I'm studying the following tri-diagonal matrix

$$ X = \begin{pmatrix} 0 & x_0 & 0 & 0 &\cdots & 0 & 0 & 0 \\\ x_0 & 0 & x_1 & 0 &\cdots & 0 & 0 & 0 \\\ 0 & x_1 & 0 & x_2 &\cdots & 0 & 0 & 0 \\\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots & \vdots& \vdots \\\ \vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \vdots& \vdots \\\ \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \ddots& \vdots \\\ 0 & 0 & 0 & 0 &\cdots & x_{N-1} & 0 & x_N \\\ 0 & 0 & 0 & 0 & \cdots & 0 & x_N & 0 \end{pmatrix} $$

where the entries are given by the sequence $x_n = \frac{\sqrt{(N+1-n)(n+1)}}{N+1}$

Numerically I find that the eigenvalues are given by $\lambda_k = 1-\frac{2k}{N+1}$ for $k\in[0,N+1]$.

Also I find numerically that the orthogonal transformation $V$ diagonalizing the above matrix $D = V^T X V$ (such that the columns of $V$ are the eigenvectors of $X$) can be chosen to be symmetric: $V = V^T$.

I have been trying to find/come up with a proof for the above two statements, however, so far unsuccessful. I do have an exact expression for the eigenvector $\vec{v}_0 = c_i\vec{e}_i$ corresponding to $\lambda_0=1$, being $$ c_i = \frac{1}{\sqrt{2^{N+1}}}\sqrt{\begin{pmatrix} N+1 \\ i \end{pmatrix} } $$

Also $X = V^TDV$ implies a recurrence relation on the eigenvectors $\vec{v}_i$ of $X$, being $\lambda\vec{v}_k = x_{k-1}\vec{v}_{k-1} +x_k\vec{v}_{k+1}$ (this is equivalent the statement $V=V^T$). Unfortunately this has not helped me so far.

Any help on proving either of the above statements would be greatly appreciated!

Thanks.