Proof that real symmetric negative matrix is negative definite I have a reasonably simple symmetric $p \times p$ matrix $H$, where the $(j,k)$th element is given by
$$h_{j,k} = -\sum_{i=1}^n \frac{a_i}{b_i^2} x_{ij} x_{ik}$$
and we know that all $a_i \geq 0$ (with at least one strictly positive) and $x_{ij} \geq 0$ (again, with at least one for each $j$ being strictly positive).
So every element of the matrix $H$ is strictly non-positive, and the diagonal is strictly negative.
I am fairly certain that the matrix is negative definite in general, but confirming that it fits one of the different criteria has proven beyond my ability to generalise to $p > 3$.
Is there a very simple way to confirm that a matrix with this structure is (or is not) negative definite?
Thanks
 A: We can assume that all entries $a_i$ are positive, since those terms with $a_i = 0$ do not contribute to the sum.
To determine whether the matrix is negative definite, just compute, for an arbitrary column vector $y = (y_1, \dots, y_p)^T$:
$$
\sum_{j,k = 1}^p y_j y_k h_{j,k} = 
\sum_{i = 1}^n \sum_{j,k = 1}^p y_j y_k (- \frac{a_i}{b_i^2} x_{ij}x_{ik}) =
- \sum_{i = 1}^n  \frac{a_i}{b_i^2} (\sum_{j = 1}^px_{ij}y_j) 
(\sum_{k = 1}^p x_{ik} y_k) = - \sum_{i = 1}^n  \frac{a_i}{b_i^2} |z_i|^2 = - \sum_{i = 1}^p w_i^2 
$$
with $z_i = \sum_{j=1}^n x_{ij}y_j$ and $w_i = \sqrt{a_i}/|b_i| z_i$ for $i = 1, \dots, n$. Clearly this is never positive and it is zero only if the column vector $w = (w_1, \dots, w_n)^T$ is zero. 
Now $w$ is of the form $w = DXy$ where $X$ is the $n \times p$ matrix with entries $x_{ij}$ and $D$ is the diagonal $n \times n$ matrix with entries $\sqrt{a_i}/|b_i|$ along the diagonal. Then if the rank of $DX$ is $p$, we know that $w = 0$ only if $y = 0$. So the matrix $(h_{j,k})$ is negative definite in this case. On the other hand, if the rank of $DX$ is $< p$, there exists $y \ne 0$ such that $w = DXy = 0$, and the matrix $(h_{j,k})$ is negative semidefinite. 
Since $D$ now has full rank $n$, the matrices $DX$ and $X$ have the same rank. The rank of $X$ may be determined as usual. 
