I'm reading a book about semi-conductors, and when figuring out how dopants affect the energy-levels, one wishes to find the eigenvalues (and vectors) of a NxN tridiagonal matrix of the form
$$\begin{pmatrix}
E_1 & -A \\
-A & \ddots & -A \\
& -A & (E_1-F) & \ddots \\
& & \ddots & \ddots & -A \\
& & & -A & E_1
\end{pmatrix}$$
for $E_1,A,F \in \mathbb{R}$. That is, it's Toeplitz except for the value in the middle of the matrix being $E_1-F$ instead of $E_1$. The book mentions that this eigenvalue can be solved by hand, but doesn't do it. I've been trying and failing to solve it (except for $F=0$), and was hoping that someone here could shed some light on the problem!
Best Answer
Your matrix is real, symmetric, tridiagonal, and Toeplitz+rank1. It is also centrosymmetric, so its eigenvectors are either symmetric or anti-symmetric with respect to the middle element.
Without the modification at the center, the solution is well known: $$ \lambda_k = E_1-2A\cos \frac{k\pi}{N+1}\qquad v_{kj} = \sqrt{\frac{2}{N+1}}\sin\frac{kj\pi}{N+1}\qquad k = 1,\ldots,N$$ It is easy to see the eigenvectors are alternatingly even and odd, as required by centrosymmetry. The solution in the PDF is basically a hand calculation, and the ansatz that the eigenvectors are sinusoidal is justified on physical grounds (spatially homogeneous Green's functions).
With the (rank-1) modification to the center element, only the even eigenpairs are affected. Let's empirically investigate this first. For $E_1=1$, $A=0.1$, and $F=0.2$, we get the following plots:
You can see that the lowest eigenvalue has fallen, and the other eigenvalues are bunched in pairs (because the antisymmetric ones didn't move, while the symmetric ones moved but have the same overall behavior). The lowest eigenvector is now localized at the center (a trap state), while the remaining eigenvectors are still sinusoidal away from the center.
Applying the rank-1 modification process described here, we get that the new eigenvalues are solutions to $$ \sum_{j=1}^N \frac{\sin^2 \frac{j\pi}{2}}{E_1-2A\cos \frac{j\pi}{N+1}-\lambda} = \frac{N+1}{2F} $$ You can see that when $j$ is even, the term in the numerator of the sum disappears, and so $\lambda$ must be equal to the original eigenvalue (the left side of the denominator).
I assume you are interested in the lowest eigenvalue (since that is the trap state that falls outside the ordinary energy bands). In this case, each term of the sum can be approximated using the midband value: $$ \frac{N+1}{2}\frac{1}{E_1-\lambda_1} \approx \frac{N+1}{2F} $$ So the lowest eigenvalue is approximately $\lambda_1 = E_1-F$. This only works assuming $F \gg A$.