Hadamard Product and Eigendecomposition

Question

I just found this related question in here Q1.

Given a positive definite matrix $\mathbf{A}$, consider its eigendecomposition $(\mathbf{A}\mathbf{V} = \mathbf{V}\mathbf{D})$. Consider an arbitrary matrix $\mathbf{B}$,

Is there any closed-form expression of the eigendecomposition of $\mathbf{A} \circ \mathbf{B}$?

$\circ$ denotes the Hadamard product.

Edition 29/05/2015

Since considering an arbitrary matrix $\mathbf{B}$ does not lead to a closed-form expression, let us consider that all elements of $\mathbf{B}$ can be defined such as $[\mathbf{B}]_{m,n}= e^{i\theta_{m,n}}$ where $\theta_{m,n} \in [0, 2\pi]$ for $m = 1, \ldots , M$ $n = 1, \ldots , N$. Of course, $\mathbf{A}$ has the same dimensions of $\mathbf{B}$.

Thank you in advance

Answer 1

There won't be a nontrivial closed form in general. A trivial "closed-form" is mentioned below.

The Hadamard product $A \circ B$ is a principal submatrix of the Kronecker product $A\otimes B$. Thus, let $A=U\Lambda U^{-1}$ and $B=VDV^{-1}$ (I assumed $B$ is diagonalizable and not fully arbitrary as required in the question). Then, we have \begin{equation*} A \circ B = E^*(U\Lambda U^{-1} \otimes V D V^{-1})E = E^*[(U\otimes V)(\Lambda\otimes D)(U\otimes V)^{-1}]E. \end{equation*} Thus $A\circ B$ is a compression of the matrix on the right hand side. This is about as far as we get.