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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!!!

nsigma
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  • One slight simplification: call your matrix $A$ and let $B=DAD^{-1}$, where $D=\operatorname{diag}(1,i,i^2,\ldots,i^N)$. Then $B$ is real symmetric. It is the entrywise absolute value of $A$. – user1551 Jul 02 '19 at 18:48

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(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 $N+1+S=\{1,3,5,\ldots,2N+1\}$.

user1551
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