# Can the limit of a sequence of matrices of fixed rank have higher rank?

Consider the $n\times n$ matrices given by $A_n=1/n*I_{n\times n}$. For every finite $n$ each $A_n$ has rank $n$ but the limit is the zero matrix, and has rank zero. So the limit of a sequence of rank $n$ matrices need not be of rank $n$. Now I am considering the reverse question.

Consider a sequence $A_k \in \mathbb{R}^{m \times n}$ (or $\mathbb{C}^{m \times n}$). Is it possible to construct a sequence such that each $A_k$ has some rank $d \le \min(m,n)$ but: $\operatorname{rank} \left( \lim \limits _{k \to \infty} A_n \right) >d$?

If not how would one go about proving that matrices cannot jump rank?

• Indeed the rank cannot jump to higher values. To show this for square matrices, one can start with their characteristic polynomials. – Did Apr 18 '18 at 14:21
• The rank is lower semicontinuous on the set of all matrices. It can only jump down. – Hans Engler Apr 18 '18 at 14:31

The set of matrices of rank (at most) $d$ is closed in the set of $m\times n$ matrices.
You can generalize the proof that the set of singular matrices is closed, by considering the set of matrices of rank less than $d$ as the set $f^{-1}({0})$, for $f\colon \mathbb{R}^{m\times n} \to \mathbb{R}$ defined by $$f(A) = \sum_{A'} \lvert\det(A')\rvert$$ where the sum if over all $(d+1)\times (d+1)$ sub-matrices. Then $f$ is continuous, and $f(A)$ is non-zero if and only if $A$ has rank at least $d+1$.