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 Post subject: Invertible Matrix
PostPosted: Thu Jun 09, 2016 9:22 pm 
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Let \( a =\dfrac{2\pi}{n} \). Prove that the matrix
$$ \begin{bmatrix}
1 & 1 &\cdots &1 \\
\cos a& \cos 2a &\cdots &\cos na \\
\cos 2a&\cos 4a &\cdots &\cos 2na \\
\vdots & \vdots & \ddots &\vdots \\
\cos(n-1)a &\cos 2(n-1)a &\cdots &\cos(n-1)na
\end{bmatrix}$$ is invertible.

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 Post subject: Re: Invertible Matrix
PostPosted: Thu Jun 09, 2016 9:24 pm 
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For \(a=\frac{2\pi}{n}\,,\quad n\in\mathbb{N}\) and for every \(k=0,1,\ldots,n-1\) the numbers \[{\rm{e}}^{m\frac{2k\pi}{n}\,{\rm{i}}}\,,\quad m=1,2,\ldots n-1\] are different \(n\)-th roots of \(1\). So \begin{align*}
\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{align*} We have that \begin{align*}
|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{align*} Adding the equations \begin{align*}
|{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{align*} we have that \[|{A_n}|=n!\,|{A_{1}}|=n!\,.\] So the matrix \(A_n\) is invertible for every \(n\in\mathbb{N}\).

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