# 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.

• Is each $\delta_{ij}$ of size $n\times n$? And is your $L$ the matrix with elements $\ell_{ij}$ (thus also of size $n\times n$) ? And the $e_i$'s are any vectors? – H. H. Rugh Aug 6 '16 at 13:10
• @H.H.Rugh thanks for your comment. The size of each $\delta_{ij}$ will differ according to the size of vector $e_i$. For example, consider $e_1=[1, 0]^T$ and $e_2=[1,0,1]^T$, then $\delta_{11}$ will be $2\times 2$ whereas $\delta_{12}$ will be $2\times 3$. As for $L$, this matrix contains scalar elements $l_{ij}$. For our example, let's consider $L=\begin{bmatrix} 1&-1\\-1 & 1\end{bmatrix}$. – Has Aug 6 '16 at 18:38
• Ok, makes sense. Just checked some numerics with the e1 and e2 you gave, but different L. You get 2 evals non-zero but the relationship with spec(L) is not (yet) clear. – H. H. Rugh Aug 6 '16 at 20:03
• Thank you professor @H.H.Rugh. Can we consider $$\begin{bmatrix} l_{11}\delta_{11} & \ldots & l_{1n}\delta_{1n} \\ \vdots & \ddots & \vdots \\ l_{n1}\delta_{n1} & \ldots & l_{nn}\delta_{nn} \end{bmatrix}$$ as a Hadamard product? We know that if $\delta_{ij}$ were scalars, that would be possible. But for this matrix, can we say the same? – Has Aug 7 '16 at 5:21
• I don't think so. But it may share some properties with Hadamard products because of the special form for the $\delta$-matrices – H. H. Rugh Aug 7 '16 at 5:25

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).
• Thank you professor @H.H. Rugh . But, can you kindly explain more? I was trying to find a relationship between $L$ and $\Delta$. I would really appreciate it if you could shed some light on this. Thanks again – Has Aug 8 '16 at 19:54
• Not quite sure what to say. But in the example you mentioned in a previous comment we have $e_1^T e_1=1$ and $e_2^T e_2=2$. The matrix $L$ you give then becomes $H=[1 -2; -1 2]$ ($i$th column multiplied by $e_i^Te_i$). The non-zero evals of your augmented matrix are the same as the non-zero evals of $H$ (which you may check numerically as well). Do you want more details in my calculation of $H$? – H. H. Rugh Aug 9 '16 at 8:13
• I am guessing by $\hat L$ you are referring to $$\begin{bmatrix} l_{11}\delta_{11} & \ldots & l_{1n}\delta_{1n} \\ \vdots & \ddots & \vdots \\ l_{n1}\delta_{n1} & \ldots & l_{nn}\delta_{nn} \end{bmatrix}$$ Is that right? As a result, you say that the nonzero eigenvalues of this matrix and $$H = \left( \begin{matrix} \ell_{11} (e_1^T e_1) & \cdots & \ell_{1n} (e_n^T e_n) \\ \vdots & & \vdots \\ \ell_{n1} (e_1^T e_1) & \cdots & \ell_{nn} (e_n^T e_n) \end{matrix} \right)$$ are the same. – Has Aug 9 '16 at 18:22
• I am so sorry, but I can't understand "The augmented matrix$\hat L$ maps $\mathbb{C}^N$into ${\rm im} \Lambda$ and we have$\hat{L} \Lambda c = \Lambda H c$". What is $c$? I am really sorry :( but I couldn't understand the map you have defined. – Has Aug 9 '16 at 18:26