useful. In case it further simplifies the problem, the matrix elements $\phi_{ij}$ are also real and symmetric. Thanks again for the comment. – omellette Apr 29 at 1:30