Invertible matrix

Prove that the matrix
$$\begin{bmatrix} 1 & 1 &\cdots  &1 \\  \cos a& \cos 2a &\cdots  &\cos na \\  \cos 2a&\cos 4a  &\cdots  &\cos 2na \\  \vdots & \vdots  & \ddots  &\vdots  \\ \cos(n-1)a &\cos 2(n-1)a  &\cdots  &\cos(n-1)na \end{bmatrix}$$

whereas \( a= \dfrac{2\pi}{n} \) is invertible.

Solution:



For \( a=\dfrac{2\pi}{n} , n \in \mathbb{N} \) and for every \( k=0, 1, 2, \dots, n-1 \) the numbers:
$$e^{m\frac{2k\pi}{n}i} , \; m=1, 2, \dots, n-1$$

are different \( n \) th roots of \( 1 \). Hence:

$$\begin{aligned} \mathop{\sum}\limits_{k=0}^{n-1}{{\rm{e}}^{m\frac{2k\pi}{n}\,{\rm{i}}}}=0\quad&\Rightarrow\quad\mathop{\sum}\limits_{k=0}^{n-1}{\Re\bigl({{\rm{e}}^{m\frac{2k\pi}{n}\,{\rm{i}}}}\bigr)}=0\\ &\Rightarrow\quad\mathop{\sum}\limits_{k=0}^{n-1}{\cos\bigl({m\tfrac{2k\pi}{n}}\bigr)}=0\\ &\Rightarrow\quad\mathop{\sum}\limits_{k=0}^{n-1}{\cos({mka})}=0\quad (1)\,,\quad m=1,2,\ldots n-1 \end{aligned}$$

Therefore:
$$ \begin{aligned} |A_n|&=\begin{vmatrix} 1 & 1 &\cdots&1 &1 \\ \cos a& \cos 2a &\cdots &\cos (n-1)a &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a &\cos 2na \\ \vdots & \vdots & \ddots &\vdots &\vdots \\ \cos(n-2)a &\cos 2(n-2)a &\vdots &\cos (n-2)(n-1)a &\cos(n-2)na \\ \cos(n-1)a &\cos 2(n-1)a &\cdots &\cos (n-1)(n-1)a &\cos(n-1)na \end{vmatrix}\\\\ &\stackrel{R_{n}\to\sum_{k=1}^{n}{R_k}}{=\!=\!=\!=\!=\!=\!=\!=\!=}\begin{vmatrix} 1 & 1 &\cdots&1 &1 \\ \cos a& \cos 2a &\cdots &\cos (n-1)a &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a &\cos 2na \\ \vdots & \vdots & \ddots &\vdots &\vdots \\ \cos(n-2)a &\cos 2(n-2)a &\vdots &\cos (n-2)(n-1)a &\cos(n-2)na \\ \mathop{\sum}\limits_{k=0}^{n-1}{\cos({ka})} &\mathop{\sum}\limits_{k=0}^{n-1}{\cos({2ka})} &\cdots &\mathop{\sum}\limits_{k=0}^{n-1}{\cos({(n-1)ka})} &\mathop{\sum}\limits_{k=0}^{n-1}{\cos({nka})} \end{vmatrix}\\\\ &\stackrel{(1)}{=\!=}\begin{vmatrix} 1 & 1 &\cdots &1 &1 \\ \cos a& \cos 2a &\cdots &\cos (n-1) a &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a &\cos 2na \\ \vdots & \vdots & \ddots &\vdots &\vdots\\ \cos(n-2)a &\cos 2(n-2)a &\cdots &\cos (n-2)(n-1)a &\cos(n-2)na \\ 0 &0 &\cdots & 0 &\mathop{\sum}\limits_{k=0}^{n-1}{1} \end{vmatrix}\\\\ &=\begin{vmatrix} 1 & 1 &\cdots &1 &1 \\ \cos a& \cos 2a &\cdots &\cos (n-1) a &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a &\cos 2na \\ \vdots & \vdots & \ddots &\vdots &\vdots\\ \cos(n-2)a &\cos 2(n-2)a &\cdots &\cos (n-2)(n-1)a &\cos(n-2)na \\ 0 &0 &\cdots & 0 & n \end{vmatrix}\\\\ &=n\begin{vmatrix} 1 & 1 &\cdots &1 &1 \\ \cos a& \cos 2a &\cdots &\cos (n-1) a &\cos na \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a &\cos 2na \\ \vdots & \vdots & \ddots &\vdots &\vdots\\ \cos(n-2)a &\cos 2(n-2)a &\cdots &\cos (n-2)(n-1)a &\cos(n-2)na \\ 0 &0 &\cdots & 0 & 1 \end{vmatrix}\\\\ &=n\begin{vmatrix} 1 & 1 &\cdots &1\\ \cos a& \cos 2a &\cdots &\cos (n-1) a \\ \cos 2a&\cos 4a &\cdots &\cos 2(n-1)a \\ \vdots & \vdots & \ddots &\vdots\\ \cos(n-2)a &\cos 2(n-2)a &\cdots &\cos (n-2)(n-1)a \end{vmatrix}\\\\ &=n\,|A_{n-1}| \end{aligned}$$

So:
$$\begin{aligned} |{A_n}|&=n|{A_{n-1}}|\\ n|{A_{n-1}}|&=n(n-1)|{A_{n-2}}|\\ \vdots\quad &\quad\quad \quad\vdots\\ n(n-1)\cdots5\cdot4\,|{A_{3}}|&=n(n-1)\cdots4\cdot3\,|{A_{2}}|\\ n(n-1)\cdots4\cdot3\,|{A_{2}}|&=n(n-1)\cdots3\cdot2\,|{A_{1}}| \end{aligned}$$

and by adding the equations we get that:
$$|{A_n}|=n!\,|{A_{1}}|=n!$$

leading us to that the matrix is indeed invertible for all \( n \in \mathbb{N} \).