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This is a question about convergence of eigenvalues which essentially came up in studying the spectrum of St.-Liouville operators. We want to look at matrices that agree in most of their entries and want to investigate whether this implies convergence of the eigenvalues.

We start with two matrices $$ A_1:=\left[ \begin {array}{cc} 3.5&- 0.5\\ - 0.5& 0.75 \end {array} \right] $$ with eigenvalues $$\lambda_{1,1} := 0.661912511160047 \quad \lambda_{1,2}:=3.58808748883995 $$ and the matrix $$ B_1:=\left[ \begin {array}{ccc} 3.5&- 0.5&-1/4\,\sqrt {2} \\ - 0.5& 0.25&-1/2\,\sqrt {2}\\ - 1/4\,\sqrt {2}&-1/2\,\sqrt {2}&- 0.5\end {array} \right] $$ with eigenvalues $$\mu_{1,0}:=-0.9958877876 \quad \mu_{1,1}:= 0.6554756723 \quad \mu_{1,2}:=3.590412115.$$

We observe that $\lambda_{1,1} \approx \mu_{1,1} $ and $\lambda_{1,2} \approx \mu_{1,2}$.

Now, we extend our matrices to larger dimensions, denoting them as $A_{i},B_{i}$ ,in the following way: So we get $A_i$ from $A_1$ by doing the following:

(i) we use $A_1$ as the $A_i[n-1:n,n-1:n]$ submatrix of $A_i$. The elements down the diagonal are found from bottom to top by successive iterations in steps of two: So $A_i(n-2,n-2) = A_i(n-1,n-1) + 5$,

$A_i(n-3,n-3) = A_i(n-2,n-2) + 7$

$A_i(n-4,n-4) = A_i(n-3,n-3) + 9$ and so on.

(ii)Down the first subdiagonal all entries are $-0.5$ and

(iii)down the subsubdiagonal all entries are $-0.25$.

All other entries are zero!

For the $B_i$ we use the same extension, but use the different basis matrix

$$ B_i[n-2:n,n-2:n]:=\left[ \begin {array}{ccc} 3.5&- 0.5&-1/4\,\sqrt {2} \\ - 0.5& 0.25&-1/2\,\sqrt {2}\\ - 1/4\,\sqrt {2}&-1/2\,\sqrt {2}&- 0.5\end {array} \right] . $$

Notice that due to the fact that we use THE SAME iterative scheme to define $A_i$ and $B_i$ we get very similar matrices $A_i,B_i$. So we get for example

$$A_4:=\left[ \begin {array}{ccccc} 24.5&- 0.5&- 0.25&0&0 \\ - 0.5& 15.5&- 0.5&- 0.25&0\\ - 0.25&- 0.5& 8.5&- 0.5&- 0.25\\ 0&- 0.25&- 0.5& 3.5& - 0.5\\ 0&0&- 0.25&- 0.5& 0.75\end {array} \right] $$

with eigenvalues $$\lambda_{4,5 }:= 24.5307920815531 \quad \lambda_{4,4}:= 15.5136493593423 \quad \lambda_{4,3}:= 8.51760322347614 \quad \lambda_{4,2}:=3.54058988050425 \quad\lambda_{4,1}:=0.647365455124154$$

and

$$ B_4:= \left[ \begin {array}{cccccc} 24.5&- 0.5&- 0.25&0&0&0 \\- 0.5& 15.5&- 0.5&- 0.25&0&0\\ - 0.25&- 0.5& 8.5&- 0.5&- 0.25&0\\ 0&- 0.25&- 0.5& 3.5&- 0.5&-1/4\,\sqrt {2}\\ 0&0&- 0.25&- 0.5& 0.25& -1/2\,\sqrt {2}\\ 0&0&0&-1/4\,\sqrt {2}&-1/2\,\sqrt {2}&- 0.5\end {array} \right] $$

This matrix has the eigenvalues $$\mu_{4,1} = 0.6473654185 \quad \mu_{4,2} =3.540589910 \quad \mu_{4,3} =8.517603211 \quad \mu_{4,4} =15.51364936 \quad \mu_{4,5} =24.53079208,\mu_{4,0}=-0.9999999836$$

Obviously, the eigenvalues of $A_4$ and $B_4$ are extremely close together. Though, $B_4$ has an additional eigenvalue $\mu_{4,0}$ without a partner in the spectrum of $A_4$.


