1
$\begingroup$

Consider the following $(3 \times 3)$ matrix:

$K_3 = \left( \begin{array}{ccc} a & -1 & 0 \\ -1 & a+1 & -1 \\ 0 & -1 & a \end{array} \right)$

The question has a quantum physics context, so we'll assume that $a$ is such that $K$ is hermitian. This matrix has eigenvalues $-1 + a$, $a$, and $a+1$. Now consider growing the matrix to a $(4\times4)$ in the following way:

$K_4 = \left( \begin{array}{ccc} a & -1 & 0 & 0 \\ -1 & a+1 & -1 & 0 \\ 0 & -1 & a+1 & -1 \\ 0 & 0 & -1 & a \end{array} \right)$

Its eigenvalues are $-1 + a$, $1+a$, $1-\sqrt{2} + a$, and $1+\sqrt{2} + a$. There seems to be at least some structure in this. I was wondering whether there's a way to predict what the eigenvalues of increasingly bigger matrices of this form will be. Specifically, is it possible to find the eigenvalues of:

$K_n = \left( \begin{array}{ccccccc} a & -1 & 0 & \dots & 0 & 0 & 0 \\ -1 & a+1 & -1 & \dots & 0 & 0 & 0 \\ 0 & -1 & a+1 & \dots & 0 & 0 & 0 \\ 0 & 0 & -1 & \dots & -1 & 0 & 0 \\ 0 & 0 & 0 & \dots & a+1 & -1 & 0 \\ 0 & 0 & 0 & \dots & -1 & a+1 & -1 \\ 0 & 0 & 0 & \dots & 0 & -1 & a \end{array} \right)$

I can imagine the expressions become rather monstrous if it's at all possible, so taking something like $a = 1$ is fine for my purposes, if it simplifies matters.


With the help of mookid I managed to find a general form for the characteristic polynomials of $K - Ia$. Defining $P_n$ as the characteristic polynomial of the $(n \times n)$ matrix $K - Ia$, we have:

$P_n = -\lambda Q_{n-1} - Q_{n-2}$

Where $\lambda$ are the requested eigenvalues and $Q_n$ are the characteristic polynomials of the same matrix, where the zero on the $nn$-th entry has been replaced by a one:

$Q_n = (1-\lambda) Q_{n-1} - Q_{n-2}$

With boundary conditions $Q_1 = -\lambda$ and $Q_2 = \lambda^2 - \lambda - 1$.

Quoting wikipedia, with regards to invertibility of tridiagonal matrices: "Closed form solutions can be computed for special cases such as symmetric matrices with all off-diagonal elements equal". I'm hoping that means there also exists an analytic solution for the roots of these polynomials (and hence the eigenvalues for arbitrary $n$). Does anyone know if this is possible, and/or how to go about finding the solution?

$\endgroup$
0
$\begingroup$

You just have to go for the $a=0$ case; in general, the eigenvalues are $a+\lambda, \lambda$ eigenvalue of $\left( \begin{array}{ccccccc} 0 & -1 & 0 & \dots & 0 & 0 & 0 \\ -1 & 1 & -1 & \dots & 0 & 0 & 0 \\ 0 & -1 & 1 & \dots & 0 & 0 & 0 \\ 0 & 0 & -1 & \dots & -1 & 0 & 0 \\ 0 & 0 & 0 & \dots & 1 & -1 & 0 \\ 0 & 0 & 0 & \dots & -1 & 1 & -1 \\ 0 & 0 & 0 & \dots & 0 & -1 & 0 \end{array} \right)$

Now refering to wikipedia, the sequences of characteristic polynomials is

$$P_2 = X^2 - 1, P_3 = -X^3+X \\ P_{n+1} = X\times P_n - (-1)\times(-1)\times P_{n-1} =-XP_n - P_{n-1}$$

$\endgroup$
2
  • $\begingroup$ Thanks! That's a really useful relation. What exactly is $X$ in your equations? I find: $P_2 = \lambda^2 - 1$, $P_3 = -\lambda^3 +\lambda^2 + 2\lambda$, and $P_n = -\lambda P_{n-1} - P_{n-2}$, where $\lambda$ are the eigenvalues of $K - Ia$. $\endgroup$ – Timsey Jun 25 '14 at 14:18
  • $\begingroup$ There's one small problem with this method. To calculate $P_{n+1}$, you need $P_n$. Here $P_n$ isn't the characteristic polynomial of the matrix $K_n - Ia$, but of a matrix that has a 1 as its $nn$-th entry (instead of a zero). Defining $Q_n$ as the characteristic polynomials of these matrices, we find: $P_n = -\lambda Q_{n-1} - Q_{n-2}$, where $Q_n = (1 - \lambda) Q_{n-1} - Q_{n-2}$. Starting values are $Q_1 = - \lambda$ and $Q_2 = \lambda^2 - \lambda - 1$. $\endgroup$ – Timsey Jun 25 '14 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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