Alternative proof for continuity of matrix inversion

Question

I am to show that

$\mathrm{inv}: \mathrm{GL}_{n \times n}(\mathbb{R}) \to \mathrm{GL}_{n \times n}(\mathbb{R}); A \mapsto A^{-1}$

is a continuous function. I have shown this by showing that the deteriminant is continuous, and that forming the adjunct is continuous, in which case one can apply the formula

$A^{-1} = \frac{1}{\mathrm{det}(A)}\mathrm{adj}(A)$.

This is all fine, but I would love to see a more intuitive proof that doesn’t simply rely on showing that some formula is composed of continuous functions. I tried finding some sort of relation between the operator norms of $A$ and $A^{-1}$ but couldn’t find anything that holds in the general case. For example, if something along the lines of

$\min_{x \in \mathbb{S}^{n-1}}{(Ax)} \times \max_{x \in \mathbb{S}^{n-1}}{(A^{-1}x)} = 1$

were true, one could construct an epsilon-delta proof via that. However, the above expression while inutive, appears to be false (it seems to hold only if the chosen x happen to be eigenvectors unless my attempts to verify it in mathematica were somehow wrong). Does anyone know of a more intuitive proof of the statement than the one I outlined above?

Answer 1

Let $A_0\in GL({\mathbb R}^n)$ be given, and put $${1\over\|A_0^{-1}\|}=:\alpha>0\ .$$ It follows that for all $x$ we have $$|x|=|A_0^{-1}A_0 x|\leq{1\over\alpha}|A_0 x|$$ and therefore $|A_0x|\geq\alpha|x|$. Assume that $\|A-A_0\|<{\alpha\over2}$. Then $$|Ax|\geq|A_0x|-|(A-A_0)x|\geq{\alpha\over 2}|x|\ ,$$ hence such an $A$ is again regular and has an inverse $A^{-1}$ of norm $\leq{2\over\alpha}$. From $$A^{-1}-A_0^{-1}=A^{-1}(A_0-A)A_0^{-1}$$ we now deduce $$\|A^{-1}-A_0^{-1}\|\leq\|A^{-1}\|\>\|A_0-A\|\>\|A_0^{-1}\|\leq{2\over\alpha^2}\|A-A_0\|\ .$$ This shows that ${\rm inv}$ is even Lipschitz continuous at $A_0$.

Answer 2

Hint:

For a small matrix $\delta$ (such that $\|A^{-1}\delta\|<1$), $$(A+\delta)^{-1}=(A(I+A^{-1}\delta))^{-1}=(I+A^{-1}\delta)^{-1}A^{-1}=(I-A^{-1}\delta+(A^{-1}\delta)^2-\cdots)A^{-1}$$ (the series can be made to converge) so that

$$\|(A+\delta)^{-1}-A^{-1}\|\le\|A^{-1}\|^2\,\|I-A^{-1}\delta+\cdots\|\,\|\delta\|$$

is small as well.

Answer 3

In finite dimension, also works the implicit function theorem. The matrix product $(X,Y)\longmapsto XY$ is bilinear and continuous, so $C^1$. The inverse is given by the equation $XA = I$ Check that $$\frac{\partial(XA - I)}{\partial A}$$ is inversible and this proves that $A\longmapsto X = A^{-1}$ is $C^1$.