Show the continuity of $g(A,B) = A^{-1} B$.

Question

Define $g : M_{n \times n} \times M_{n \times n}$ by $g(A, B) = A^{-1} B$. I want to prove that $g$ is continuous for all $(A,B)$ such that $A$ is invertible. My idea is to show the inequality: $$||g(A, B) - g(C, D)|| = ||A^{-1}B - C^{-1}D|| \\= ||A^{-1}B - A^{-1} D + A^{-1} D-C^{-1} D||\le||A^{-1}||||B-D|| + ||D||||A^{-1} - C^{-1}|| < \cdots < \epsilon $$ given that $||(A,B) - (C,D)|| < \delta$. How can I fill out $< \cdots <$ ? or is this approach wrong? If it is wrong, can you give some suggestion?

Answer 1

I see at least two different ways to prove $g$ continuity. In both cases, notice that $g(A,B) = f(h(A),B)$ where $h : A \mapsto A^{-1}$ and $f : (A,B) \mapsto A\cdot B$.

Taking into account the finite dimension

$f$ is continuous as the entries of the product $A \cdot B$ are polynomials from the entries of $A,B$. As polynomials are continuous, $f$ is continuous.

$h$ is continuous as the entries of $A^{-1}$ are rational fractions of $A$ with the denominator equal to $\det A$ not vanishing when $A$ is invertible. As rational fractions are continuous where the denominator is not vanishing, $h$ is continuous when $\det A \neq 0$.

We can then conclude as the composition of continuous functions is a continuous function.

Without taking into account the finite dimension

The space of real matrices is a Banach algebra. And we can then apply known results of Banach algebras. In particular the continuity of $f$ and $h$.