Let $A\_k$ be a sequence of real, rank $r$, $n$ x $m$ matrices such that $A\_k$ converges to a rank $r$ matrix $A$. Let $v\_k, u\_k$ be sequences of vectors such that $u\_k\rightarrow u$ and $A\_k v\_k=u\_k$. I would like to know if it is possible to show that there exist a vector $v$ such that $Av=u$. Appreciate any help.



Sequence of constant rank matrices
I think it is best to settle this problem geometrically, that is if you think of matrices as linear maps from $\mathbb R^m$ to $\mathbb R^n$. The images of these maps are $r$-dimensional linear subspaces of $\mathbb R^n$. Let $X\_k$ denote the image of $A\_k$, then $u\_k\in X\_k$, and you want to prove that the limit vector $u$ belongs to the image of $A$.


Let $X$ denote the set of all possible limits of sequences such that the $k$th element of the sequence belongs to $X\_k$ for every $k$ (for example, $\{u\_k\}$ is one such sequence, hence $u\in X$). Clearly $X$ is a linear subspace of $\mathbb R^n$, and it contains the image of $A$. And the dimension of $X$ is no greater than $r$. Indeed, suppose that $X$ contains $r+1$ linearly independent vectors $w\_1,\dots,w\_{r+1}$. Each $w\_i$ is a limit of a sequence of vectors $u\_k^{(i)}\in X\_k$. For a sufficiently large $k$, the vectors $u\_k^{(1)},\dots,u\_k^{(r+1)}$ are linearly independent because the set of linearly independent $(r+1)$-tuples is open. This contradicts the fact that $\dim X\_k=r$.


Since $X$ is a linear subspace of dimension at most $r$ and it contains the ($r$-dimensional) image of $A$, it must coincide with that image. Since $u\in X$ by definition, it follows that $u$ belongs to the image of $A$, q.e.d.


