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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.

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  • $\begingroup$ So for $N=0$ you think the matrix has eigenvalues $\pm 1?$ $\endgroup$ Mar 23, 2016 at 20:01
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    $\begingroup$ For $N=0$ the matrix is $\begin{pmatrix} 0 & 1\\ 1&0\end{pmatrix}$ so that checks out. $\endgroup$
    – Nick Cook
    Mar 23, 2016 at 22:43
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    $\begingroup$ I'm surprised at the close votes. This looks like a perfectly reasonable question to me, probably well above MO average. $\endgroup$ Mar 24, 2016 at 0:11
  • $\begingroup$ unfortunately, I misread the question so clicked 'close', but it seems like an interesting question... $\endgroup$
    – Suvrit
    Mar 24, 2016 at 1:36

1 Answer 1

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The Matrix $X$ I was looking at was nothing more than the angular momentum operator $L_x$ in the spin $l=(N+1)/2$ representation, see angular momentum on wikipedia.

It can be written as $L_x = 1/2(L_+ + L_-)$ where $L_+$ and $L_-$ obey $[L_z,L_\pm] = L_\pm$ and $L_z$ is the diagonal matrix with integer spaced entries ranging from $-l$ tot $l$. From the knowledge of spin representations (or $SL(2,\mathbb{C}$) if you which) one can easily find the spectrum of $L_x$.

Also, the matrix diagonalizing $X$ is just $V = e^{\pi i/2 L_y}$ which in itself is not symmetric but from it a matrix $V' = VD$ can be constructed which also diagonalizes $X$ and which is symmetric. Here $L_y = i/2(L_+ - L_-)$.

I would like to thank all who have put thought in my question!

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