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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?

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    Hint: The determinant function is continuous. – T. Eskin May 13 '12 at 13:57
  • @ThomasE. Ah excellent hint, but what metric do you use on the set of $n \times n$ matrices? As I'm trying to show that $|\det A - \det C| < \epsilon$ right? –  May 13 '12 at 14:09
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    Ah actually on second thoughts it's obviously continuous as it's a polynomial function of the entries of the matrix –  May 13 '12 at 14:18
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    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
  • Well, you'll have to make the set M(n,n) of all nxn matrices a metric space (or a topological space) first, to talk of openness. What I suggest is that, you use the metric: d(A,B)= rank(A-B). It is a metric because rank(A-B)>=0 and rank(A-B)=0 => A=B; rank(A-B)=rank(B-A) and rank(A-C)=rank(A-B+B-C)<=rank(A-B)+rank(B-C); Now, all that you'll have to show is that the set of all non-singular matrices of order n is open, in the metric space (M(n,n),rank). – Somabha Mukherjee May 13 '12 at 14:33
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    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
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    It's important to think of the matrices as having a given topology. It's quite easy to cook up a topology on $\operatorname{Mat}_{n \times n}(\mathbb R)$ that makes this set open, but that's not really the question! – Dylan Moreland May 13 '12 at 14:55
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    Summing up, the set is open and the question is closed. – Georges Elencwajg May 13 '12 at 15:47

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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!

Thomas
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Somabha Mukherjee
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    The metric you propose is discrete so any questions about openness become kind of trivial... – Jason DeVito - on hiatus 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