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Is the set of all $n\times n$ matrices with determinant $1$ an open ,dense connected subset of $M(n,\mathbb R) $ i.e set of all matrices over $\mathbb R$?

I know it will be a closed subset of $M(n,\mathbb R)$,not bounded and hence not compact.But not sure about the others

Learnmore
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2 Answers2

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Suppose $A$ has determinant one, then $p(s) = \det (A-sI)$ is a non-constant polynomial, hence we can find arbitrarily small $s$ such that $p(s) \neq 1$. Hence the space is not open.

The function $A \mapsto \det A$ is continuous, hence, for example, $2I$ has a neighbourhood in which $\det A \neq 1$.

One way of showing that the space is connected (in fact, path connected) is to show a path between $I$ and the matrix.

copper.hat
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  • density is clear,but I cant get the idea for open ness – Learnmore Nov 12 '14 at 07:52
  • If it was open, then you could find a small ball around each element such that the determinant was one. The first sentence above shows that this is not possible. – copper.hat Nov 12 '14 at 07:54
  • thanks @copper for explaining – Learnmore Nov 12 '14 at 07:57
  • You are welcome. I thought I had a clean way of showing connectedness, but it became very messy (I forgot I had to deal with complex eigenvalues). Basically you can use the (real) Jordan form and some messy paths to show that there is a path from any matrix with determinant one and $I$. – copper.hat Nov 12 '14 at 08:04
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If it were open and closed, then $M(n,\mathbb R)$ would not be connected – but it is.

Watson
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Hagen von Eitzen
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