Does convergence to singular matrix imply divergence of condition numbers? Consider a sequence of invertible matrices $(A_n)$ in $\mathrm{GL}_d(\mathbb{R})$ that converges to a singular matrix $A$ in $\mathbb{R}^{d\times d}$.
$$\tag{1}
\text{Is it necessary that the sequence $(\kappa(A_n))$ of condition numbers diverges to $\infty$?}
$$
(Here, $\kappa(A_n):=\|A_n^{-1}\|\|A_n\|$ for any fixed norm $\|\cdot\|$ on $\mathbb{R}^{d\times d}$.)
As stated, the implication $(1)$ is clearly false: The sequence of diagonal matrices $A_n:=\mathrm{diag}(\alpha_n, \alpha_n)$ with $\alpha_n\rightarrow 0$ is a counterexample.
However: Do you know if $(1)$ holds when we additionally require each $A_n$ to have unit rows?
 A: If $A_n$ is guaranteed to have at least one row with 2-norm at least $1$ and $\|\cdot\|$ is any matrix norm, then the answer is yes.
First, note that if $A_n$ has a row with 2-norm at least $1$ for all $n$, it follows that the Frobenius norm of $A_n$ satisfies $\|A_n\|_F \geq 1$. By the equivalence of norms over finite dimensional vector spaces, it follows that there exists a $c > 0$ such that $\|A_n\| \geq c$ for all $n$. So, it suffices to show that if $A_n \to A$ with $A$ singular, then $\|A_n^{-1}\| \to \infty$.
To see that $\|A_n^{-1}\|\to \infty$, suppose to the contrary that it doesn't. It follows that there exists a subsequence $(A_{n_k})_{k \in \Bbb N}$ and an upper bound $M>0$ such that $\|A_{n_k}^{-1}\| \leq M$ for all $k \in \Bbb N$. Because $(A_{n_k}^{-1})_{k \in \Bbb N}$ is a bounded sequence, it has a convergent subsequence.
Letting $m_j = n_{k_j}$, we can say that $(A_{m_j}^{-1})$ has some limit $B$. It follows that
$$
AB = \left(\lim_{j \to \infty} A_{m_j}\right)\left(\lim_{j \to \infty} A_{m_j}^{-1}\right) = \lim_{j \to \infty} A_{m_j} A_{m_j}^{-1} = I,
$$
which is to say that $B$ is the inverse of $A$, contradicting our premise.
