Calculating the eigenvalues of a matrix How to find the eigenvalues of
$$\begin{bmatrix}
0 & 1   &   &  &\\
k & 0   & 2 &  &\\
  & k-1 & 0 & 3 &\\
  &     &\ddots&\ddots&\ddots\\
  &     &      &2 & 0 &k \\
  &     &      & & 1 &0
\end{bmatrix}$$
I tried recurrence equation, but it doesn't work to find the characteristic polynomial. Any hint or solution are welcomed, thanks for your help!
PS:The missing parts are all zeroes.
 A: Let your matrix be $A_{k+1}$ and let $P_{k+1}$ be the $(k+1)\times(k+1)$ matrix
$$
\begin{pmatrix}1&1\\&\ddots&\ddots\\&&\ddots&1\\ &&&1\end{pmatrix}.
$$
Then $P_{k+1}^{-1}A_{k+1}P_{k+1}=B_{k+1}:=\begin{pmatrix}-k&0\\ ke_1&A_k+I_k\end{pmatrix}$, where $e_1=(1,0,0,\ldots,0)^T$ (you may verify that $A_{k+1}P_{k+1}=P_{k+1}B_{k+1}$). Therefore, if $\sigma(\cdot)$ denotes the spectrum of a matrix, we have the recurrence relation $\sigma(A_{k+1})=\{-k\}\cup\left(1+\sigma(A_k)\right)$. So,
\begin{align*}
\sigma(A_2)&=\{-1,1\},\\
\sigma(A_3)&=\{-2,0,2\},\\
\sigma(A_4)&=\{-3,-1,1,3\},\\
&\vdots\\
\sigma(A_{k+1})&=\{-k,\,-k+2,\,-k+4,\,\ldots,\,k-4,\,k-2,\,k\}.
\end{align*}
Edit: There may be a better and more revealing proof. From WolframAlpha, some eigenvectors of $A_{k+1}$ are comprised of signed binomial coefficients (see the results for the cases $k=2$, $k=3$, $k=4$ and $k=5$), so $A_{k+1}$ may possess some interesting mathematical properties that are yet to be discovered.
Edit 2: It turns out that this matrix is called a Kac matrix or a Clement-Kac-Sylvester matrix, and was proposed by Clement (1959) as a matrix for test purposes. See a related question here.
Edit 3: According to Taussky and Todd (1991) (thanks to J.M. for the reference), two elementary proofs of the above result had been given by Muir and Metzler (1933, 1960) and Mazza (1923). Muir and Metzler's proof is strikingly similar to mine. Essentially, it says that if
$$
F_{k+1}=\begin{pmatrix}1&-1\\&\ddots&\ddots\\&&\ddots&-1\\ &&&1\end{pmatrix},
$$
then $F_{k+1}^{-1}A_{k+1}F_{k+1}=\begin{pmatrix}k&0\\ ke_1&A_k-I_k\end{pmatrix}$. In contrast, Mazza's proof essentially says that if
$$
G_{k+1}=\begin{pmatrix}1&0&-1\\&\ddots&\ddots&\ddots\\&&\ddots&\ddots&-1\\ &&&\ddots&0\\ &&&&1\end{pmatrix},
$$
then $G_{k+1}^{-1}A_{k+1}G_{k+1}=
\left(\begin{array}{c|c}
\begin{array}{cc}0&k\\ k&0\\ \end{array}
&{\Large 0}_{2\times(n-2)}\\
\hline
B_{k-1}&A_{k-1}\\
\end{array}\right)$ where $B_{k-1}=\begin{pmatrix}0&k-1\\ 0&0\\ \vdots&\vdots\\ 0&0\end{pmatrix}$.
Eigenvectors of $A_{k+1}$ are also well-studied. See J.M.'s comments (thanks to J.M. again).
