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 Post subject: Invertible MatrixPosted: Thu Jun 09, 2016 9:22 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 828
Location: Larisa
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 MatrixPosted: Thu Jun 09, 2016 9:24 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 443
Location: Ioannina, Greece
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*}
\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}|\,.
|{A_n}|&=n|{A_{n-1}}|\\
n|{A_{n-1}}|&=n(n-1)|{A_{n-2}}|\\
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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