Prove $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ defined by $f(x):=x^{-1}$ is continuous

Question

I have to prove that $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ defined by $f(x):=x^{-1}$ is continuous. What I have so far:

Consider $Inc \circ f: GL(n,\mathbb{R})\rightarrow \mathbb{R}^{n^2}$, where $Inc$ is the inclusion map from $GL(n,\mathbb{R})$ into $\mathbb{R}^{n^2}$. Throughout we take the standard topology on $\mathbb{R}^{n^2}$ and the relative topology on $GL(n,\mathbb{R})$. I've managed to prove that $Inc$ is continuous, $Inc \circ f$ continuous $\iff f$ continuous and that $GL(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$.

How can I use this information to prove that $f$ is continuous?

Answer 1

Use $$ A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1} $$ and the sub-multiplicativity of (most of) the standard norms.

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Since that goes too fast, use the Neumann series as in $$ B^{-1}=A^{-1}(I-X)^{-1}=A^{-1}(I+X+X^2+X^3+…) $$ leading to $$ \|B^{-1}-A^{-1}\|\le\frac{\|X\|}{1-\|X\|} $$ where $X=I-BA^{-1}$ and thus $\|X\|\le\|A^{-1}\|·\|A-B\|$. Which is valid as long as $\|B-A\|\le\|A^{-1}\|^{-1}$.

Answer 2

Hint: Use Cayley-Hamilton theorem to show that $A^{-1}$ is a polynomial of $A$.