In high energy physics we oftentimes encounter the following problem. For a given parametrized matrix $\{H_{ij}(\Lambda)\}$, we know that in the limit $\Lambda\to\infty$ some of its entries become large:

$$ \text{For some $(i,j)$: }H_{ij}(\Lambda)\to\infty\text{ as }\Lambda\to\infty. $$

We also know that in the limit $\Lambda\to\infty$ the lowest eigenvalues in the spectrum of $\{H_{ij}(\Lambda)\}$ — which we are most interested in — tend to converge to particular values (while the larger eigenvalues may potentially diverge).

Have we had access to an "infinite-precision" computer, we could just diagonalize the matrix for increasing values of $\Lambda$ extrapolate to $\Lambda\to\infty$. In reality, this is impractical.

I am wondering if there exist any well-known ways to deal with such a situation, at least for some specific cases (such as divergent elements staying on the diagonal and matrix being sparse).

  • $\begingroup$ Isn't a basis where the entries are not divergent, even though the eigenvalues are not bounded above. $\endgroup$
    – nicoguaro
    Commented Jun 9, 2022 at 22:35
  • $\begingroup$ I'm afraid that finding such a basis may be as hard as finding these eigenvalues. $\endgroup$
    – mavzolej
    Commented Jun 9, 2022 at 22:51
  • $\begingroup$ Write the eigenvalues as a function of $\Lambda$, and then do extrapolation techniques or Taylor expansion in $1/\Lambda$. $\endgroup$ Commented Jun 10, 2022 at 23:08
  • $\begingroup$ How can we do this analytically for a matrix of size larger than $4\times 4$? In reality, the matrix is VERY large, even though its entries as analytic functions of $\Lambda$ are known. Doing this numerically this also is not an option as choosing the $\Lambda$-divergent entries to be much larger than others leads to computational issues. $\endgroup$
    – mavzolej
    Commented Jun 11, 2022 at 5:48
  • $\begingroup$ Is the matrix symmetric? What do you mean by "lowest" otherwise? $\endgroup$ Commented Jun 12, 2022 at 15:32


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