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Consider a linear system $$Ax=b\qquad (*)$$ and a sequence of perturbed linear systems $$(A+\delta A_n)x=b+\delta b_n. \qquad (n)$$ Suppose that all the linear systems are consistent (i.e., have solutions).

My question is: Let $\overline{x}$ be a solution to $(*)$. Suppose that $\delta A_n \rightarrow 0$ and $\delta b_n \rightarrow 0$. Does there exist a sequence of solutions $\overline{x}_n$ such that each $\overline{x}_n$ is a solution to the linear system $(n)$, satisfying $\overline{x}_n \rightarrow \overline{x}$?

A possible helpful result is that for any solution $x_n$ to the linear system $(n)$, there exists a solution $x$ to the linear system $(*)$ such that $$\frac{\|x_n - x\|}{\|x\|}\leq \frac{\|A^\sharp\|\|A\|}{1-\|A^\sharp\|\|\delta A_n\|}\left(\frac{\|\delta b_n\|}{\|b\|}+\frac{\|\delta A_n\|}{\|A\|}\right)\quad \text{if} \ \|A^\sharp\| \| \delta A_n\| <1,$$ where $A^\sharp$ denotes the (weighted) Moore-Penrose inverse or the Drazin inverse. However, the existence of a solution in this result is for the linear system $(*)$ rather than the linear system $(n)$. The above result also holds if we replace $(x, x_n, A)$ with $(x_n, x, A+\delta A_n)$ respectively by considering the $( * )$ as a perturbation of $(n)$. However, if so, $x_n$ and $(A+\delta A_n)^\sharp$ are not guaranteed to be bounded and thus we cannot have a upper bound of $\|x_n-x\|$ tending to $0$. (Correct?)

Anyone can help? Thank you!

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Consider $A= (1, 0 ; 1,0)$ and $b= (1 ; 1)$. Let $x$ be any solution of this, so $x= (1; t)$ for some $t$.

Now, perturbe this to $(1, 0 ; 1 - 1/n,1/n^2)$ and $b$ unchanged, so pertruebed by $0$ (but one could also impose some nontrivial perturbation). The perturbation(s) tends to zero. But the (only) solution is $(1; n)$. Thus $x_n$ does not converge, and one seees that the $x_n$ and the inverse (in this case it is actually invertible, but one could expand the system to avoid this) of the perturbed system are not bounded.

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Very nice!! Felix. –  Felix Goldberg May 18 '12 at 11:42
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