Continuity of matrix inversion

Question

Show that the set $U \subset \mathbb{R}^{n^{2}}$ of matrices $A$ with $\det(A) \neq 0$ is open. Let $A^{-1}$ be the inverse of the matrix $A$. Show that the mapping $A \mapsto A^{-1}$ is continuous from $U$ to $U$.

My solution to the first part is that $\det(A)$ can be expressed as a polynomial in the entries of $A$, and since polynomial functions are continuous we have that the determinant function is continuous from which we can say the given set is indeed open. Any thoughts on the next part?

Answer 1

First: $f=\det$ is continuous, and the set $\operatorname{GL}_n(\mathbb{R})$ of all invertible matrices can be written as $$ \operatorname{GL}_n(\mathbb{R}) = f^{-1}(\mathbb{R}^\ast) $$ Since $\mathbb{R}^\ast := \mathbb{R}-\{0\}$ is open and $f$ is continuous, $f^{-1}(\mathbb{R}^\ast)$ is open.

Now, for showing $g\colon A\in \operatorname{GL}_n(\mathbb{R}) \mapsto A^{-1}$ is continuous, use the fact that $\det$ is continuous, that $A\mapsto\operatorname{adj} A$ is continuous (not hard to show), and that $$ A^{-1} = \frac{1}{\det A}\operatorname{adj} A. $$

Answer 2

You're on the right track. You have been given a way to define a neighborhood. He's told you in the problem to think of $n\times n$ matrices as $\Bbb R^{n^2}$.

HINT: Do you have a concrete formula for the inverse of a nonsingular matrix? Say, something involving cofactors?

Answer 3

For funsies because this popped up on the front page...

Another way to see inversion is continuous (smooth, analytic) is to note that $$ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n $$ holds in any Banach algebra for $\|x\|<1$. In particular, if $A=I-B$ for small $B$, then $$ A^{-1}=\sum_{n=0}^{\infty}B^n. $$ You can translate any matrix to a neighborhood of the identity, invert, and translate it back, i.e. if $A=C(I-B)$, then $$ A^{-1}=C^{-1}\sum_{n=0}^{\infty}B^n. $$ This is clearly continuous (smooth, analytic) for fixed $C$ and varying $B$ in a neighborhood of the identity.