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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.

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up vote 5 down vote accepted

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.

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Very nice. – Mariano Suárez-Alvarez Mar 5 '10 at 21:06

It suffices to show that you can replace the original sequence $v_k$ by a new one that is bounded.

You should start by figuring out what can go wrong, i.e, how is it possible for the sequence $v_k$ to be unbounded?

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That's true. (you mean $v_k$) – Shake Baby Mar 5 '10 at 17:56
Yes, thanks. I've fixed it. – Deane Yang Mar 5 '10 at 18:02

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