Let $E = \{A \in M_n(\mathbb R): \text{rank}(A) = k\}$. I would like to determine the connected components of $E$. There is a similar question asked here, but the underlying field is $\mathbb C$.

For $A \in E$, let $A = U\Sigma V^T$ be the singular value decomposition. If we don't require $U, V$ to be orthogonal, then \begin{align*} A = M \begin{pmatrix} I_{k \times k} & 0 \\ 0 & 0 \end{pmatrix} N =: M\hat{I}N, \end{align*} where $M,N \in GL_n(\mathbb R)$. But $GL_n(\mathbb R)$ has two connected components defined by the sign of determinants, does this mean there are four connected components in $E$ by enumerating all the possibilities of sign of determinants of $M,N$?

If the dimension $n$ is odd, then I think there should be two connected components. Let $\phi: GL_n(\mathbb R) \times GL_n(\mathbb R) \to M_n(\mathbb R)$ be given by $S \times T \mapsto S \hat{I} T$. We can view $E$ as the continuous image of $\phi$. Since $A = M \hat{I} N = (-M) \hat{I} (-N)$ and $\det(-M) = (-1)^{n}\det(M) = -\det(M)$, then the four cases mentioned above should correspond to two connected images.

  • $\begingroup$ @ChrisCuster: That's a different question. As mentioned above, $GL_n(\mathbb R)$ has two connected components for sure. But I am not sure $E$ has four connected components since it is possible some cases will coincide. $\endgroup$ – user1101010 Jul 26 '18 at 2:07
  • $\begingroup$ Right. The op already knew that. My mistake. I am thinking maybe two... $\endgroup$ – Chris Custer Jul 26 '18 at 2:10
  • $\begingroup$ I think if $k<n$, $E$ is connected. Because you can do Gaussian elimination to get your matrix to be diagonal. If you have at least one zero row, you can make it equal the first row, then make the first row have nonzero entry $1$, then make the "scrap" row zero again. This can all be done continuously, and shouldn't change the rank at any step. Doing it for each of the first $k$ rows gets you to a canonical rank $k$ matrix. $\endgroup$ – Steve D Jul 26 '18 at 6:27

Let us write $E_k=\{A\in \mathcal M_n(\mathbb R) : \textrm{rank}(A)=k \}$ and and $I_k$ the $k\times k$ identity matrix.

  • If $k=0$, $E_k=\{0\}$.
  • If $k=n$, then $E_k=GL_n(\mathbb R)$ and you already know that $GL_n(\mathbb R)$ has two connected components given by the sign of the determinant : $$GL_n(\mathbb R) = \{A\in \mathcal M_n(\mathbb R) : \det(A)>0\}\sqcup \{A\in \mathcal M_n(\mathbb R) : \det(A)<0\}$$
  • If $k<n$, then one can use the usual decomposition, as mentened by the OP : for any matrix $A$ of rank $k$, there exist $P,Q\in GL_n(\mathbb R)$ such that : $$A=PJQ \quad \text{ with } \ J=\begin{pmatrix} I_k & 0 \\ 0 & 0\end{pmatrix}$$ Let's write $\varepsilon=\det(P)$ and $\varepsilon'=\det(Q)$. By the first case, one has two continuous paths $t\in [0;1] \mapsto P_t\in GL_n(\mathbb R)$ and $t\in [0,1] \mapsto Q_t\in GL_n(\mathbb R)$ satisfying : $$\left\{\begin{array}{l} P_0=P \\ P_1=\begin{pmatrix}1 & & & \\ & \ddots & & \\ & & 1\\ & & & \varepsilon \end{pmatrix} \end{array}\right. \quad \text{ and } \quad \left\{\begin{array}{l} Q_0=Q \\ Q_1=\begin{pmatrix}1 & & & \\ & \ddots & & \\ & & 1\\ & & & \varepsilon' \end{pmatrix} \end{array}\right.$$ Now, consider the continuous path $\gamma:t\in [0,1] \mapsto \mathcal M_n(\mathbb R)$ given by : $$\gamma(t)=P_tJQ_t.$$ For any $t\in [0,1]$, $\gamma(t)\in E_k$, $\gamma(0)=A$ and $\gamma(1)=J$: as a consequence, there is only one connected component in $E_k$, it is a connected set.
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    $\begingroup$ Great answer! Might be worthwhile to call out explicitly why this works when $k<n$: you push the determinate to the bottom, where there's a zero in $J$. $\endgroup$ – Steve D Jul 26 '18 at 13:06

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