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Let $M(m,n,r)$ be the matrix space of real matrices $m\times n$ with $rank \leq r$.

Is $M(m,n,r)$ an open set? or closed set? or Does it have some property?

Regards

Juan Pablo Soto Quirós
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    If $r<\min(n,m)$, it's a closed subset with empty interior. If $r\ge\min(n,m)$, it's the whole space. –  Jun 10 '15 at 05:06

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Here's a slick argument for the closedness of $M(m,n,r)$ using singular values. Let $W:[0,1]\to M(m,n,r)$. Then, the eigenvalues of $W^*(x)W(x)$ vary continuously with $x$. (Kato's book has this for the finite dimensional setting. Link here.) Those eigenvalues are the singular values of $W$. We can use these to calculate the rank. So consider a sequence $W_n\to W$. Since singular values are continuous, we have $$ s_i(W_n)\to s_i(W) $$ So if $s_i(W_n)=0$, then $s_i(W)=0$ also. And in a metric space, checking sequences is good enough for checking closure.

Empty interior is easy to check when $r < \min(m,n)$. We can use singular values again to make it easy. Consider a $W$ with rank $r<\min(m,n)$. It has a singular value decomposition $$ W = U\Sigma V $$

Where $\Sigma$ has a $0$ along it's diagonal because $r<\min(m,n)$. Let $T$ be the matrix $U\Sigma' V$ where $\Sigma'$ is $\Sigma$ with the diagonal $0$ replaced with an $\varepsilon$.

Zach Stone
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