Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Like the title says how would you show that the set of matrices such that $\det A \neq 0$ is open?

I can't even see where to start! As I can't envisage how I would find a matrix 'ball' of diameter $\epsilon$ for every element of the set?

share|improve this question
6  
Hint: The determinant function is continuous. –  Thomas E. May 13 '12 at 13:57
1  
Ah actually on second thoughts it's obviously continuous as it's a polynomial function of the entries of the matrix –  user26069 May 13 '12 at 14:18
1  
Another hint: It is easier in this case to use the open-set definition of continuity than the $\epsilon$-$\delta$ definition. –  Neal May 13 '12 at 14:25
2  
You can identify real $n\times n$-matrices with $\mathbb{R}^{n\times n}$, which has natural topology that makes the determinant continuous. –  Michael Greinecker May 13 '12 at 14:53
2  
Summing up, the set is open and the question is closed. –  Georges Elencwajg May 13 '12 at 15:47
show 3 more comments

marked as duplicate by Did, Dylan Moreland, Davide Giraudo, Jason DeVito, Brandon Carter May 13 '12 at 15:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

Make the set $M(n,n)$ of all $n\times n$ matrices a metric space first, by taking $d(A,B)=\text{rank}(A-B)$. This is a metric, since $$\begin{align} \text{rank}(A-B)&\geq 0, \text{rank}(A-B)=0 \Rightarrow\\ &A=B ; \end{align}$$ $$ \begin{align} \text{rank}(A-B) &= \text{rank}(B-A) \text{ and } \text{rank}(A-B) \\ &=\text{rank}((A-C)+(C-B)) \\ &\leq \text{rank}(A-C)+\text{rank}(C-B), \Rightarrow \\ &(M(n,n),\text{rank}) \text{ is a metric space.} \end{align} $$ Now, you'll have to show that the set of all non-singular i.e. full-rank matrices is open. Because, if $A$ is a full-rank matrix, taking $\epsilon=1/2$, we have: $\text{rank}(A-X)<1/2 \Rightarrow \text{rank}(A-X)=0 \Rightarrow A=X$ i.e. $X$ is of full-rank. Thus any point(matrix) of the set of all nonsingular matrices, is an interior point of it. So, unless the metric is specified, the problem is meaningless!

share|improve this answer
2  
The metric you propose is discrete so any questions about openness become kind of trivial... –  Jason DeVito May 13 '12 at 14:53
    
Unless the toplogy or the metric is specified, I can't see how to proceed. Still, there's a hint: If det(A)=0, then there is a d>0, such that det(A+eI) is non-zero for all e in (0,d). This is because you may take d to be the smallest characteristic root of A and remember that a matrix is singular iff 0 is an eigenvalue of it. –  Somabha Mukherjee May 13 '12 at 15:07
    
@SomabhaMukherjee It is standard when a topology on a Euclidean space is not specified to give it the Euclidean topology. In this case, I think the question is clearly asking for a proof in the case that you use the topology on $\mathbb{R}^{n \times n}$ –  Chris Janjigian May 13 '12 at 17:41
add comment