# 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 think that your idea is already an overkill. :) May 6, 2015 at 23:50
• I think the adjoint method you suggest is the way to go. Simple and direct. May 7, 2015 at 0:10

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.

• It's not globally Lipschitz, though: the inverses of $2^{-n} I$ and $2^{-n-1}I$ are far away from each other. The local Lipschitz constant depends on $\| A^{-1} \|$ and $\| B^{-1} \|$ not being too huge.
– Ian
May 7, 2015 at 0:05
• You're right, the constant depends on $A$ and $B$. Let me edit. May 7, 2015 at 0:05
• thank you, this is what i was looking for !
– L F
May 7, 2015 at 3:53
• Since the map is an antiautomorphism, it suffices to show it is continuous at the identity; not that this changes much in the proof. May 7, 2015 at 9:43

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 $\bigcup xU = G$.

• thank you, this is so overkill but i can't use diffeomorphims yet :(
– L F
May 7, 2015 at 3:52
• It isn't really valid to assume a Lie group structure to prove this. A Lie group is by definition a group object in the category of smooth manifolds, so the inversion map is necessarily smooth by definition.
– ಠ_ಠ
May 15, 2015 at 5:22
• I'm not working with an arbitrary Lie group here, but rather $G = GL_n(\mathbb{R})$ specifically. The exponential map is an explicit power series rather than an abstraction, and you can show it's a local diffeomorphism by a direct computation. I don't even use the general group operation--- as opposed to inversion--- anywhere. May 15, 2015 at 5:30

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.