Hello dear,

Please clarify what you mean by "but different elements in the secondary diagonals, which are randomly arranged." It would be useful to give examples of the type of matrix you are referring to.

e.g. the upper secondary diagonal would be

t1, t2, t1, t3, t2, t1, t1, etc i.e. randomly arranged

Dear Constatinos, thank you for this nice question; the answer is positive:

• If A is an n×n tridiagonal Hermitian matrix with n even and all diagonal elements 0, then its characteristic polynomial is biquadratic. It says, that if λ is an eigenvalue, then necessarily - λ is eigenvalue of A, with the same multiplicity.
• If n is odd, 0 is an eigenvalue of A and [det (A-λI)]/λ is biquadratic.
• Second step consists on the fact, that: If λ is an eigenvalue of A, then cλ is eigenvalue of A+cI.

Dear Viera,

Thanks a lot for the enlightenment! :)

For your first and second bullet,

I have checked with n=3 and n=4 (and main diagonal elements zero).

For your third bullet maybe there is a typographical error:

if λ is an eigenvalue of A, then c+λ is an eigenvalue of A+cI

Best Regards, Thank you again for your kind explanation,

Constantinos Simserides

Dear Constantinos, you´re right, the sign + dropped out.

As for my contribution: the latest bullet (with +, naturally) is a known property of any kind of matrices. For the first two bullets - it was a joy to do this little discovery!

He he, thanks a lot again! All the best! :)

Please, let's continue the discussion, if possible. :)

The order of the chacteristic polynomial is n = the order of the matrix. Hence, increasing n, we have to solve more complicated characteristic equations. By now, I have checked analytically n=3, n=4, n=5, and things are as expected, but now that I have arrived at n=6, I have a characteristic equation of the form

λ^6 - θ_1 λ^4 + θ_2 λ^2 - θ_3 =0,

where θ_i are positive, which does not seem biquadratic... I was expecting to find, that, maybe it would contain only terms with λ^6, λ^3 and sth constant... And, increasing further n, we will have more complicted equations. Now?

Lets' call χ(n) the characteristic polynomial of order n. Doing this little algebra, by the way, I have proved that, e.g. x(6) = -λ x(5) - t_5^2 x(4) and similarly for lower n. Maybe, an inductive formula of this type can be proven. t_5 is the first element of the secondary diagonal.

In other words, the secondary diagonal from top left to bottom right contains t_5 t_4 t_3 t_2 t_1.

I think that we have found the answer. The property holds.

It has been included in our article which has been uploaded to arXiv:

arXiv:1901.06273v1 [cond-mat.soft] 17 Jan 2019

Quasi-periodic and fractal polymers: Energy structure and carrier transfer

M. Mantela, K. Lambropoulos, M. Theodorakou, and C. Simserides