Consider a $n\times n$ matrix: \begin{equation} X=\begin{bmatrix} a &1-a & & \\ & a &1-a & \\ & & \ddots &\\ & & & 1-a\\ & & & a \end{bmatrix} \end{equation} where $a\in(0,1)$. Is it any way to calculate its singular values?
Here is my thought: I know it's easy to check all of its eigenvalues equal to $a$, but it seems non-trivial to calculate eigenvalues of $XX^\top $ (which is equivalent to square of singular values of $X$).
We can calculate the eigenvalues and eigenvectors of tridiagonal matrix directly. Set $A=XX^{\top}\in\mathbb{R}^{n\times n}$, and $$ A=\begin{pmatrix} \alpha&\beta&&\\ \beta&\alpha&\beta&\\ &\ddots&\ddots&\ddots\\ &&\beta&\alpha \end{pmatrix}, $$ where $$ \alpha=a^2+(1-a)^2,\quad \beta=a(1-a). $$
For $Ax=\lambda x$, where $x\neq0$, $x=(x_1,\cdots,x_n)^{\top}\in\mathbb{R}^n$, we have $$ (\ast)\qquad\beta x_{k+1}+(\alpha-\lambda)x_k+\beta x_{k-1}=0,\quad k=1,\cdots,n. $$ Here $x_0=x_{n+1}=0$.
The solution of $(\ast)$ can be written as $$ x_k=C_1\mu_1^k+C_2\mu_2^k,\quad k=1,\cdots,n, $$ where $\mu_1,\mu_2$ are two roots of the characteristic equation w.r.t. $(\ast)$ $$ \beta\mu^2+(\alpha-\lambda)\mu+\beta=0. $$ It can be easily verified that $\mu_1\neq\mu_2$. So $$ (\ast\ast)\qquad\mu_1+\mu_2=-\frac{\alpha-\lambda}{\beta},\quad \mu_1\mu_2=1. $$ Since $x_0=x_{n+1}=0$, we have $$ C_1+C_2=0,\quad C_1\mu_1^{n+1}+C_2\mu_2^{n+1}=0, $$ and $$ \left(\frac{\mu_1}{\mu_2}\right)^{n+1}=1\quad\Rightarrow\quad \frac{\mu_1}{\mu_2}=e^{2\mathrm{i}j\pi/(n+1)},\quad j=1,\cdots,n. $$ Note that it have $n$ pair roots $\mu_1^{(j)},\mu_2^{(j)}$. Along with $(\ast\ast)$ we can get all eigenvalues of $A$ $$ \lambda^{(j)}=\alpha+2\beta\cos\frac{j\pi}{n+1},\quad j=1,\cdots,n, $$ and corresponding eigenvector is $$ x^{(j)}=(\sin\frac{j\pi}{n+1},\sin\frac{2j\pi}{n+1},\cdots,\sin\frac{nj\pi}{n+1})^{\top}. $$
For singular value of $X$, just take a square root.
