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$M_n$ is a $n\times n$ matrix with $M_{n+1}=\begin{pmatrix}M_n & a_n \\ b_n^T & c_n\end{pmatrix}$ and $a_n, b_n, c_n \to 0$ for $n\to \infty$. Is this sufficient to state $$ \lim_{n\to\infty}(M_n^{-1}) = (\lim_{n\to\infty}M_n)^{-1} ?$$

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    So presumably all of the $M_n$'s are assumed to be invertible, but what do the limits mean, and in what sense is $\lim_{n\to\infty}M_n$ supposed to be invertible?2011-01-03
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    @Jonas Meyer: I mean the limit $M:=lim_{n\to\infty}M_n$ as a linear operator with a discrete spectrum (like used for example in Quantum Mechanics for the harmonic oscillator)2011-01-04
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    @Tobias: Thank you, but unfortunately for me I am not familiar enough with the mathematical models of quantum mechanics to infer the meaning of those limits or in what sense $\lim_{n\to\infty}M_n$ is supposed to be invertible. The limits are unclear to me unless I know on what space the linear operators are acting, and what sense of convergence is used for the particular type of operators you're considering on this space. For starters, are they acting on Hilbert space (\ell^2)? If you wouldn't mind, I would appreciate definitions, or a reference.2011-01-07
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    @Jonas Meyer: sorry, I meant to stay as general as possible, but yes I assume Hilbert spaces, i.e. $l_n^2$ and in the limit $\mathcal L^2$2011-01-07
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    @Tobias: Thank you. I would still like to know what the limits mean, please. I.e., what sort of convergence is involved? What sorts of operators? Closed, densely defined? Is the inverse $(\lim_{n\to\infty}M_n)^{-1}$ then defined on the range of the operator $\lim_{n\to\infty}M_n$ (which is assumed to be 1-to-1)?2011-01-07
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    As @Jonas pointed out, this does not seem to make much sense as asked. Take $a_{n} = b_{n} = 0$. Then in whatever sense *I* can understand the limit $M = \lim_{n \to \infty} M_{n}$, the operator $M$ will be compact and thus very far from invertible.2011-02-25
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    @TheoBuehler: Wouldn't $a_n=b_n=0$ result in a diagonal $M_n$? I'm afraid I don't know about the implications of a compact operator however...2011-02-25
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    Yes, that's precisely the intention. Take $c_{n} = \frac{1}{n}$. The formal inverse of $M$ has $c_{n} = n$. This operator is hopelessly non-continuous and only defined on a rather small subspace of $\ell^{2}$.2011-02-25
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    @TheoBuehler: Oh I see... So $c_n\to 0$ is actually a bad thing here, would requiring $0<|c_n|<\infty$ for most $n$ render this into a more valid question then?2011-02-25
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    Yes, that's the point. Just speaking of the diagonal case: If you want $M$ and the inverse to be defined on all of $\ell^{2}$, you must have $0 < r \leq |c_{n}| \leq R$ (one says $c_{n}$ is bounded and bounded away from zero). The spectrum will then be in the annulus with radii $r$ and $R$ and that of the inverse in the annulus with radii $R^{-1}$ and $r^{-1}$. I don't know anything sensible to say about about the non-diagonal case. Let me just remark that generally matrices are a rather unwieldy tool for representing operators on a Hilbert space.2011-02-25
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    @Theo, @Tobias, if it is possible please formulate the answer from the comments. As far as I understand the question is not answerable in the current form, so an answer indicating the problem would be nice. I do not want my bounty to go to waste :)2011-03-02
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    @mpiktas: I'm going to add an answer with some more info than my comments tomorrow.2011-03-02

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