# Lowest eigenvalues of a matrix with divergent entries

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).

• Isn't a basis where the entries are not divergent, even though the eigenvalues are not bounded above. Commented Jun 9, 2022 at 22:35
• I'm afraid that finding such a basis may be as hard as finding these eigenvalues. Commented Jun 9, 2022 at 22:51
• Write the eigenvalues as a function of $\Lambda$, and then do extrapolation techniques or Taylor expansion in $1/\Lambda$. Commented Jun 10, 2022 at 23:08
• 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. Commented Jun 11, 2022 at 5:48
• Is the matrix symmetric? What do you mean by "lowest" otherwise? Commented Jun 12, 2022 at 15:32