Linear Independence for Vectors of Cosine Values

Question

For every integer $n\geq3$ define an $n\times n$ matrix $A$ using the following row vector as the first row: $(c_0,c_1,c_2,\ldots, c_{n-1})$ where $c_k=\cos (2\pi k/n)$.

The subsequent rows are obtained as cyclic shifts. This matrix seems to be always of rank 2.

Actually I was working with some families of representations of dihedral groups, arising in another context and was computing the degrees of the representations. That calculation implied this matrix should be of rank 2. But definitely there must be some direct way of doing it.

Can anyone tell me how to accomplish it?

Answer 1

The identity

$\cos((n+1)x) =2\cos(x)\cos(nx)-\cos((n-1)x) $ shows that each column is a linear combination of the previous two columns, so the rank is 2.

This also holds for $\sin$.

Answer 2

A typical cyclic shift is $$(c_k,c_{k+1},\ldots,c_{k+n-1})$$ noting that $c_n=c_0$, $c_{n+1}=c_1$ etc. This is the real part of the complex vector $$(\zeta^k,\zeta^{k+1},\ldots,\zeta^{k+n-1})$$ where $\zeta=\exp(2\pi i/n)$. Thus \begin{align} (c_k,c_{k+1},\ldots,c_{k+n-1}) &=\frac12(\zeta^k,\zeta^{k+1},\ldots,\zeta^{k+n-1}) +\frac12(\zeta^{-k},\zeta^{-k-1},\ldots,\zeta^{-k-n+1})\\ &=\frac{\zeta^k}2(1,\zeta,\ldots,\zeta^{n-1}) +\frac{\zeta^{-k}}2(1,\zeta^{-1},\ldots,\zeta^{-n+1}). \end{align} Each row is a linear combination of the two vectors $(1,\zeta,\ldots,\zeta^{n-1})$ and $(1,\zeta^{-1},\ldots,\zeta^{-n+1})$ so the matrix has rank $\le2$.

Answer 3

A third way to realize this is to treat it as matrix representation of convolution. Such a matrix is circulant and has always an eigensystem of complex exponentials. So we can even explicitly calculate the linear combination with Euler's formulas for relation between complex exponential and sin and cos:

$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}\\\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$