linear algebra - 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 $u_kin X_k$, 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 $uin X$). 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 $uin X$ by definition, it follows that $u$ belongs to the image of $A$, q.e.d.

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