0
$\begingroup$

Possible Duplicate:
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?

  • 0
    gah, you beat me to it :)2012-01-17

2 Answers 2

4

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.

  • 0
    after thinking about this some more, it seems what you have said above is the only reasonable way to think about this problem2012-01-17
1

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.