Condition for $\|X_n^{-1}\|$ to be bounded?

Question

Let $\lambda_n = \|X_n^{-1}\|$, where $X_n$ is a non-singular $p\times p$ square matrix and $\|A\| = \sup_{|x| = 1}|Ax|$, with $|\cdot|$ the Euclidian norm. Is there a sufficient condition so that the following holds?

$$\limsup_{n\to\infty} \lambda_n < \infty$$

My intuition is that if the $X_n$ are uniformly nonsingular, i.e.

$$\exists \delta > 0, N; |\det(X_n)| > \delta, \forall n > N$$

then indeed the statement is true.

Answer 1

The uniform nonsingular is not a sufficient condition. See example below.

Consider the diagonal matrix $D_n$ having diagonal coefficients $(1/n,n, 1, \dots, 1)$. Then $D_n^{-1}$ is the diagonal matrix with coefficients $(n, 1/n, 1, \dots, 1)$.

Consequently:

1. $\vert \det D^{-1}_n \vert = 1$ for all $n \in \mathbb N$. In particular $D_n$ is not singular.
2. $D_n^{-1}e_1= ne_1$, where $e_1$ is the first vector of the canonical basis. Hence $\lambda_n \ge n$ and the sequence $(\lambda_n)$ is unbounded.