Eigenvalues of an imperfect circulant matrix

Question

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!

Answer 1

$$\begin{bmatrix} c & d & 0 & \dots & 0 & 0\\ 0 & a & b & \dots & 0 & 0\\ 0 & 0 & a & \dots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \dots & a & b\\ b & 0 & 0 & \dots & 0 & a \end{bmatrix} = \mathrm U + b \, \mathrm e_n \mathrm e_1^\top =: \mathrm M$$

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,

$$\det (s \mathrm I_n - \mathrm M) = \color{blue}{(s-a)^{n-1} (s-c) - b^{n-1} d}$$