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How many connected components does the space $\{A \in M_n(\mathbb{R}) \mid A^2 \neq 0\}$ have?

I tried writing out the conditions that arise from $A^2 \neq 0$ explicitly for a general $A \in M_n(\mathbb{R})$ however I do not understand how the surfaces that they generate intersect in the space.

kabenyuk
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Lorenzo Catani
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  • I don't think this question is duplicate, since the condition $A^2\neq0$ does not entail equality $\operatorname{det}(A)\neq0$. Moreover, it seems to me that the set $P={A\in M_n(\mathbb{R})\mid A^2\neq0}$ is linearly connected. It suffices to prove that any matrix $A$ for which $A^2\neq0$ can be connected with matrix $I$ by a continuous path in which there are no matrices whose square is $0$. It is possible to pass from $A$ to any matrix conjugate to it. Here there is no obstacle associated with changing the sign of the determinant. – kabenyuk Jan 21 '23 at 14:41
  • @kabenyuk My bad. I have reopened the question. – José Carlos Santos Jan 21 '23 at 14:55

1 Answers1

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The answer is: for $n = 1$ there are exactly two components and for $n > 1$ this set is connected. I suppose that there are a lot of ways to prove it, and below is the shortest that I can come up with. It relies on only one non-trivial fact, namely that matrices with positive determinant constitute a connected set.

Assume that $n > 1$. Let $S = \{A \in \mathbb R^{n \times n}: A^2 \ne 0\}$, $\mathrm{GL}_+ = \{A \in \mathbb R^{n \times n}: \det A > 0\}$. It is well-known that $\mathrm{GL}_+$ is connected and its closure coincides with the set of all matrices with nonnegative determinant. Therefore, $\mathrm{GL}_+$ is connected and dense in $S_+ = \{A \in S: \det A \ge 0\}$. Thus, $S_+$ is connected. Similarly $S_- = \{A \in S: \det A \le 0\}$ is also connected. The connectedness of $S$ follows from the fact that $S_+$ and $S_-$ have non-empty intersection (here is the only moment were it is important that $n > 1$).

Matsmir
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