Motivated by the concept of [diagonally dominated matrices](https://en.wikipedia.org/wiki/Diagonally_dominant_matrix#cite_note-3) we consider the space $S$ of all complex $n\times n$ matrices with $|a\_{ii}|>\sum\_{j\neq i} |a\_{ij}|$, for every $i$. [Every element of $S$ is invertible](https://mathoverflow.net/questions/244042/is-this-a-full-rank-matrix)



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> Is $S$ a connected subset of $GL\_{n}(\mathbb{C})$?What is its fundamental group?
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One can consider the same question for real matrices with positive determinant(to avoid disconnecte-ness)



A certain subset of general linear group
We can construct a homotopy retract of this to the set of diagonal matrices with nonzero diagonal entries by scaling nondiagonal entries to 0, and from there to a torus by scaling the diagonal entries to norm 1. Therefore, it is homotopy equivalent to $(S^1)^n$, and is connected.


For real matrices, the retract above still works, and takes you to $\{-1,1\}^n$. The subset with positive determinant is the subset with product 1, and is not connected for $n>2$. I think the connected component you may want is the subset where all eigenvalues have positive real part, but have not checked.


