Suppose I have two matrices $A \in \mathbb R^{m \times n}$ and $D \in \mathbb R^{m \times m}$, where $D$ is diagonal, and $m > n$. Is there a simple set of necessary and sufficient conditions for when the matrix $A^T D A$ is positive definite?
A few observations so far:
- A necessary condition is that $\text{rank}(A) = n$. If $\text{rank}(A) < n$, then $A$ has a nontrivial kernel, so there exists a non-zero vector $u \in \mathbb R^m$ such that $u^T A^T D A u = 0$.
- A sufficient condition (assuming $A$ has full rank) is that all of the diagonal entries of $D$ are positive.
One possibility I'm stuck on is Weyl's inequality. Notice that $A^T D A$ can be written as a linear combination of outer products of the columns $A_i$: $$ A^T D A = \sum_{i=1}^n D_{ii} A_i A_i^T = \sum_{i: D_{ii} > 0} D_{ii} A_i A_i^T + \sum_{i : D_{ii} \le 0} D_{ii} A_i A_i^T $$ Let $\lambda_1(M) \ge \lambda_2(M) \ge \cdots \ge \lambda_m(M)$ denote the eigenvalues of a matrix $M \in \mathbb R^{m \times m}$. By Weyl's inequality, for any $i = 1, \dots, n$, we know \begin{equation} \lambda_i\left( \sum_{i: D_{ii} > 0} D_{ii} A_i A_i^T \right) + \lambda_m \left( \sum_{i : D_{ii} \le 0} D_{ii} A_i A_i^T \right) \le \lambda_i(A^T D A) \end{equation} and \begin{equation} \lambda_i \left( \sum_{i : D_{ii} \le 0} D_{ii} A_i A_i^T \right) + \lambda_m\left( \sum_{i: D_{ii} > 0} D_{ii} A_i A_i^T \right) \le \lambda_i(A^T D A) \end{equation} To establish a lower bound on the eigenvalues of $A^T D A$, we would need to choose $i = m$. But the smallest eigenvalue of each $D_{ii} A_i^T A_i$ for $D_{ii} > 0$ is zero, while the smallest eigenvalue of each $D_{ii} A_i^T A_i$ for $D_{ii} \le 0$ is non-positive. Thus it seems these inequalities are useless for establishing $A^T D A$ as positive definite.
Does anyone know of some stronger results?