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The question is as follows: $\DeclareMathOperator{\rank}{rank}$

Is the set $S_r = \{A \in \Bbb R^{n \times n}: \rank(A) \leq r\}$ closed in $\Bbb R^{n \times n}$ in the Euclidean topology?

I have a feeling that this is true, but I got stuck looking for a convincing argument. Certainly this is true for $r = n-1$. Perhaps it can be shown that $S_{r-1}$ is a closed subset of $S_r$?

No neat tricks are coming to mind, but I would think that there must be one. This seems like the kind of thing that has a canonical answer, so links are welcome.

Ben Grossmann
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2 Answers2

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Consider the application $$ f_k(M) = (\det M_{-i_1 \dots -i_k; -j_1 \dots -j_k})_{1\le i_1 <\dots < i_k\le n;1\le j_1 <\dots < j_k\le n} $$ where $ M_{-i_1 \dots -i_k; -j_1 \dots -j_k}$ is the matrix $M$ without the lines $i_a$ and without the columns $j_b$.

Hence $$\text{rank }M \le r\iff M\in \bigcap_{k = r+1}^n f_k^{-1}(\{0\}) $$

Hence this set is closed.

mookid
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Hint: $\operatorname{rank}(A)\le r$ if and only if the determinant of any $k\times k$ submatrix of $A$ vanishes for $k>r$.

EPS
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