2
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $x \mapsto -{1 \over x^2}$.)

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.