## 2 Explicit Element Geometric Transformation
### Analysis of the Transformation
Theorem 2: _If the transformation parameter \(\lambda>0\), the eigenvalue \(\mu_{1}\) has the largest absolute value among all the eigenvalues of matrix \(A\)._  
Proof of Theorem 2: Since \(\lambda>0\), if \(\sum\limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi m}{n}=|\sum\limits_{m= 1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi m}{n}|\geq|\sum\limits_{m=1}^{ceil( \frac{n-1}{40})}\sin\frac{2\pi jm}{n}|\) is true for each value of \(j\) in the set \(j\in\{2,3,...,n-1\}\), we can easily get \(|\mu_{1}|\geq|\mu_{j}|\). Notice that \(f(n,n-j)=\sum\limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi(n-j)m}{n}=- \sum\limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi jm}{n}=-f(n,j)\), then the only condition needed to be satisfied is that \(|\sum\limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi m}{n}|\geq|\sum \limits_{m=1}^{ceil(\frac{n-1}{40})}\sin\frac{2\pi jm}{n}|\) is true for each value of \(j\) in the set \(j\in\{2,3,...,ceil(\frac{n-1}{20})\}\), which is illustrated in Figure 4 by plotting the absolute value of function \(f(n,j)\) for different pair \((n,j)\). Thus, it holds that \(|\mu_{1}|\geq|\mu_{j}|\) is true for each value of \(j\) in the set \(j\in\{2,3,...,n-1\}\). Inspired by the proof in Reference [28], the diagonalization of \(A\) will be used to explore the essence of the proposed transformation. Thus, the matrix composed of the associated eigenvectors in the columns, given by \(T=1/\sqrt{n}R\), \(R=(\
u_{0},\
u_{1},...,\
u_{n-1})\) is used, where the eigenvector \(\
u_{j}\) of matrix \(A\) is given by \(\
u_{j}=(1,\omega_{j},\omega_{j}^{2},...,\omega_{j}^{n-1})^{T}\)[31]. It holds that \(A=TDT^{*}\), where \(D\) is the diagonal matrix composed of the eigenvalues of matrix \(A\) and \(T^{*}\) is the conjugate transpose of matrix \(T\). With the decomposition of the \(n\times n\) identity matrix \(I\) in the form of \(I=\sum\limits_{k=0}^{n-1}I_{k}\), where \(I_{k}\) denotes the \(n\times n\) matrix with the only nonzero entry \((I_{k})_{k,k}\)=1, the transformation matrix after \(l\) cycles of the modified stretching operation can be written as:  
\[A^{l}=(TDT^{*})^{l}=TD^{l}T^{*}=\sum\limits_{k=0}^{n-1}\mu_{k}^{l}T_{k}T^{*}= \sum\limits_{k=0}^{n-1}\frac{\mu_{k}^{l}}{n}RI_{k}R^{*}. \tag{11}\]  
Therefore, the corresponding polygonal element after \(l\) cycles of the modified stretching operation can be decomposed in the following manner:  
\[p^{l}=A^{l}p^{0}=\sum\limits_{k=0}^{n-1}\mu_{k}^{l}\underbrace{\frac{1}{n}RI_ {k}R^{*}p^{0}}_{e_{k}}=\sum\limits_{k=0}^{n-1}\mu_{k}^{l}e_{k}. \tag{12}\]  
The so-called eigenpolygons \(e_{k}\) do depend only on the initial polygon element \(p^{0}\). Examples of such decompositions are depicted in Figure 5 for a triangular, quadrilateral, pentagonal and hexagonal element, where the element nodes are marked with different colors in order to denote the element orientation. Figure 4: The absolute value of function \(f(n,j)\) for different pair \((n,j)\). Figure 5: Eigenpolygon decomposition of a triangular, quadrilateral, pentagonal and hexagonal element. The first two eigenpolygons and their corresponding eigenvalues play an important role in determining the characters of element geometric transformation. Due to \(\mu_{0}=1\), the first eigenpolygon is given by:  
\[e_{0}=\frac{1}{n}RI_{0}R^{*}p^{0}=\frac{1}{n}Ip^{0}=\frac{1}{n}(\sum_{k=0}^{n-1 }p_{k}^{0})(1,1,\cdots,1)^{T}, \tag{13}\]  
which means all nodes of the polygonal element are placed at the centroid of the initial polygonal element \(p^{0}\) during the transformation. The second eigenpolygon represents the counterclockwise oriented regular polygonal element, due to the expression:  
\[e_{1}=\frac{1}{n}RI_{1}R^{*}p^{0}=\frac{1}{n}\left( \begin{array}{cccc}1& \omega_{1}^{1}&\cdots&\omega_{1}^{n-1}\\ \omega_{1}^{1}&\omega_{1}^{2}&\cdots&1\\ \vdots&\vdots&\ddots&\vdots\\ \omega_{1}^{n-1}&1&\cdots&\omega_{1}^{n-2}\end{array}