Eigenvalues and eigenvectors of a tridiagonal matrix with one nonzero diagonal coefficient

Question

I'm looking for the eigenvalues and eigenvectors of the $n\times n$ matrix, denoted $D_n=(d_{i,j})_{1\le i,j\le n}$ defined as $$ d_{i,j} = \left\{\begin{array}{ll} 1 &\text{if } (|i-j|=1)\text{ or } (i=j=n),\\ 0 &\text{otherwise.} \end{array} \right. $$ I tried to find the characteristic polynomial of $D_n$, denoted $\chi_{n}$, and I found that $$ \chi_n(X) = (X-1)\tilde{\chi}_{n-1}(X)-\tilde{\chi}_{n-2}(X) $$ where $\tilde{\chi}_p$ is the characteristic polynomial of the $p\times p$ matrix, denoted $\tilde{D}_n=(\tilde{d}_{i,j})_{1\le i,j\le n}$ defined as $$ \tilde{d}_{i,j} = \left\{\begin{array}{ll} 1 &\text{if } |i-j|=1,\\ 0 &\text{otherwise.} \end{array} \right. $$ It doesn't seem to lead to an easy solution. Indeed, one can easily find the eigenvalues of $\tilde{D}_n$ with this kind of method but I do not see how to find the eigenvalues of $D_n$ with this method.

Answer 1

If $(\lambda,x)$ is an eigenpair, then we have the recurrence relation $x_k - \lambda x_{k-1} + x_{k-2} = 0$ for $k=3,\ldots,n$ and also the "boundary conditions" $x_2 = \lambda x_1$ and $x_{n-1}+x_n = \lambda x_n$.

Since nonzero multiples of eigenvectors are eigenvectors, the recurrence relation entails that we may take $x_k=e^{i(k\theta+\phi)}+e^{-i(k\theta+\phi)}$ for some real arguments $\theta$ and $\phi$, where $$ e^{i\theta} = \frac{\lambda+i\sqrt{4-\lambda^2}}2. $$ So, the boundary condition $x_2 = \lambda x_1$ gives \begin{align*} e^{i(2\theta+\phi)}+e^{-i(2\theta+\phi)} &=\lambda (e^{i(\theta+\phi)}+e^{-i(\theta+\phi)})\\ &= (e^{i\theta}+e^{-i\theta})(e^{i(\theta+\phi)}+e^{-i(\theta+\phi)})\\ &=e^{i(2\theta+\phi)}+e^{-i(2\theta+\phi)}+e^{i\phi}+e^{-i\phi}. \end{align*} Hence $e^{i\phi}=-e^{-i\phi}$ and we may further let $x_k=e^{ik\theta}-e^{-ik\theta}$. The second boundary condition $x_{n-1}+x_n = \lambda x_n$ now gives \begin{align*} e^{i(n-1)\theta}-e^{-i(n-1)\theta}+e^{in\theta}-e^{-in\theta} &=\lambda(e^{in\theta}-e^{-in\theta})\\ &=(e^{i\theta}+e^{-i\theta})(e^{in\theta}-e^{-in\theta})\\ &=e^{i(n-1)\theta}-e^{-i(n-1)\theta}+e^{i(n+1)\theta}-e^{-i(n+1)\theta}. \end{align*} Thus $e^{i(n+1)\theta}-e^{-i(n+1)\theta}=e^{in\theta}-e^{-in\theta}$, i.e. $\sin\left((n+1)\theta\right)=\sin(n\theta)$. Hence either $\theta=2m\pi$ or $(n+1)\theta=(2m+1)\pi-n\theta$ for some integer $m$. Since $\theta$ is not an integer multiple of $\pi$ (or else $x=0$), we must have $$ \theta=\frac{2m+1}{2n+1}\pi $$ for some $m$ such that $2n+1$ doesn't divide $2m+1$. Hence the eigenvalues of $D_n$ are given by $\lambda=2\Re(e^{i\theta})=2\cos\left(\frac{2m+1}{2n+1}\pi\right)$ for $m=0,1,\ldots,n-1$.