So what I want to do is the following:

By the iterative definition of these matrices we get sequences $(A_i)_i$ and $(B_i)_i$ with eigenvalue sequences $(\lambda_{i,1})_{i \ge 1}$,$(\lambda_{i,2})_{i \ge 1}$,$(\lambda_{i,3})_{i \ge 2}$,$(\lambda_{i,4})_{i \ge 3}$ and so on and eigenvalues $(\mu_{i,0})_{i \ge 1}$,$(\mu_{i,1})_{i \ge 1}$,$(\mu_{i,2})_{i \ge 1}$,$(\mu_{i,3})_{i \ge 2}$ and $(\mu_{i,4})_{i \ge 3}$.. .

I want to show that $\mu_{i,0} \rightarrow -1$ and all the other eigenvalues converge to their partner value, so $\lambda_{i,k} \rightarrow c_k \in \mathbb{R}$(for i approaching infinity, hence going over to larger extended matrices) and accordingly $\mu_{i,k} \rightarrow c_k.$

Numerical simulations actually suggest that this happens ( I calculated up to $A_{150}$ and $B_{150}$ where I reached a pretty good convergence to the values already strongly suggested by $A_4$ and $B_4$, but I am not able to show it.) I have never seen something similar before.

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This is only a partial answer, but if you instead keep the diagonal constant, what you have is (essentially) a sequence of Toeplitz matrices, determinants of such matrices satisfy linear recursions. I gave a combinatorial proof of this statement here, but this has been observed earlier in more generality, see this paper.

In your generalization, I still think determinants (and also characteristic polynomials) satisfy linear recursions (now with constant polynomial coefficients), and once you have a linear recursion, it is "easy" to convince yourself that you also get some type of convergence, see Beraha, Kahane, Weiss, Limits of zeroes of recursively defined polynomials.

This is most likely the explanation of what you see here.

EDIT: Now that I think about it a bit more, in [1], it is shown than one can essentially (under some restrictions) skip any finite set of initial/terminal rows and columns in the matrices, and get the same asymptotics of eigenvalues. Therefore, "the limit set of eigenvalues" (defined in the paper), for the two sequences of matrices are the same. That means, if $\lambda^*$ is a limit point of eigenvalues of matrix series $A_n$, then it is also a limit point of some subsequence of eigenvalues of $B_n$, and vice versa. This is over the complex numbers, since I do not require that the matrices are symmetric, and hence, there is no natural ordering of the eigenvalues.

What you observe seem to be a refinement of this statement in the case the matrices are symmetric (since you can observe which subsequences that are "close" as there is a natural indexing of the eigenvalues).

Edit 2: The length of the recurrence is in your case should be $\binom{5}{3} = 10$, (the number of diagonals total, over the number of diagonals on and below the main diagonal). This is a generalization/analogue of the "well-known" three-term recurrence for orthogonal polynomials.

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    $\begingroup$ Yes, I believe the reference sciencedirect.com/science/article/pii/S0196885804000119 I gave treats this case. $\endgroup$ – Per Alexandersson Aug 5 '14 at 15:43
  • $\begingroup$ The recurrence should be of length 10 in your case (If I am not mistaken) and it does not start immediately, but the start degree of the polynomials must be sufficiently large. $\endgroup$ – Per Alexandersson Aug 5 '14 at 15:50
  • $\begingroup$ On the "edges", you really can't fit the formula, but inside on the main diagonal, it seems that you can choose $\nu=1, \lambda=\mu = -3/-0.25$, and off-diagonal entries give $\nu=1, \lambda=\mu = 0$. As mentioned, there is good reason to believe that the first/last column/row do not affect the limit behavior. $\endgroup$ – Per Alexandersson Aug 5 '14 at 16:17
  • $\begingroup$ Oh, haha. you are right, I was a bit quick; and seems like the determinants do not satisfy linear equations, (although maybe one can argue that they are close to some sequence that satisfy a linear equation). $\endgroup$ – Per Alexandersson Aug 5 '14 at 16:42

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