# invertible matrices connected or not

The question asks "Is the set of all $3\times3$ real invertible matrices connected?"

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?

I know that $3\times3$ matrices can be viewed as homeomorphic to $\mathbb{R}^9$, but how do we define the topology on this subspace (i.e. set of all $3\times3$ matrices)? What will an open set look like in this topology?

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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 '14 at 21:39
Oh, thanks. Can you describe what will an open set look like in the space of matrices? – ask Aug 15 '14 at 21:40

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? – ask Aug 15 '14 at 21:45
No, he means the multiplicative group of real numbers. So $\mathbb{R}-0$ – Seth Aug 15 '14 at 21:58
Oh, Thanks for your hints! – ask Aug 15 '14 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 '14 at 7:57
@IncnisMrsi: I have learned the notation $K^*$ for the multiplicative group of a field $K$ (this notation is used in some books). I try to change it to $K^\times$, thanks. – Martin Brandenburg Aug 16 '14 at 9:40

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? – ask Aug 16 '14 at 5:07
I added more detail to the answer above. – copper.hat Aug 16 '14 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 '14 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 '14 at 16:29
The explicit inverse is given by Cramer's rule (in the finite-dim case). – Martin Brandenburg Aug 16 '14 at 19:25

$$\det(\mathbb{R}\setminus\{0\})^{-1}=\underbrace{\det(\langle-\infty;0\rangle)^{-1}}_{A}\cup\underbrace{\det(\langle0:\rangle)^{-1}}_{B}$$

Since $\det()$ is continous, then $A,B$ are two open sets, non empty, disjoints in clousure, so $\det(\mathbb{R}\setminus\{0\})^{-1}$ is disconnected.

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With regard to Luis Felipe VillavicencioLopez request for a credible source: a proof can be found in Frank W. Warner, Foundations of Differentiable Manifolds and Lie Groups, Theorem 3.68 (page 131 in my copy).

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