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Is there a way to obtain eigenvalues of three anti-diagonal $n \times n$ Hankel matrix $$ H=\left( \begin{array}{ccccc} L & 0& ... & 0& b& c \\ 0 & 0& ... & b& c& d \\ 0 & 0& ... & c& d& 0 \\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots \\ b&c&d&...&0&0\\ c&d&0&...&0&L\\ \end{array} \right) $$ in closed form?

Is it possible in the case $L=0$? In my specific case $a,b,c$ are imaginary and $L$ is real.

I can do it for a specific case $b=d$ by noticing that eigenvalues are the same (up to complex conjugation) as in Toeplitz matrix $$ H=\left( \begin{array}{ccccc} L+c & b& 0&\dots \\ b&c&b&\dots\\ \vdots&\ddots&\ddots\\ \dots & 0&b & c+L \end{array} \right) $$ for which formula is given in https://www.emis.de/journals/AMEN/2005/040903-7.pdf.

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  • $\begingroup$ The matrix is almost a circulant matrix which would have a basis of complex exponentials as eigenvectors and the FFT of the most commonly occurring row vector as eigenvalues. Maybe that can help. $\endgroup$ – mathreadler Feb 28 '17 at 9:58

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