A conjecture about the eigenvalues of symmetric pentadiagonal Toeplitz matrix Is there a way to find out the exact eigenvalues and eigenvectors of a real symmetric pentadiagonal Toeplitz $n\times n$ matrix with the form given below?
$$
\begin{pmatrix}
a & b & c & 0 & \cdots & \cdots & 0 \\
b & a & b & c & 0 & \cdots & 0 \\
c & b & a & b & c & \ddots & \vdots \\
0 & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots \\
0 & \ddots & \ddots & \ddots & \ddots & b & c \\
\vdots & \ddots & 0 & c & b & a & b \\
0 & \cdots & 0 & 0 & c & b & a
\end{pmatrix}
$$
In particular, I'm dealing with this conjecture (verified by numerical diagonalization):
The difference of the smallest two eigenvalues are of the order $1/n^2$ if $f(k)$ has one minimum, or $1/n^3$ if $f(k)$ has two minima, where $f(k)=a+2b\cos(k)+2c\cos(2k)$ and $k\in[0,2\pi)$.
 A: The following papers should be helpful:


*

*A note for an explicit formula for the determinant of pentadiagonal
and heptadiagonal symmetric Toeplitz matrices, Mohamed Elouafi, Applied Mathematics and Computation 219 (2013) 4789–4791

*An eigenvalue localization theorem for pentadiagonal
symmetric Toeplitz matrices, Mohamed Elouafi, Linear Algebra and its Applications 435 (2011) 2986–2998
A: I wasn't able to figure out an exact formula for eigenvalues and eigenvectors for a pentadiagonal
Toeplitz matrix.  Knowing them for a tridiagonal Toeplitz, though, is very helpful.  (Those can be
found, for example, in this paper on the sensitivity of the spectrum of a tridiagonal Toeplitz
matrix.)
As you've already discovered, the eigenvalues are approximately $a+2\cos(\omega)+2\cos(2\omega)$ where $\omega=(\frac{\pi}{n+1})*i$ where $i$ is the index of the eigenvalue.  In fact, $A=Q\Lambda Q^T+E$ where $A$ is the matrix you're interested in, $\Lambda$ is a diagonal matrix with approximate eigenvalues as entries along the diagonal, $Q_{ij}=\sin((\frac{\pi}{n+1})ij)$ is the matrix whose columns are the approximate eigenvectors, and $E$ is a matrix of all zeros except a $c$ in the upper left and lower right entries.  The form of the eignevectors comes from the tridiagonal case.  However, there are some important caveats, and they flow from the fact that $E$ might not be small in the 2-norm (relative to $A$).
Two interesting special cases are with $b=1$ and $a=c=0$, and $c=1$ and $a=b=0$.  (I'll assume in the rest that $a=0$ because $a$ only represents a shift in the eigenvalues: the diagonal is a multiple of the identity and shares eigenvectors with every matrix.)  With $b=1$ and $c=0$, the above eigenpairs are exact: $c=0$ means $E=0$, and we're in the tridiagonal case.  With $b=0$ and $c=1$, things are more complicated.  When $A$ is of even size, $n=2m$, all eigenvalues are of multiplicity 2.  The eigenvalues are of $b=1$ and $c=0$ of size $m$, with each eigenvalue repeated once.  The eigenvectors of $A$ are related too: they're those of $b=1$ and $c=0$ of size $m$ interleaved with zeros; the two eigenvectors for each eigenvalue are staggered with respect to one another.  When $A$ is of odd size, $n=2m+1$, the eigenvalues are the union of those for $b=1$ and $c=0$ of size $m$ and size $m+1$.  The eigenvectors are again interleaved with zeros.
I'm assuming by "smallest" eigenvalue you mean "most negative".  If you mean "closest to 0", adjust accordingly below.
The case of even-sized $A$ with $b=0$ and $c=1$ means repeated eigenvalues, and the difference between the two smallest eigenvalues is exactly 0.  With $b=\epsilon$, $c=1$, and $\epsilon\ll 1$, the difference between the two smallest eigenvalues is $O(\epsilon/n^2)$.  If $\epsilon$ is independent of $n$, this fits your conjecture.  However, if you make $\epsilon=1/n^2$ or $\epsilon=\exp(-n)$, then the difference will shrink much faster than your conjecture.
With odd-sized $A$ and $b=\epsilon$, $c=1$, and $\epsilon\ll 1$, the difference between the two smallest eigenvalues is $O(1/n^3)$.  With an $A$ of either parity and $b=1$, $c=\epsilon$, and $\epsilon\ll 1$, the difference between the two smallest eigenvalues is $O(1/n^2)$.
Are there any other restrictions on the matrix size you're implicitly assuming?  Or on the size of the entries $a$, $b$, and $c$?
A: A general solution for the eigenvalues of such symmetric diagonal matrix
A=\begin{bmatrix} 
a_0 & a_1 & a_2& a_3 &......&..&..&a_{n-1} \\
a_{n-1} & a_0 & a_1& a_2& a_3 &.....&...&a_{n-2}\\
a_{n-2}& a_{n-1} &a_0& a_1& a_2& a_3&...&a_{n-3}\\
a_{n-3} & a_{n-2}& a_{n-1}& a_0& a_1& a_2&....&a_{n-4}\\
......&.....&.....&....&....&....&.....&a_0&\\
\end{bmatrix}
(n by n) can be expressed as (if $a_r=a_{n-r}$)
$E_k=a_0+2a_1\cos(\frac{2k\pi}{n})+2a_2\cos(\frac{4k\pi}{n})+2a_3\cos(\frac{6k\pi}{n})+....L,$ k=1,2,3...n,
where $L=2a_s\cos(\frac{2sk\pi}{n})$ if n=2s+1 and $L=(-1)^ka_{s}$ if n=2s.
In this particular matrix, put $a_r=0$ for $r\ge 3$
