If $f(A)\to A^{-1}$, prove that $f$ is continuous.

Let $f \colon GL_{n}(\mathbb{R})\to GL_{n}(\mathbb{R})$ be a function which maps $A\mapsto A^{-1}$. Prove that $f$ is continuous.

$GL_{n}(\mathbb{R})=\det^{-1}(\mathbb{R}\setminus\{0\})$ is the set of invertible matrix of order $n\times n$ with real coefficients.

• Should I do $A^{-1}=\dfrac{\operatorname{adj}(A)}{\det(A)}$ and use it like a rational polynomial?

• Are there a overkill way to prove this? Any hint?

• Thanks MSE!

I guess we can show that the map from GL$_n\mathbb{R} \to M_n(\mathbb{R})$ given by $A \mapsto A^{-1}$ is continuous with respect to the operator norm $||\cdot ||$. To prove this, fix some $A \in GL_n(\mathbb{R})$. Whenever $||B-A||$ is small enough, $||B^{-1}||$ is bounded away from infinity (see below), and then just notice that $$||A^{-1}-B^{-1}||= ||B^{-1}(B-A)A^{-1}|| \leq ||B^{-1}|| \cdot ||B-A|| \cdot ||A^{-1}|| \leq \alpha \cdot ||B-A|| $$ where $\alpha$ is a constant which depends on the maximum of $||B^{-1}||$ on the prescribed neighborhood of $A$.

If you want explicit bounds on $||B^{-1}||$, one can use the formula $B^{-1} = A^{-1} \sum_0^{\infty} (A-B)^nA^{-n}$ (which is true when $||B-A||< \frac{1}{||A^{-1}||}$) in order to show that $||B^{-1}|| \leq \frac{||A^{-1}||}{1-||B-A||\cdot||A^{-1}||}$, so that $||B^{-1}|| \leq 2||A^{-1}||$ whenever $||B-A|| < \frac{1}{2||A^{-1}||}$.

This proof is overkill. In fact it works in any Banach space, not just finite-dimesional ones.

Put $\mathfrak{g} = \mathfrak{gl}_n(\mathbb{R})$ and $G = GL_n(\mathbb{R})$. The exponential map $\exp:\mathfrak{g} \to G$ is a diffeomorphism near $1\in G$. Since the diagram \begin{align*} \begin{array}% \mathfrak{g} & \stackrel{g\to -g}{\longrightarrow} & \mathfrak{g} \\ \downarrow{\exp} & & \downarrow{\exp} \\% G & \stackrel{g\to g^{-1}}{\longrightarrow} & G% \end{array} \end{align*} commutes, it follows that $g\to g^{-1}$ is smooth on a neighborhood $U$ of $1$. But $g \to (gx)^{-1} = x^{-1}g^{-1}$ is then clearly also smooth on $U$ for any $x\in GL_n(\mathbb{R})$, so the map $g \to g^{-1}$ is smooth on $\bigcup xU = G$.

An easy way to see that inversion is continuous (and in fact smooth) is to use the matrix inversion algorithm by elementary row operations that you probably learned in your first linear algebra course: Gauss-Jordan elimination. Using this algorithm, we obtain $A^{-1}$ from $A$ by matrix multiplication by elementary matrices. Since matrix multiplication is smooth, and the composite of smooth maps is smooth, then in fact the inversion is smooth.