I am looking at a sequence of matrices $M_n \in \mathbb{R}^{p\times p}$ that satisfy the following properties:
- $M_n = A_n^T B_n A_n$, where $A_n \in \mathbb{R}^{n\times p}$ is full rank and $B_n \in \mathbb{R}^{n\times n}$ is symmetric positive definite for all $n$
- All $p$ diagonal elements of $M_n$ diverge as $n \rightarrow \infty$
Through simulations, it appears that $M_n^{-1}$ approaches the zero $p \times p$ matrix. Can I deduce this from these two properties alone? Property 2 only implies that the largest eigenvalue of $M_n$ diverges, not necessarily the smallest (which is what we need).
Suppose the top $p$ rows of $A_n$ form the identity matrix, the rest being $0$. $B_n$ is any positive definite $n \times n$ matrix. Then $M_n$ is the top left $p \times p$ block of $B_n$, which can be any $p \times p$ positive definite matrix. So you seem to be asserting that if $M_n$ is a sequence of positive definite $n \times n$ matrices whose diagonal elements go to $\infty$ as $n \to \infty$, $M_n^{-1} \to 0$. This is certainly false. Try the case $p=2$ with $$M_n = \pmatrix{n & n-1\cr n-1 & n\cr},\ M_n^{-1} = \pmatrix{\frac{n}{2n-1} & \frac{1-n}{2n-1}\cr \frac{1-n}{2n-1} & \frac{n}{2n-1}\cr} $$