Sequence of matrices: finding product and inverse

Question

Let $A_n \to A$ and $B_n \to B$ as $n \to \infty $, where $A_n$ and $B_n$ are invertible square matrices of same order. Does there exist any sequence of matrices to converge to $AB$ and $A^{-1}$?

Answer 1

Yes, we have $A_nB_n \to AB$ and $A_n^{-1} \to A^{-1}$.

Let $\|\cdot\|$ be the operator norm. Since $(A_n)_n$ converges, it is in particular bounded so there exists $M > 0$ such that $\|A_n\| \le M, \forall n \in \mathbb{N}$. We have

\begin{align} \|A_nB_n - AB\| &= \|A_nB_n - A_nB + A_nB - AB\| \\ &\le \|A_n(B_n - B)\| + \|(A_n - A)B\| \\ &\le \|A_n\|\|(B_n - B)\| + \|(A_n - A)\|\|B\| \\ &\le M\|(B_n - B)\| + \|(A_n - A)\|\|B\| \\ &\xrightarrow{n\to\infty} 0 \end{align}

so $A_nB_n \to AB$.

Pick $n_0 \in \mathbb{N}$ such that for $n \ge n_0$ we have $\|A_n - A\| \le \frac1{2\|A^{-1}\|}$. For such $n$ we have

$$\|A_n^{-1}\| - \|A^{-1}\| \le \|A_n^{-1} - A^{-1}\| = \|A_n^{-1}(A_n - A)A^{-1}\| \le \|A_n^{-1}\|\|A_n - A\|\|A^{-1}\| \le \frac12\|A_n^{-1}\|$$

so $\|A_n^{-1}\| \le 2\|A^{-1}\|$. Therefore

$$\|A_n^{-1} - A^{-1}\| \le \|A_n^{-1}\|\|A_n - A\|\|A^{-1}\| \le 2\|A^{-1}\|^2\|A_n-A\| \xrightarrow{n\to\infty} 0 $$ so $A_n^{-1} \to A^{-1}$.