# How to prove that given set is a connected subset of the space of matrices?

Let $M$ be the space of all $m\times n$ matrices. And $C=\{X\in M|\operatorname{rank}(X)\leq k\}$ where $k\leq \min\{m,n\}$. Check whether the set $C$ is:

1. Closed
2. Connected
3. Compact
4. Open

What are some other good properties of the set $C$,for example is it a manifold?

Clearly the set $C$ is closed if someone is interested a good proof can be found here, hence $C$ is not open. Also as $C$ is unbounded therefore not compact. How to check whether the set $C$ is connected or not?

#### Solutions Collecting From Web of "How to prove that given set is a connected subset of the space of matrices?"

If $M \in C$, then $\lambda M \in C$ for all $\lambda$, hence $M$ and $0$ are path connected. It follows that $C$ is connected.

Pick a non zero $M \in C$, then $\lambda M\in C$ for all $\lambda$, hence $C$ is not bounded, hence not compact.

If $k=\min\{m,n\}$, then see easily that $C$ is open and closed, so suppose $k < \min\{m,n\}$.

Let $\mu_1,…,\mu_p$ be the linear maps that correspond to all the $k+1$ minors of a $m \times n$ matrix, and let
$\phi(M) = \min(\det \mu_1(M),…,\det \mu_p(M))$. Then we see that
$M \in C$ iff $\phi(M) = 0$, and since $\phi$ is continuous, we see that
$M = \phi^{-1}\{0\}$ is closed.

Now suppose $M \notin C$, that is, $M$ has rank $k+1$ or greater. Then
${1 \over k} M \notin C$, and ${1 \over k} M \to 0 \in C$, hence the complement $C^c$ is not closed. It follows that $C$ is not open.