Eigenvalues of an imperfect circulant matrix For the circulant matrix, for example,
$$\begin{bmatrix}
a & b & 0 & & 0 \\
0 & a & b & \cdots & 0 \\
0 & 0 & a & & 0 \\
  & \vdots & &\ddots & \vdots\\
b & 0 & 0 & \cdots & a
\end{bmatrix},$$
we can obtain its eigenvalues analytically using the property of circulant matrix. Is it still possible to get the eigenvalues of the following "impefect circulant matrix" analytically? 
$$\begin{bmatrix}
c & d & 0 & & 0 \\
0 & a & b & \cdots & 0 \\
0 & 0 & a & & 0 \\
  & \vdots & &\ddots & \vdots\\
b & 0 & 0 & \cdots & a
\end{bmatrix},$$
Thank you!
 A: $$\begin{bmatrix}
c & d & 0 & \dots & 0 & 0\\
0 & a & b & \dots & 0 & 0\\
0 & 0 & a & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & a & b\\
b & 0 & 0 & \dots & 0 & a
\end{bmatrix} = \mathrm U + b \, \mathrm e_n \mathrm e_1^\top =: \mathrm M$$
where $\rm U$ is an $n \times n$ upper triangular and bidiagonal matrix. Using the matrix determinant lemma, the characteristic polynomial of matrix $\rm M$ is
$$\begin{aligned} \det (s \mathrm I_n - \mathrm M) &= \det \big( (s \mathrm I_n - \mathrm U) - b \, \mathrm e_n \mathrm e_1^\top \big) \\ &= \det (s \mathrm I_n - \mathrm U) - b \, \mathrm e_1^\top \mbox{adj} \big( s \mathrm I_n - \mathrm U \big) \,\mathrm e_n\\ &= \det (s \mathrm I_n - \mathrm U) - b \, \mathrm e_n^\top \mbox{adj}^\top \big( s \mathrm I_n - \mathrm U \big) \,\mathrm e_1 \\ &= \det (s \mathrm I_n - \mathrm U) - b \, \mbox{cofactor}_{n,1}\big( s \mathrm I_n - \mathrm U \big) \end{aligned}$$
Since $\rm U$ is upper triangular,
$$\det (s \mathrm I_n - \mathrm U) = (s-a)^{n-1} (s-c)$$
The $(n,1)$-th cofactor is
$$(-1)^{n+1} (-b)^{n-2} (-d) = (-1)^{2n} b^{n-2} d = b^{n-2} d$$
and, thus,
$$\det (s \mathrm I_n - \mathrm M) = \color{blue}{(s-a)^{n-1} (s-c) - b^{n-1} d}$$
