What are the eigenvalues/eigenvectors of the matrix $A=\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)\_{k,l=1}^n$ when $n$ is odd?



The eigenvalues of the matrix $\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)_{k,l=1}^n$
Here, we verify the observation of @BrendanMcKay that the eigenvalues are $n$ with multiplicity $(n+1)/2$ and $-n$ with multiplicity $(n-1)/2$.


Note that your matrix is skew-circulant, and it is known (see e.g. the references [here](https://repositorium.sdum.uminho.pt/bitstream/1822/62618/1/real-cscs-final-zhang.pdf)) that the eigenvectors of a skew-circulant matrix are $\mathbf{v}\_j\in \mathbb{C}^n$ of the form $(\mathbf{v}\_j)\_k = \exp((2j+1)k\pi i/n)$, where $j,k$ range from $0$ to $n-1$ for convenience. Note that $(\mathbf{v}\_j)\_0=1$, so it suffices to compute the inner product of the first row of your matrix with each $\mathbf{v}\_j$ to determine the eigenvalues. Writing out the exponential, we find that the eigenvalue associated to $\mathbf{v}\_j$ is thus
\begin{equation\*}
\sum\_{k=0}^{n-1} \frac{\cos((2j+1)k\pi/n)}{\cos(k\pi/n)}+i\sum\_{k=0}^{n-1} \frac{\sin((2j+1)k\pi/n)}{\cos(k\pi/n)}.
\end{equation\*}


Now, notice that the second sum vanishes by symmetry. The "zeroth" term is clearly $0$, while the $k$th and $(n-k)$th terms for $k\geq 1$ cancel because $\cos(x)$ is odd with respect to $x=\pi/2$, while
\begin{align\*}
\sin((2j+1)(n-k)\pi/n)&=\sin((2j+1)\pi-(2j+1)k\pi/n)\\
&=\sin(\pi-(2j+1)k\pi/n)\\
&=\sin((2j+1)k\pi/n).
\end{align\*}
Therefore, it suffices to show that the first term is indeed equal to $(-1)^jn$. I'm guessing this is standard, but I am not too great at trigonometric manipulations, so below is an elementary argument.


We do this by induction. For $j=0$, this is trivial as each term in the sum is $1$. Now suppose we have shown this for some $j\geq 0$ and we wish to prove it for $j+1$. Observe that
\begin{align\*}
\cos((2(j+1)+1)k\pi/n) &= \cos((2j+1)k\pi/n+2k\pi/n)\\
&=\cos((2j+1)k\pi/n)\cos(2k\pi/n)-\sin((2j+1)k\pi/n)\sin(2k\pi/n)\\
&=\cos((2j+1)k\pi/n)(2\cos^2(k\pi/n)-1)-2\sin((2j+1)k\pi/n)\sin(k\pi/n)\cos(k\pi/n),
\end{align\*}
where we use standard trigonometric identities. Dividing by $\cos(k\pi/n)$ and summing, the desired sum is
\begin{equation\*}
2\sum\_{k=0}^{n-1}\left[\cos((2j+1)k\pi/n)\cos(k\pi/n)-\sin((2j+1)k\pi/n)\sin(k\pi/n)\right]-\sum\_{k=0}^{n-1}\frac{\cos((2j+1)k\pi/n)}{\cos(k\pi/n)}.
\end{equation\*}
The latter term is indeed $(-1)^{j+1}n$ by the inductive hypothesis, so it suffices to show that the first term vanishes. But the first sum is exactly
\begin{equation\*}
2\sum\_{k=0}^{n-1} \cos((2j+2)k\pi/n),
\end{equation\*}
where we again use trigonometric identities. This is zero, as it is the real part of the polynomial $1+x+x^2+\ldots+x^{n-1}$ evaluated at a (nontrivial) $n$th root of unity. This completes the induction.


Hope this helps!


