Eigenvalues of a Hermitian tridiagonal matrix. I want to know how to calculate the eigenvalues of the following Hermitian tridiagonal $(N+1)\times(N+1)$ matrix,
$$
A=\begin{pmatrix}
N+1&i\sqrt{N}\\
-i\sqrt{N}&N+1&i\sqrt{2(N-1)}\\
&-i\sqrt{2(N-1)}&\ddots&i\sqrt{3(N-2)}\\
&&-i\sqrt{3(N-2)}&\ddots&\ddots\\
&&&\ddots&N+1&i\sqrt{N}\\
&&&&-i\sqrt{N}&N+1
\end{pmatrix}
$$
that is $a_{kk} = N+1$, $a_{k,k+1} = i\sqrt{k(N-k+1)}$.
From another method to treat the problem (it is from physics), I get the eigenvalues are 
$$
\lambda = 1,3,5,\ldots,2N+1.
$$
Any help is appreciated!
My answer: I find $A-(N+1)I$ is the same as the matrix of $J_y$ with using eigenstates  of $J_z$  as basis ($J_y,J_z$ are angular momentum operators in quantum mechanics). Then use the result of representations of $\mathfrak{su}_2$, we can get the eigenvalues of $A-(N+1)I$ are $-N,-N+2,\dots,N-2,N$. 
Really thanks for help!!! 

(For convenience, all matrices below are zero-indexed.)
Let $B=A-(N+1)I$. Then $B$ is a tridiagonal complex skew-symmetric matrix. When $0\le k<N$, we have
\begin{align}
b_{k,k+1}&=i\sqrt{(k+1)(N-k)},\\
b_{k+1,k}&=-b_{k,k+1}.
\end{align}
(Note that the above formula is different from the one in the OP because our matrix is now zero-indexed.) Let $D=\operatorname{diag}(d_0,d_1,\ldots,d_N)$ where
$$
d_k=\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}.
$$
The product of super-diagonal entries of $B$ in the above is considered empty when $k=0$, so that $d_0=1$ by convention. Then $C=DBD^{-1}$ is a tridiagonal matrix such that
\begin{aligned}
c_{k,k+1}
=\frac{d_k}{d_{k+1}}b_{k,k+1}
&=\frac{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
b_{k,k+1}
=k+1,\\
c_{k+1,k}
=\frac{d_{k+1}}{d_k}b_{k+1,k}
&=\frac{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
b_{k+1,k}\\
&=\frac{\frac{1}{(k+1)!}\prod_{i=0}^kb_{i,i+1}}
{\frac{1}{k!}\prod_{i=0}^{k-1}b_{i,i+1}}
(-b_{k,k+1})\\
&=\frac{-b_{k,k+1}^2}{k+1}=N-k.
\end{aligned}
In other words, $C$ is the Kac matrix
$$
C=\begin{pmatrix}
0&1\\
N&0&2\\
&N-1&\ddots&3\\
&&N-2&\ddots&\ddots\\
&&&\ddots&0&N\\
&&&&1&0
\end{pmatrix}.
$$
The eigenvalue problem for Kac matrix was solved in this answer. The spectrum of $C$ is $S=\{-N,\,-(N-2),\,\ldots,\,N-2,\,N\}$. As $B$ is similar to $C$ and $A=B+(N+1)I$, the eigenvalues fo $A$ are given by $N+1+S=\{1,3,5,\ldots,2N+1\}$.
