I have a symmetrix matrix, $M = M^T$ which has negative values everywhere i.e., $M_{i,j}<0$ for all $i, j$. Furthermore, I know that we have $M_{i, i} \leq M_{i, j} \forall i,j$ and additionally that $M_{i, i} \leq M_{j, i} \forall i,j$ i.e., that the largest (most negative) values in each row and column are on the diagonal.
What can I say about the definiteness of $M$?
EDIT: the conditions I have above weren't quite what I had in mind. Let me rephrase, in the situation I mean.
I have $I$ features in a database, which counts how often each feature is active. The matrix $M$ has values:
$$ M_{i,j} = \begin{cases} -N_i,\quad i=j\\ -N_{i,j}\quad otherwise\end{cases} $$ where $N_i$ is the number of days that $i$ is active and $N_{i,j}$ is the number of days that both $i$ and $j$ are active. I want to know whether this matrix is negative semi-definite.
This matrix satisfies the conditions above, but they aren't strict enough. For example,
$$ M = -\begin{bmatrix} 10 & 10 & 5 \\\ 10 & 10 & 10 \\\ 5 & 10 & 10\end{bmatrix} $$ satisfies the conditions I mentioned above. However, it could not have been generated from data. The second row implies that every time the second feature is active, the first and third features are also active. The first row implies that the third feature is not always active if the first feature is second, which is a contradiction.
I'm not entirely sure how to encode this constraint, but I feel as though this matrix should come out to be negative semi-definite.
Let $D$ be the set of all days in the record and $A=(a_{id})$ be the matrix such that $a_{id}=1$ if feature $i$ is active on day $d$ or $a_{id}=0$ otherwise. Then $a_{id}a_{jd}$ is equal to $1$ if both features $i$ and $j$ are active on day $d$, or $0$ otherwise. It follows that $m_{ij}=-\sum_{d\in D}a_{id}a_{jd}$, i.e. $M=-AA^\top$. Hence $M$ is negative semidefinite.