eigenvalues of a generalization of Hadamard product matrix I have a Laplacian matrix ($L$), which is positive semi-definit. Then I have this matrix 
$$\Delta=\begin{bmatrix} \delta_{11} & \ldots & \delta_{1n} \\
\vdots & \ddots & \vdots \\
\delta_{n1} & \ldots & \delta_{nn}
\end{bmatrix}$$
in which
$\delta_{ij}=e_ie_{j}^{T}$, and  $e_i$ is a  vector. 
I need to form the following matrix, and analyze the eigenvalues. 
$$\begin{bmatrix} l_{11}\delta_{11} & \ldots & l_{1n}\delta_{1n} \\
\vdots & \ddots & \vdots \\
l_{n1}\delta_{n1} & \ldots & l_{nn}\delta_{nn}
\end{bmatrix}$$
where $l_{ij}$ is a scalar, but $\delta_{ij}$ is a matrix. This certainly looks like the Hadamard product, but the problem is that the dimensions of $L$ and $\Delta$ don't necessarily match. 
Any ideas about the eigenvalues?
 A: Denote by $\hat{L}$ the augmented matrix of size $N\times N$
  where $N=d_1+\cdots + d_n$.
  Define the map $\Lambda : {\Bbb C}^n \rightarrow {\Bbb C}^N$
  by $\Lambda (c)= \left[\begin{matrix} c_1 e_1 \\
        \vdots\\ c_n e_n \end{matrix} \right]$.
The augmented matrix $\hat{L}$ maps ${\Bbb C}^N$ into ${\rm im} \Lambda$ 
  and we have $\hat{L} \Lambda c = \Lambda H c$ where
  $$  H = \left( \begin{matrix}
      \ell_{11} (e_1^T e_1) & \cdots & \ell_{1n} (e_n^T e_n) \\
      \vdots & & \vdots \\
      \ell_{n1} (e_1^T e_1) & \cdots & \ell_{nn} (e_n^T e_n) 
      \end{matrix}
      \right)
  $$
  is an $n\times n$ Hadamard matrix in the usual sense. It has the same non-zero spectrum
  as $\hat{L}$. Oddly enough it is not symmetric? 
As an example consider 
  $$ e_1=
     \left[\begin{matrix} 1  \\ 0  \end{matrix} \right],
     e_2=
     \left[\begin{matrix} 1  \\ 0 \\ 1 \end{matrix} \right],
     Y_1=
     \left[\begin{matrix} y_{11} \\ y_{12} \end{matrix} \right],
     Y_2=
     \left[\begin{matrix} y_{21} \\ y_{22} \\ y_{23} \end{matrix} \right],
     Y=
     \left[\begin{matrix} Y_1 \\ Y_2 \end{matrix} \right] \in {\Bbb C}^5 .$$
  The last is a column vector of length $5=2+3$.
  In a similar vein we write 
     $$ X=
 \left[\begin{matrix} X_1 \\ X_2 \end{matrix} \right] \in {\Bbb C}^5 .$$
  Acting upon $X$ with our  $5 \times 5$ matrix $\hat{L}$ in question we
  get $Y=\hat{L} X$ where for each of the 'principal' components
    $$ Y_i =
   \sum_{j=1}^2 \ell_{ij} (e_i e_j) X_j =
   e_i \left( \sum_{j=1}^2 \ell_{ij}  (e_j X_j)\right)
   \in {\rm Span} \{e_i\}.$$
   So each $Y_i$ is proportional to $e_i$. In other words, $Y$ is in
   the image of the map 
     $$ \Lambda : c =
 \left[\begin{matrix}  c_1 \\ c_2 \end{matrix} \right]
 \in {\Bbb C}^2 \mapsto 
 \left[\begin{matrix}  e_1 c_1 \\ e_2 c_2 \end{matrix} \right]
 \in {\Bbb C}^5. $$
   For the non-zero eigenvalues it is enough to see how
   $\hat{L}$ acts upon this image.
  So let $X_j= e_j c_j$. Then
  $$ Y_i = 
 (\hat{L} X)_i=
 (\hat{L} \Lambda c)_i=
 e_i \sum_{j=1}^2 \ell_{ij} (e_j e_j) c_j =
 e_i \sum_{j=1}^2 H_{ij} c_j,$$
     where $H$ is the 2 by 2 matrix (in our present example):
  $$  H = 
  \left( \begin{matrix}
  \ell_{11} (e_1^T e_1) &  \ell_{12} (e_2^T e_2) \\
  \ell_{21} (e_1^T e_1) &  \ell_{22} (e_2^T e_2) 
  \end{matrix}
  \right)
  =
  \left( \begin{matrix}
  \ell_{11}  &   2 \ell_{12} \\
  \ell_{21}  &  2 \ell_{22}  
  \end{matrix}
  \right)
  $$
 Whenever  you have matrices verifying
 $\Lambda H c = \hat{L} \Lambda c$ and 
 ${\rm im} \Lambda = {\rm im} \hat{L}$ then $H$ and $\hat{L}$ will have the
 same non-zero eigenvalues (fairly easy to check).
