Finding eigenvalues in almost tridiagonal matrix I need to find the eigenvalues of an $n\times n$ symmetric tridiagonal matrix $A$, except it has $1$s on $A_{1n}$ and $A_{n1}$.
The diagonal entries are all $4$, while superdiagonal and subdiagonal entries are all $1$.
$$A=\pmatrix{
4&1&&&&&&1\\
1&4&1&&&&&\\
&1&4&1&&&&\\
&&1&4&\ddots&&&\\
&&&1&\ddots&1&&\\
&&&&\ddots&4&1&\\
&&&&&1&4&1\\
1&&&&&&1&4\\
}$$
If there were zeroes in the corners $A_{1n}$ and $A_{n1}$, I would use the formula:
$$ a - 2 \sqrt{bc} \, \cos(k \pi / {(n+1)}),$$
for $k=1,...,n$.
But how can I find eigenvalues in this case?
 A: For further reference, taking $a=4,b=c=1$ in your formula (given in (https://en.wikipedia.org/wiki/Tridiagonal_matrix)), the eigenvalues of tridiagonal matrix $A'$ being obtained from $A$ by setting $A'_{n1}=A'_{1n}=0$, are:
$$\tag{*}\mu_k=4 + 2 \cos(\tfrac{2\pi k}{n+1})\ \ \ \ \text{for} \ \ \ \ k=1,\cdots n.$$
$A$, being a circulant matrix [with circulant "message" $410...01$ associated with polynomial $f(x):=4+1x+1x^{n-1}$], has the following eigenvalues:
$$\tag{1}\lambda_k=f(\omega_k) \ \ \text{where} \ \ \omega_k=\exp(\tfrac{2 i\pi k}{n}) \ \ \text{(k - th root of unity)}$$
(see (https://en.wikipedia.org/wiki/Circulant_matrix)).
$$\lambda_k=4 + 1 \omega_k+ 1 (\omega_k)^{n-1}=4+\exp(\tfrac{2 i\pi k}{n})+\exp(\tfrac{2 i\pi k (n-1)}{n})=4+\exp(\tfrac{2 i\pi k}{n})+\exp(\tfrac{-2 i\pi k}{n}).$$
Thus 
$$\tag{2}\lambda_k=4 + 2 \cos(\tfrac{2\pi k}{n})\ \ \ \ \text{for} \ \ \ \ k=0,1,\cdots n-1.$$
A striking fact is that formula (2) is almost identical to formula (*), except that  denominators $n+1$ have been replaced by denominators $n$. 
Thus, for large values of $n$, the induced perturbation is small...
Nice, isn't it ? 
Let us now give an example and show that all this can be considered in connection with the Discrete Fourier Transform (DFT).

An example: If $n=4$, matrix 
$$A=\begin{pmatrix}4  &   1  &   0  &   1 \\
     1  &   4  &   1  &   0 \\
     0  &   1  &   4  &    1\\
     1  &  0  &  1  &  4 \end{pmatrix}$$
has eigenvalues 
$$\lambda_0=6,\lambda_1=4,\lambda_2=2,\lambda_3=4$$
that comply with formulas (2).
A connection with the Discrete Fourier Transform (DFT). 
Let us proceed with the same example. The DFT of "message" $V=(1,4,1,0)$ is given by a matrix-vector multiplication, with the Fourier matrix whose coefficients are:
$$\tag{3}F_{KL}=e^{f KL} \ \ \text{with} \ \ f :=- i \tfrac{2\pi}{n}, \ \text{here with} \ n=4.$$
(please note the minus sign), i.e.,
$
\underbrace{\begin{pmatrix}
    e^{f0\times0}  & e^{f0\times1}   &   e^{f0\times2}  &   e^{f0\times3} \\
    e^{f1\times0}  & e^{f1\times1}   &   e^{f1\times2}  &   e^{f1\times3} \\
     e^{f2\times0}  & e^{f2\times1}   &   e^{f2\times2}  &  e^{f2\times3} \\
    e^{f3\times0}  & e^{f3\times1}   &   e^{f3\times2}  &   e^{f3\times3}  \end{pmatrix}}_{F}$
$\begin{pmatrix}1 \\
     4\\
     1\\
     0 \end{pmatrix}=
\left(\begin{array}{rrrr}1  & 1  &  1  & 1 \\
    1  & -i   &   -1  &   i \\
    1  & -1   &  1  &    -1 \\
    1 & i   &  -1  &   -i \\ \end{array}\right)
\begin{pmatrix}4 \\
     1\\
     0\\
     1 \end{pmatrix}=\begin{pmatrix}6 \\
       4\\
     2\\
     4 \end{pmatrix}$
finding back in this way the eigenvalues given above.
This example can be extended to an $n \times n$ matrix $A$, where the matrix-vector multiplication shown above is equivalent to the linear combination 
$$\tag{4}4F_0+1F_1+1F_{n-1}$$
(where $F_k$ is the $k$th column of $F$). And we find back expression (2). 
For  a little more about the connection with Fourier Transform, see the answer to this question (Circulant matrix : eigenvector).
