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The question asks "Is the set of all 3 by 3 real invertible matrices connected or not?"

My intuitive idea is that we can establish a separation consisting of matrices with positive and negative determinant respectively, whose union will be the whole set but with no intersections. However I am not sure how to show intersection of one set with the closure of the other set is empty, according to the definition of being disconnected. (My guess is that the closure for real matrices with negative determinant is just itself union matrices with 0 determinant)

So what will be a rigorous proof of this, using only tools in point-set topology and knowledge in linear algebra?

And I know that 3 by 3 matrices can be viewed as a subspace of R^9, but how do we define the topology on this subspace (i.e. set of all 3 by 3 matrices)? I mean what will an open set look like in this subspace topology?

Thx!

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The space of real $n \times n$ - matrices is homeomorphic to $\mathbb{R}^{n^2}$. It is not just a subspace. –  Martin Brandenburg Aug 15 at 21:39
    
Oh, thanks. Can you describe what will an open set look like in the space of matrices? –  12345 Aug 15 at 21:40

2 Answers 2

  1. The image of a connected subset under a continuous map is connected.
  2. The determinant is a surjective continuous map $\mathrm{GL}_n(\mathbb{R}) \to \mathbb{R}^*$.
  3. The space $\mathbb{R}^*$ is not connected.

Now conclude. :)

By the way, one can show that $\mathrm{GL}_n(\mathbb{R})^+ = \{A \in \mathrm{GL}_n(\mathbb{R}) : \det(A)>0\}$ is connected, and likewise $\mathrm{GL}_n(\mathbb{R})^-$, so that $\mathrm{GL}_n(\mathbb{R})$ has two connected components. But $\mathrm{GL}_n(\mathbb{C})$ is connected.

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Sorry, I am not familiar with your R∗. Is this the real number union infinity, as described here mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html? –  12345 Aug 15 at 21:45
    
No, he means the multiplicative group of real numbers. So $\mathbb{R}-0$ –  Seth Aug 15 at 21:58
    
Oh, Thanks for your hints! –  12345 Aug 15 at 22:19
    
A cute proof (from some control paper I have forgotten) that $\mathrm{GL}_n(\mathbb{C})$ is connected is to use the path $t \mapsto e^{(1-t) \ln A + t \ln B}$. –  copper.hat Aug 16 at 7:57
    
@Martin Brandenburg: why you don’t use the ${\mathbb R}^\times$ notation for the multiplicative group and write asterisk instead? Asterisks are used indiscriminately everywhere, by everyone, and for a multitude of unrelated senses. –  Incnis Mrsi Aug 16 at 9:38

The set of invertible matrices is open.

To see this, first note that if $\|\cdot\|$ is a sub-multiplicative norm, and $\|X\| <1$, then the series $\sum_{n=0}^\infty \|X^n\|$ is bounded and so the matrix $Y=\sum_{n=0}^\infty (-1)^nX^n$ is well-defined. It is easy to check that $Y(I+X) = I$, and so the matrix $\|I+X\|$ is invertible. To emphasise, if we 'perturb' the identity $I$ by an amount $X$, then $I+X$ is invertible whenever $\|X\|<1$.

Now suppose $A$ is invertible. Then $A+X = A(I+A^{-1} X)$, hence if $\|A^{-1} X\| < 1$, we see that $A+X$ is invertible. In particular, if $\|X\| < {1 \over \|A^{-1}\| }$, then $A+X$ is invertible. Hence the set of invertible matrices is open. (I am implicitly using the fact that all norms on finite-dimensional spaces are equivalent, so the particular norm used doesn't matter.)

If the set of invertible matrices was connected, then since open, this would imply path connected as well.

Hence we could connect $A$ to $-A$ by some path, but since you are dealing with a $3 \times 3$ matrix, we have $\det (-A) = - \det A$, hence at some point the path would pass through zero (by continuity).

(A more detailed analysis, see How many connected components does $\mathrm{GL}_n(\mathbb R)$ have? for example, shows that there are exactly two connected components.)

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Can you explain how you determine the set of invertible matrices is open? –  12345 Aug 16 at 5:07
    
I added more detail to the answer above. –  copper.hat Aug 16 at 5:43
    
$\mathrm{GL}_n = \{\det \neq 0\}$ is clearly open since $\det$ is continuous. Your proof is more involved, but it works also for Banach algebras. –  Martin Brandenburg Aug 16 at 8:54
    
@MartinBrandenburg: You are right, that would have been much simpler. I think having an 'explicit' inverse is useful, but definitely overkill here. –  copper.hat Aug 16 at 16:29
    
The explicit inverse is given by Cramer's rule (in the finite-dim case). –  Martin Brandenburg Aug 16 at 19:25

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