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I am looking for analytic expressions for the eigenvalues of matrices of the form

$$A = \begin{bmatrix} 6 & -4 & 1 & 0 & 0 & 0 & 0 \\ -4 & 6 & -4 & 1 & 0 & 0 & 0 \\ 1 & -4 & 6 & -4 & 1 & 0 & 0 \\ 0 & 1 & -4 & 6 & -4 & 1 & 0 \\ 0 & 0 & 1 & -4 & 6 & -4 & 1 \\ 0 & 0 & 0 & 1 & -4 & 5 & -2 \\ 0 & 0 & 0 & 0 & 1 & -2 & 1 \\ \end{bmatrix}$$

for arbitrary dimensions. The displayed $(7 \times 7)$ dimension is chosen for expositional convenience only.

The matrix arises as follows. Let $U$ denote the lower triangular matrix with ones on and below the diagonal. In principle, I am interested in the eigenvalues of $B = (U^2)'U^2$. It seems easiest, however, to find the reciprocals as the eigenvalues of $A = B^{-1}$. Clearly, the matrix $A$ is almost equivalent to a symmetric pentadiagonal Toeplitz matrix, with the exception of the last two rows and columns. Perhaps it is possible to first find the eigenvalues of

$$A^* = \begin{bmatrix} 6 & -4 & 1 & 0 & 0 \\ -4 & 6 & -4 & 1 & 0 \\ 1 & -4 & 6 & -4 & 1 \\ 0 & 1 & -4 & 6 & -4 \\ 0 & 0 & 1 & -4 & 6 \\ \end{bmatrix}$$

although I am not sure how to extend these to $A$. Some background information that may be useful:

The $(n \times n)$ matrix $U'U$ (recall that we're interested in $(U^2)'U^2$) has eigenvalues whose reciprocals are given by

$$\lambda_i^{-1} = 4\sin^2\left(\frac{(2i-1)\pi}{2(2n+1)}\right),$$

as demonstrated in Akesson and Lehoczky (1998). One can numerically verify that

$$\lambda_i^{-2} = 16\sin^4\left(\frac{(2i-1)\pi}{2(2n+1)}\right)$$

seems to be a very accurate approximation of the reciprocal of the $i$-th eigenvalue of $(U^2)'U^2$, but it is not an exact solution. A circulant version of the matrix $A^*$ appears on page 3 of Wang and Long (2020), where it is stated that its eigenvalues are $16\sin^4\left(\frac{i\pi}{n}\right)$. Finally, theorem 2 in Elouafi (2011) provides a general characterization for the eigenvalues of symmetric pentadiagonal Toeplitz matrices in terms of the zeros of particular functions, but I struggle to obtain analytic expression for the eigenvalues of $A^*$ from this theorem (let alone for $A$).

I suspect that directly solving the system of equations $Av = \lambda v$ might correspond to solving a fourth order homogenous linear difference equation under suitable initial value conditions, although I am not sure this is a feasible route to follow.


References

  • Akesson and Lehoczky (1998) Discrete eigenfunction expansion of multi-dimensional Brownian motion and the Ornstein-Uhlenbeck process

  • Elouafi (2011) An eigenvalue localization theorem for pentadiagonal symmetric Toeplitz matrices

  • Wang and Long (2020) Multiple solutions of fourth-order functional difference equation with periodic boundary conditions

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    $\begingroup$ "analytic expressions of the matrix" you mean of its eigenvalues? $\endgroup$ Commented Sep 11, 2021 at 5:15
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    $\begingroup$ Am I seeing it right that the general formula for the $\left(i,j\right)$-th entry of your matrix $A^*$ is $\left(-1\right)^{i-j} \dbinom{4}{2-i+j}$ ? In that case, you can get rid of the $\left(-1\right)^{i-j}$ factor by conjugating by an appropriate diagonal matrix. The resulting matrix has been studied; its determinant is a particular case of (1.2) in Christian Krattenthaler, arXiv:math/9902004v3. $\endgroup$ Commented Sep 11, 2021 at 5:17
  • $\begingroup$ @darijgrinberg Interesting observation! It indeed seems that your equation generates $A^*$. However, that does not seem immediately helpful to calculate the eigenvalues, as we would need the determinant of $A^* - \lambda\mathbb{I}$, which is not generated by that nice formula anymore. Am I missing something? $\endgroup$
    – E_Wijler
    Commented Sep 11, 2021 at 11:58
  • $\begingroup$ Correct; I'm just hoping it gives some context. $\endgroup$ Commented Sep 11, 2021 at 13:51

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