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Suppose we take the set of all $n\times n$ real matrices with entries in $[0,1]$ in Euclidean space. Let $N_\epsilon$ be the $\epsilon$ neighborhood of the set of all singular matrices in this space, i.e. $N_\epsilon = \cup_{A:\det(A)=0} B_\epsilon(A)$.

Is there a simple way to compute or upper-bound the volume of $N_\epsilon$ in terms of $n$ and $\epsilon$?

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    $\begingroup$ What is the norm with respect to which you take $B_{\varepsilon}$? $\endgroup$
    – erz
    Sep 25, 2018 at 4:18
  • $\begingroup$ Let's say the trace inner product, but really I'm looking more for how one might answer this type of question. $\endgroup$
    – zumbaya
    Sep 25, 2018 at 4:59
  • $\begingroup$ Isn't the volume simply infinite? There are infinitely many singular matrices of pairwise distance $\epsilon$, each one having a neighborhood with a volume of an $(n\times n)$-dimensional ball of radius $\epsilon$ (I assume this is how volume is defined in your case). $\endgroup$
    – M. Winter
    Sep 25, 2018 at 8:19
  • $\begingroup$ @M.Winter: The OP is restricting to matrices with entries in $[0,1]$. $\endgroup$
    – Mark Meckes
    Sep 25, 2018 at 11:06
  • $\begingroup$ The set of singular matrices defines an hypersurface, the volume goes to zero linearly with epsilon $\endgroup$
    – Martin de Borbon
    Sep 25, 2018 at 17:27

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