Consider the set of invertible $n \times n$-matrices $GL_n(\mathbb{R}) = \{A \in M_{n \times n}(R) \mid A\text{ is invertible}\}$. I now want to prove that the transformation

$$f: A \mapsto A^{-1}$$

is a homeomorphism from $GL_n(\mathbb{R})$ to itself.

Thanks in advance. I'm not very used to these constructions. It seems that $f$ bijective, simply because of the uniqueness of an inverse matrix, is that so? And if $f$ itself is continuous, then $f^{-1}$ is, because $f = f^{-1}$. But how can I show that f is continuous in the first place?


If $f(A) = f(B)$ then $A^{-1} = B^{-1}$ and and so $I= A B^{-1}$ and so $B = A$. Hence $f$ is injective.

Since $f(f(A)) = A$, we see that $f$ is surjective.

All that remains is continuity.

One quick way to see this is to note that we can write $f(A) = {1 \over \det A} \operatorname{adj} A$. The determinant and each entry in the adjugate are polynomials in the entries of $A$, hence smooth. Since $\det A \neq 0$, we see that $f$ is continuous.

If one is willing to throw a multiplicative norm into the mix, we can see that $f$ is differentiable as well: Suppose $\|H\| < {1 \over \|A^{-1}\|}$, then $(A+H)^{-1} = ((I+HA^{-1})A)^{-1} = A^{-1} (I+HA^{-1})^{-1}$, and since $(I+X)^{-1} = \sum_{k=0}^\infty (-1)^k X^k$ for $\|X\|<1$, we have $f(A+H)-f(A) = -A^{-1} H A^{-1} + \sum_{k=2}^\infty (-1)^k (H A^{-1})^k$. A little work shows that the summation is bounded by $K \|H\|^2$ for some $K$, from which we see that $Df(A)(H) = -A^{-1} H A^{-1}$. (cf. the derivative of $x \mapsto {1 \over x}$ is $x \mapsto -{1 \over x^2}$.)

  • $\begingroup$ Your argument for surjectivity prooves injectivity aswell. $\endgroup$ – Tim B. May 20 '15 at 17:29
  • $\begingroup$ @LeBtz: Good point. $\endgroup$ – copper.hat May 20 '15 at 17:36

The entries of $A^{-1}$ are rational functions, with non-zero denominator, in the entries of $A$.


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