18
$\begingroup$

I need to prove that rank($\mathrm{A}$) is not continuous everywhere but is lower semi-continuous everywhere, where $\mathrm{A}\in \mathbb{C}^{n\times m} $

$\endgroup$
0

2 Answers 2

17
$\begingroup$

We have to show the set $S = \{A \in \mathbb{C}^{m \times n} : \texttt{rank}(A) > r \} \tag{1}$ is open for all real $r.$

We need the following fact : given a matrix $A_0 = (a_{ij}^{0})$ of rank $r_0$ there is a $\delta > 0$ such that for any matrix $A=(a_{ij})$ with $\|A - A_0\|_{\texttt{max}} < \delta$ (i,e., $ |a_{ij} - a^0_{ij}| < \delta\ \forall i,j$) we have $\texttt{rank}(A) \geq r_0.$

The proof of this is straightforward, if $r_0 = 0$ the result is immediate, if $r_0 > 0$ there is a $r_0 \times r_0$ non-singular submatrix in $A_0$, call this submatrix $A^{r_0}_0$. We have $\texttt{det}(A_0^{r^0}) \neq 0.$ By continuity of determinant there is a $\delta > 0$ such that for any $r_0 \times r_0$ matrix $B$ with $\|A_0^{r^0} - B\|_{\texttt{max}} < \delta$ we have $\texttt{det}(B) \neq 0.$ So if $A$ is any matrix with $\|A - A_0\|_{\texttt{max}} < \delta$ and $A^{r^0}$ is the submatrix of $A$ with the same indices that determine $A_0^{r^0}$ then $\|A^{r^0} - A_0^{r^0}\|_{\texttt{max}} \leq \|A-A_0\|_{\texttt{max}} < \delta$ which implies $\texttt{det}(A^{r^0}) \neq 0$ so $\texttt{rank}(A) \geq r_0.$

Now we can easily prove $(1)$. If $S$ is non-empty choose $A_0 \in S$ we have $\texttt{rank}(A_0) > r$. We can find $\delta > 0$ such that for any $A$ with $\|A-A_0\|_{\texttt{max}} < \delta$ we have $\texttt{rank}(A) \geq \texttt{rank}(A_0) > r$, so $A \in S$ and $S$ is open.

It is easy to see rank is not a continuous function : each element in the sequence of matrices $\begin{pmatrix} \dfrac{1}{n} & 0 \\ 0 & \dfrac{1}{n} \end{pmatrix}$ has rank 2, but the limit has rank 0.

$\endgroup$
1
  • $\begingroup$ So, in particular, it means that for any valid $k$, the subset $M_k\subset \mathbb{R}^{n\times n}$ of matrices with rank $> k$ is an embedded submanifold of dimension $n\times n$, right? Just checking, because it sounds weird to me $\endgroup$
    – rmdmc89
    Oct 5, 2017 at 14:24
0
$\begingroup$

Let $A\in{\mathbb C}^{n×m}$; we use the symbol $m$ to denote a square minor (a selection of $k$ rows and $k$ columns). For any minor $m$ of the matrix define a function $f_m:{\mathbb C}^{n×m}\to {\mathbb R}$ in this way: if the minor is invertible then $f_m(A) = k$ otherwise $f_m(A) = 0$. Obviously $f_m$ is lower semicontinuous. The rank is the maximum of all $f_m$, for all possible choices of $m$ (including all possible choices of $k$), so it is lower semicontinuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.