I am trying to create an element of a matrix that is a subset of a larger matrix. However, I am told that my subscripts do not match. I wanted other people's opinion as to what I am doing wrong and how to produce mathematical notation that does what I am intending. In particular, they are saying you can't use l and m as subscripts since I am using i, and j for my element. See below.
$c_{ij}$ is given by: \begin{equation} c_{ij} = \left \{ \begin{array}{ll} 1 & \mbox{if a contact relationship (link) exists between individual i and another individual j } \\ 0, & \mbox{ otherwise }\\ \end{array} \right. . \end{equation}
The elements of the similarity matrix $\mathbf{C}$ are divided up into their sub-similarity matrices based on their having the same indicator variable level.
\begin{equation} M_{ij} = \left \{ \begin{array}{ll} 1 & \mbox{if } k_i = k_j = 0 \\ -1 & \mbox{if } k_i = k_j = 1 \\ 0, \mbox{ otherwise} \\ \end{array} \right. . \end{equation}
Thus, $c_{lm}^{(0)}$ is given by:
\begin{equation} c_{lm}^{(0)} = \left \{ \begin{array}{ll} 1 & \mbox{if } c_{ij}M_{ij} = 1 \\ 0 & \mbox{if } M_{ij} = 1 \mbox{ and } c_{ij}M_{ij} \neq 1 \\ \end{array} \right. , \end{equation}
and $\mathbf{C}^{(0)} = [c_{lm}^{(0)}]_{\delta \times \delta}$, where $\delta = \mid \mathfrak{D}^0 \mid$
What I am thinking might work instead is: Let $\mathfrak{D}^{(0)}$ be the set of individuals that have covariate level zero ($k_i = 0$). Let $\mathfrak{D}^{(1)}$, be the set of individuals that have covariate level one ($k_i = 1$), where $\mathfrak{D}^{(1)} = \mathfrak{D} \setminus \mathfrak{D}^{(0)}$.
\begin{equation} Q^{(0)}_{ij} = \left \{ \begin{array}{ll} 1 & \mbox{if } c_{ij}M_{ij} = 1 \\ 0 & \mbox{if } M_{ij} = 1 \mbox{ and } c_{ij}M_{ij} \neq 1 \\ -1 & \mbox{otherwise} \\ \end{array} \right. , \end{equation}
ok, here is a modification
\begin{equation} q^{(0)}_{ij} = \left \{ \begin{array}{ll} c_{ij} & \mbox{ if } k_i = 0 \\ -1 & \mbox{ if } k_i = 1 \\ \end{array} \right. , \end{equation}
We obtain matrices $Q^{(0)} = \left[q^{(0)}_{ij}\right]_{N \times N}$, and $Q^{(1)} = \left[q^{(1)}_{ij}\right]_{N \times N} $.
Let \begin{equation} C^{(0)} = \left[c_{ij}^{(0)}\right]_{\delta \times \delta} = Q^{(0)} \left(\mathfrak{D}^{(1)} \right), \end{equation} where $Q^{(0)} \left(\mathfrak{D}^{(1)} \right)$ is the matrix with rows and columns from the set $\mathfrak{D}^{(1)}$ removed and $\delta = \mid \mathfrak{D}^0 \mid$.