Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$M_n$ 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 $$ \lim_{n\to\infty}(M_n^{-1}) = (\lim_{n\to\infty}M_n)^{-1} ?$$

share|improve this question
4  
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? –  Jonas Meyer Jan 3 '11 at 20:01
1  
@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. –  Jonas Meyer Jan 7 '11 at 3:32
2  
@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)? –  Jonas Meyer Jan 7 '11 at 8:25
4  
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. –  t.b. Feb 25 '11 at 8:45
3  
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. –  t.b. Feb 25 '11 at 9:54
show 7 more comments

2 Answers 2

Let's $ \Omega(\mathbb{R}^{n\times n})=\{ X\in\mathbb{R}^{n\times n}: \mbox{ there is } X^{-1} \} $ and
$$ \Omega(\mathbb{R}^{\mathbb{N}\times \mathbb{N}})=\left\{ X\in\bigcup_{n\in\mathbb{N}}\Omega(\mathbb{R}^{n\times n}): \mbox{ there is } n_0\in\mathbb{N} \mbox{ and there is } X^{-1}\in\Omega(\mathbb{R}^{n_0\times n_0}) \right\} $$ Note that $\Omega(\mathbb{R}^{n\times n})$ and $\Omega(\mathbb{R}^{\mathbb{N}\times \mathbb{N}})$ are vectorial spaces whit usual matrix sum and scalar product. Supose that $M_n\in\Omega(\mathbb{R}^{p_n\times p_n})$ and $c_n\in\Omega(\mathbb{R}^{q_n\times q_n})$ such that $p_n+q_n=n$ for all $n\in\mathbb{N}$ .

Proposition For all positive $k\in\mathbb{N}$ the aplication $$ \Omega(\mathbb{R}^{k\times k})\ni X\mapsto X^{-1}\in\Omega(\mathbb{R}^{k\times k}) $$ is continuous.

Proof. For see it's use block matrix inversion formula $$ \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix}^{-1} = \begin{pmatrix} \mathbf{A}^{-1}+\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & -\mathbf{A}^{-1}\mathbf{B}(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \\ -(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} & (\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B})^{-1} \end{pmatrix} $$ and induction on $k$.

PROPOSITION If $\displaystyle\lim_{n\to\infty}M_n$ converge and $\displaystyle\lim_{n\to\infty}M_n\in\Omega(\mathbb{R}^{\mathbb{N}\times\mathbb{N}})$ for all $p_n+q_n=n$ and $n\uparrow n_0\in\mathbb{N}$ then $$ \lim_{n\to \infty}( M_n)^{-1}= \big(\lim_{n\to \infty} M_n \big)^{-1}. $$

Proof. Similarly, use the block matrix inversion formula and induction on the $p_n+q_n=n$.

share|improve this answer
add comment

This considerations are a bit more than a comment, but not yet a real answer.
I look at such situation, where I also want to discuss the inverse in terms of the LDU- decomposition of $\small M_n$ Then L and U are triangular and D is even diagonal, so if $\small M_m $ is invertible at all, so is D too; L and U are also invertible.

The entries in all that matrices are constant with increasing matrix-size n, and this is also true for their inverses: increase of the matrix-size appends new entries only.

Now, the inverse $\small Q_n = M_n^{-1} $ is also $\small U^{-1} \cdot D^{-1} \cdot L^{-1} $ (let's denote $\small V=U^{-1} \qquad W=L^{-1} $ and their entries with the respective small letters v and w) and for instance the top-left element in $\small Q_n$, call it $\small q_{n:0,0} $ is the sum $\small \displaystyle \sum_{k=0}^n {v_{0,k} \cdot w_{k,0} \over d_{k,k}} $ from the first row in V and the first column in W. Now if n goes to infinity then clearly that sum can diverge even if its summands happen to converge to zero (harmonic series, for instance) and the limit-matrix "does not exist" (or "is not well defined" or "contains singularities") in such cases. So it is obvious, that there must be some more restrictions on your $\small a_n, b_n, c_n $ to make sure that you don't get into such singularities.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.