I just found this related question in here [Q1](https://math.stackexchange.com/questions/1237061/change-of-eigenvectors-by-hadamard-product).


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



Hadamard Product and Eigendecomposition
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.


