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Why do the $n \times n$ non-singular matrices form an “open” set?

I have a topological group (general linear group)

$G = \{$ invertible $n\times n$ matrices with entries in $\mathbb{R}\}$

and I am asked to show that $G$ is an open subspace of $\mathbb{R}^{n^2}$.

The question is worth very little marks so I assume that there is very little to be shown, but I'm curious if I have the right idea.

In $\mathbb{R}^{n^2}$, I assume the obvious topology is the standard one.

Also, there is a very obvious isomorphism $I$ that identifies an $n\times n$ matrix with an $n^2$ dimensional vector.

The image of $G$ under this isomorphism is all of $G$, which is open in $G$.

EDIT: the above line is of course false, as pointed out below.

So would I need to show that $I$ is a homeomorphism?

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roo
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    You aren't being asked to show that $G$ is open in $G$, but in $\mathbb{R}^{n^2}$. $G$ is not all of $\mathbb{R}^{n^2}$; some matrices are not invertible. – Chris Eagle Jan 17 '12 at 15:41
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    $G$ does not have a topology here apart from the one it gains by sitting inside of $\mathbf R^{n^2}$. – Dylan Moreland Jan 17 '12 at 15:42
  • That makes more sense than the way I was thinking. I probably need to fix what I've asked above to see if I am ready to tackle this. – roo Jan 17 '12 at 15:44
  • @DylanMoreland: What about the operator topology? If I identify an $n\times n$ matrix with a linear functional on $\mathbb{R}^{n}$ I get a norm which gives me a topology. – roo Jan 17 '12 at 15:44
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    @borninthe80s: the same "norm" can be used to define a norm on $\mathbb{R}^{n\times n}$, which means you still need to fix a topology there. But no worries: every norm in a finite dimensional vector space is comparable, and hence define the same topology anyway. – Willie Wong Jan 17 '12 at 15:56
  • Edit: I suppose that would be the same as the one it inherits from $\mathbb{R}^{n^2}$ since it is finite dimensional. – roo Jan 17 '12 at 15:57
  • gah, you beat me to it :) – roo Jan 17 '12 at 15:57

2 Answers2

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You are worrying too much. Clearly, the only topology here is the usual Euclidean topology of $\mathbb R^{n^2}$. Having said that, the result is easy:

The determinant is a continuous function on the space of $n \times n$ matrices because it is a polynomial function of the entries of a matrix. The set of invertible matrices is the inverse image under det of the open set $\mathbb R \setminus 0$, hence is open.

lhf
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  • This makes great sense, assuming I know what the topology is for $G$. I cannot say that $det(\cdot)$ is continuous unless I know what topology the domain space has. My instinct is to give it the operator topology mentioned in my comment above.

    Don't get me wrong, I'm sure you're correct.

     – roo Jan 17 '12 at 15:50
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    @borninthe80s, you're asked to show that $G$ is an open subspace of $\mathbb R^{n^2}$. It is implicit that the topology is the usual one. – lhf Jan 17 '12 at 15:53
  • so I should consider $det$ as a function from $\mathbb{R}^{n^2}$ to itself? and $G$ as a subset of the domain space? – roo Jan 17 '12 at 15:55
  • after thinking about this some more, it seems what you have said above is the only reasonable way to think about this problem – roo Jan 17 '12 at 16:00
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Your question is a special case of the following:

Lemma 1. (Neumann series) Suppose $A$ is a Banach algebra and $x \in A$ has norm less than $1$, then $e-x$ is invertible.

Proof. First note that $\sum_{n=0}^\infty x^n$ converges: $$\sum_{n=0}^\infty \lVert x^n\rVert \leq \sum_{n=0}^\infty \lVert x\rVert^n = \frac{1}{1-\lVert x\rVert}.$$

Using that $e$ and $x$ commute with all powers of $x$ we have for all $m \in \mathbb N$ that $$\left(\sum_{n=0}^m x^n\right)(e-x) = \sum_{n=0}x^n - \sum_{n=0}^m x^{n+1} = e - x^{m+1},$$ so by continuity of multiplication and the fact that $x^m \to 0$ for $m \to \infty$ we have that $(e-x)\sum_{n=0}^\infty x^n = e$, showing that $e-x$ is invertible.

Proposition 2. Suppose $A$ is a Banach algebra, then the set, $\mathcal G_A$, of invertible elements of $A$ is open in $A$.

Proof. Suppose $a \in A$ is invertible. We claim that the open ball of radius $\lVert a^{-1}\rVert^{-1}$ is contained in $\mathcal G_A$: Let $b \in A$ be such that $\lVert b-a\rVert < \lVert a^{-1}\rVert^{-1}$, then $b = a - (a-b) = (e - (a-b)a^{-1})a$. By submultiplicativity of the norm on $A$ we have $\lVert (a-b)a^{-1}\rVert \leq \lVert a-b\rVert \lVert a^{-1}\rVert < 1$. By the previous result $e-(a-b)a^{-1}$ is invertible, so we have written $b$ as the product of two invertible elements, so it too is invertible and the claim follows.

kahen
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