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I have the following Hamiltonian \begin{equation} H(x_{1},x_{2},...,x_{n})=\sum_{n=2}^{N}\frac{1}{2}(x_{n}-x_{n-1})^{2} \end{equation} And I'm interested in getting a general formula for the product of the non-zero eigenvalues of its Hessian \begin{equation} \begin{bmatrix}1 &-1 & 0 & 0 & 0\\-1&2&-1&0&0\\0&-1&2&-1&0\\0&0&-1&2&-1\\0&0&0&-1&1\end{bmatrix} \end{equation} Note that $0$ is an eigenvalue associated to $(1,1,...,1)$, so I cannot work with the determinant.

I tried to consider a change of coordinates $y_{1}=\frac{1}{\sqrt{N}}(1,1,...,1)$, $y_{2}=\frac{1}{\sqrt{2}}(1,-1,0,...0)$,... ,$y_{N}=\frac{1}{\sqrt{2}}(0,0,...,1,-1)$ where the Hamiltonian is

\begin{equation} H(y_{1},y_{2},...,y_{n})=\sum_{n=2}^{N}y_{n}^{2} \end{equation}

And from here I get that the Hessian is \begin{equation} \begin{bmatrix}0 &0 & 0 & 0 & 0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\0&0&0&0&2\end{bmatrix} \end{equation}

But there are some odd things: first of all, when I compute the eigenvalues in Maple for a $4\times4$ system I get $(0,2,2+\sqrt{2},2-\sqrt{2})$ and not $(0,2,2,2)$. Second, $(1,-1,0,...,0)$ and the other vectors are not eigenvectors of the original Hessian. Any idea of what could I be doing incorrectly or how to solve this properly? I'm only interested in the product of the eigenvalues, not really in the eigenvalues and eigenvectors per se, but if you have any idea of how to get any of them that works for me!

1 Answers 1

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Let $M_n$ be the $n\times n$ matrix with such structure and $D_n(\lambda) \stackrel{\text{def}}{=}\det(M_n-\lambda I)$ its characteristic polynomial. By applying a Laplace expansion, we get that $\{D_n(\lambda)\}_{n\geq 3}$ fulfills a simple recurrence relation, and the same holds for $\{D_n'(0)\}_{n\geq 3}$. By Vieta's theorem $\pm D_n'(0)$ is exactly what we are interested in.

It turns out that the product of the non-zero eigenvalues of $M_n$ is just $\color{red}{n}$.

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    Hi Jack, thanks for your answer! I have a few questions: When you get the relation equation, you obtain \begin{equation} a_{1}=1-\lambda, a_{2}=\lambda^{2}-3\lambda+1 \end{equation} \begin{equation} a_{j}=(2-\lambda)a_{j-1}-a_{j-1} \end{equation} \begin{equation} a_{N}=(1-\lambda)a_{N-1}-a_{N-2} \end{equation} Then, from the general equation $a_{j}$ I get a tentative $a_{n}=\alpha x_{1}^{n}+\beta x_{2}^{n}$ where $x_{i}=\frac{1}{2}[(2-\lambda)\pm\sqrt{\lambda^{2}-4\lambda}]$. How do you proceed from here? I didn't follow the derivatives thing either – 2017-01-25
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    @Sebasen: the characteristic polynomial of $M_n$ has to be a multiple of $\lambda$ since $0$ is an eigenvalue. Given the recurrence relation for such characteristic polynomials, those polynomials divided by $\lambda$ fulfill a similar recurrence relation, and you just have to understand what happens at $\lambda=0$ for the "reduced" characteristic polynomials. – 2017-01-25
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    This is indeed the same approach as the standard one for computing the determinant of tridiagonal Toeplitz matrices, with a minor tweak. – 2017-01-25
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    Sure! I wasn't thinking there. Then I can get $a_{N}$ from the previous equations, divide by $\lambda$ and the independent term is the product of the eigenvalues. Thanks! I'll try to finish the calculations and see if I get the same than you – 2017-01-25
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    @Sebasen: exactly. You're welcome! – 2017-01-25
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    I'm actually getting quite an ugly equation for $a_{N}$, how did you manage to get just $n$ as independent term if I may ask? – 2017-01-25
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    @Sebasen: there is a trick - the recurrence equation is not so pleasant, I agree, but you may check by hand that $|D_n'(0)|=n$ holds for every $n$ in the range $[3,7]$. Since the recurrence equation has a finite order, $|D_n'(0)|=n$ holds for every $n>7$ as well. – 2017-01-25
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    I got it to work, thank you again! – 2017-01-26