Is the set of invertible nxn matrices open and dense?

[datenshinoai]

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

Homework Statement

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!

 

[Mark44]

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.

 

[Office_Shredder]

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 R^{n^2}.

 

[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.