The eigenvalues of the matrix $\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)_{k,l=1}^n$

Question

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?

Answer 1

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) 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!