## 2 Explicit Element Geometric Transformation
### Analysis of the Transformation  
In this subsection, the coordinate of node \(p_{k}\) is represented by \(p_{k}=x+iy\). The reason for doing this is to take advantage of the properties of circulant determinants. Using a \(n\times n\) matrix \(A\) with zero-based indexes \(i,j\in\{0,1,...,n-1\}\) and a \(n\times 1\) vector \(p^{l}=(p_{0}^{l},p_{1}^{l},...,p_{n-1}^{l})\in C^{n}\) representing the polygonal element with \(n\) nodes after \(l\) cycles of transformation, the modified stretching operation governed by Equation (3) can be written in the form of \(p^{l+1}=Ap^{l}\), where \(A\) is a circulant Hermitian matrix given by:  
\[A=\left( \begin{array}{ccccc}a_{0}&a_{n-1}&\cdots&a_{2}&a_{1}\\ a_{1}&a_{0}&a_{n-1}&\cdots&a_{2}\\ \vdots&a_{1}&a_{0}&\ddots&\vdots\\ a_{n-2}&\cdots&\ddots&\ddots&a_{n-1}\\ a_{n-1}&a_{n-2}&\cdots&a_{1}&a_{0}\end{array} \right) \tag{5}\]  
and the component can be written as:  
\[a_{m}=\left\{ \begin{array}{ll}1&if\ m=0\\ -\lambda i&if\ n-m=1,2,...,ceil(\frac{n-1}{4.0})\\ \lambda i&if\ m=1,2,...,ceil(\frac{n-1}{4.0})\\ 0&\textit{otherwise}.\end{array} \right. \tag{6}\]  
In the case of triangular elements this yields:  
\[A=\left( \begin{array}{ccc}1&-\lambda i&\lambda i\\ \lambda i&1&-\lambda i\\ -\lambda i&\lambda i&1\end{array} \right) \tag{7}\]  
and  
\[A=\left( \begin{array}{ccccc}1&-\lambda i&0&\lambda i\\ \lambda i&1&-\lambda i&0\\ 0&\lambda i&1&-\lambda i\\ -\lambda i&0&\lambda i&1\end{array} \right) \tag{8}\]  
in the case of quadrilateral elements, respectively. With this, the basic properties of the transformation presented in the previous subsection can be analyzed for arbitrary polygonal elements. **Theorem 1**. : _The transformation preserves the centroid of a polygonal element, that is \(\frac{1}{n}\sum\limits_{k=0}^{n-1}p_{k}^{l+1}=\frac{1}{n}\sum\limits_{k=0}^{n -1}p_{k}^{l}\)._  
**Proof of Theorem 1. **  
\(\frac{1}{n}\sum\limits_{k=0}^{n-1}p_{k}^{l+1}=\frac{1}{n}\sum\limits_{k=0}^{n -1}\sum\limits_{j=0}^{n-1}A_{kj}p_{j}^{l}=\frac{1}{n}\sum\limits_{j=0}^{n-1} \sum\limits_{k=0}^{n-1}A_{kj}p_{j}^{l}\)  
\(=\frac{1}{n}\sum\limits_{j=0}^{n-1}(1+\sum\limits_{m=1}^{ceil(\frac{n-1}{4.0}) }(a_{m}+a_{n-m}))p_{j}^{l}\)  
\(=\frac{1}{n}\sum\limits_{j=0}^{n-1}(1+\sum\limits_{m=1}^{ceil(\frac{n-1}{4.0}) }(\lambda i-\lambda i))p_{j}^{l}\)  
\(=\frac{1}{n}\sum\limits_{j=0}^{n-1}p_{j}^{l}=\frac{1}{n}\sum\limits_{k=0}^{n-1} p_{k}^{l}\). Since A is a circulant Hermitian matrix, the associated eigenvalues are real-valued. Due to the circulant structure, the eigenvalues \(\mu_{k}\), \(k\in\{0,1,...,n-1\}\) of \(A\) can be easily obtained by multiplying the discrete Fourier matrix \(R\) defined by \(R_{u,v}=\omega_{uv}=exp(\frac{2\pi iuv}{n})\) with the transposed first row of\(A\)[31], where \(i\) is the imaginary unit. Hence, for arbitrary \(n\geq 3\) the corresponding eigenvalues of \(A\) are given by:  
\[\mu_{j}=a_{0}+a_{n-1}\omega_{j}+a_{n-2}\omega_{j}^{2}+\cdots+a_{1}\omega_{j}^{n- 1},\quad j=0,1,...,n-1. \tag{9}\]  
Using the fact that \(\omega_{j}^{n-m}=exp(\frac{2\pi(n-m)ji}{n})=exp(\frac{-2\pi mji}{n})=\bar{ \omega_{j}^{m}}\) and \(a_{n-m}=a_{m}\), the eigenvalues can be written as:  
\[\mu_{j}=1+\sum\limits_{m=1}^{ceil(\frac{n-1}{40})}a_{m}\omega_{j}^{n-m}+a_{n- m}\omega_{j}^{m}=1+2\lambda\sum\limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2 \pi jm}{n},\quad j=0,1,...,n-1. \tag{10}\]  
In particular, the above formula shows that \(\mu_{0}=1\) does not depend on the choice of scaling factor \(\lambda\). Moreover, it holds that \(|\mu_{1}|>|\mu_{j}|\) is true for each value of \(j\) in the set \(j\in\{2,3,...,n-1\}\).