# What what does it mean for the set of invertible nxn matrices to be open?

1. Mar 4, 2009

### datenshinoai

Real Analysis: What does it mean for the set of invertible nxn matrices to be open?

1. The problem statement, all variables and given/known data

The set of invertible nxn matrices is open in M, is it also dense?

2. The attempt at a solution

Let S = set of invertible nxn matrics, S contained in M

Let A be any invertible element in S. We want to show that there exists an open ball of ivertible elements centered at A. Let 0 < d < det^-1(A)

But I have ho idea what it means for the set of invertible nxn matrices to be open or how to proceed from here. Any help would be great!

Last edited: Mar 5, 2009
2. Mar 5, 2009

### Staff: Mentor

To talk about a set being open in M (a metric space -- you didn't say), there has to be a way to measure distance. How is the distance d between two matrices defined in M? It can't be the determinant, because that's defined only on a single matrix.

Can you give us the full explanation of this problem? The way you have stated it, it doesn't make any sense to me.

3. Mar 5, 2009

### Office_Shredder

Staff Emeritus
The standard way of setting a norm on matrices is by taking the square root of the sum of the square of its entries. Essentially, you treat the set of nxn matrices as the set Rn^2.

4. Mar 5, 2009

### Dick

There's a number of ways of defining a matrix norm. The important property they have is that the determinant function from nxn matrices to R is continuous.