How to prove that the collection of rank-k matrices forms a closed subset of the space of matrices?

Question

Let $M$ be the space of all $m\times n$ matrices. And $C=\{X\in M|rank(X)\leq k\}$ where $k\leq \min\{m,n\}$. How to proof that C is a closed subset of M?

Answer 1

We can identify $M$ with $\mathbb{R}^{mn}$, by having the matrix $X=\pmatrix{a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn}}$ correspond to the ordered tuple $(a_{11},a_{12},\ldots,a_{mn})$.

Let $r\leq\min\{m,n\}$, and let $M_r$ be the space of all $r\times r$ matrices. The map that sends an $m\times n$ matrix $X\in M$ to the $r\times r$ matrix formed from $X$ by removing all but a certain choice of $r$ rows (the $i_1,\ldots,i_r$'th rows, say) and a certain choice of $r$ columns (the $j_1,\ldots,j_r$'th colums, say), $$X=\pmatrix{a_{11} & \cdots & a_{1n}\\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn}}\mapsto \pmatrix{a_{i_1j_1} & \cdots & a_{j_1j_r}\\ \vdots & \ddots & \vdots \\ a_{i_rj_1} & \cdots & a_{i_rj_r}}$$ corresponds to the map from $\mathbb{R}^{mn}$ to $\mathbb{R}^{r^2}$ that sends $$(a_{11},a_{12},\ldots,a_{mn})\mapsto (a_{i_1j_1},a_{i_1j_2},\ldots,a_{i_rj_r}),$$ which is continuous, for any choice of the $i$'s and $j$'s.

Furthermore, the determinant map from $M_r$ to $\mathbb{R}$ that sends $Y\in M_r$ to $\det(Y)$ is continuous, because it is a polynomial in the entries of $Y$, i.e. it corresponds to a map from $\mathbb{R}^{r^2}$ to $\mathbb{R}$ that is a polynomial in $r^2$ variables.

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Suppose $k<\min\{m,n\}$. Then map $M$ to $\mathbb{R}^d$, where $d={\binom{m}{k+1}\cdot\binom{n}{k+1}}$, by sending a matrix to the ordered tuple of its $(k+1)\times(k+1)$ minors. For example, if $m=n=3$ and $k=1$, then we send $$\pmatrix{1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9}\mapsto\left( |\begin{smallmatrix} 1 & 2\\ 4 & 5\end{smallmatrix}|, |\begin{smallmatrix} 1 & 3\\ 4 & 6\end{smallmatrix}|, |\begin{smallmatrix} 2 & 3\\ 5 & 6\end{smallmatrix}|, |\begin{smallmatrix} 1 & 2\\ 7 & 8\end{smallmatrix}|, |\begin{smallmatrix} 1 & 3\\ 7 & 9\end{smallmatrix}|, |\begin{smallmatrix} 2 & 3\\ 8 & 9\end{smallmatrix}|, |\begin{smallmatrix} 4 & 5\\ 7 & 8\end{smallmatrix}|, |\begin{smallmatrix} 4 & 6\\ 7 & 9\end{smallmatrix}|, |\begin{smallmatrix} 5 & 6\\ 8 & 9\end{smallmatrix}| \right).$$

Because all $d$ of the projection maps from $M$ to $M_r$ are continuous, and because the determinant map from $M_r$ to $\mathbb{R}$ is continuous, this map is continuous.

A matrix is of rank $\leq k$, i.e. $< k+1$, if and only if all of its $(k+1)\times(k+1)$ minors are zero. Thus, $C$ is the preimage of $0\in\mathbb{R}^d$ under a continuous map, and $\{0\}$ is a closed set in $\mathbb{R}^d$ because $\mathbb{R}^d$ is Hausdorff, so therefore $C$ is a closed set in $M$.

If $k=\min\{m,n\}$, then $C=M$, and is a closed set.

Answer 2

Zev Chonoles's answer is totally fine, but since we don't need to care about redundancy, we may simply look at $$f:\mathbb R^{m\times n}\to\mathbb R^{m!n!}\;,\;\;\;A\mapsto\left(\det\left(A_{\pi_1(i)\pi_2(j)}\right)_{1\:\le\:i,\:j\:\le k}\right)_{\substack{\pi_1\:\in\:S_m\\\pi_2\:\in\:S_n}}$$ instead, where $S_n$ denotes the symmetric group. Then, $f$ maps $A$ to the determinants of the submatrix given by the first $k$ columns and rows for every possible permutation of the columns and rows of $A$. And clearly $C=f^{-1}(\{0\})$.

Answer 3

Here is a rather direct approach using QR-decomposition. Let $A_j \in C$ be a sequence converging to some matrix $A$. It remains to show that $rank(A)\le k$.

We use the QR-decomposition of $A_j$, $$ A_j = Q_j R_j, $$ with $Q_j$ orthogonal ($Q^TQ=QQ^T=I$) and $R_j$ upper triangular with $rank(R_j)\le k$. The set of orthogonal matrices is compact, so the sequence $(Q_j)$ has a convergent subsequence (still denoted by $Q_j$). Then $Q_j$ converges to an orthogonal matrix $Q$. This implies that $Q_j^{-1} \to Q^{-1}$, and $R_j = Q_j^{-1}A_j$ converges to some $R$. That $R$ has to be upper triangular. The matrices $R_j$ have rank less than $k$, hence only the first $k$ rows of $R_j$ are non-zero, so that $R$ has at most $k$ non-zero rows. This implies $rank(R)\le k$, and $rank(A) = rank(QR) \le k$.