ATRACTOR

1st example (bisection)

In this case, \(\alpha_0=\alpha_1=\frac{1}{2}\) and \(\alpha_j=0\) for \(j>1\), so that matrix \(A\) has exactly \(n\) distinct complex eigenvalues, given by \[\lambda_t\;=\;\frac{1+\omega^t}{2} \] for any \(t\in \{0,...,n-1\}\), where \(\omega\) denotes a primitive nth root of unity. Then \[\lambda_t\;=\;0\;\Longleftrightarrow\;\omega^t\;=\;-1 \;\Longleftrightarrow\; \frac{2\pi t}{n}\;=\;\pi \;\Longleftrightarrow\; 2t\;=\;n \] Thus, if \(n\) is odd, the eigenvalues are all nonzero, the determinant of \(A\) is nonzero and there is a bijection between the initial points and the new points constructed by bisection. Otherwise, if \(n\) is even, \(\lambda _{n/2}=0\) and the determinant of \(A\) is 0. Hence, in this case, the correspondence between the initial points and the new points constructed by bisection is not bijective.

2nd example (trisection)

In this case, \(\alpha_0=\frac{1}{3}\), \(\alpha_1=\frac{2}{3}\) and \(\alpha_j=0\) for \(j>1\), so that \(A\) has exactly \(n\) distinct complex eigenvalues, given by \[\lambda_t\;=\;\frac{1+2\omega^t}{3} \] for any \(t\in \{0,...,n-1\}\), where \(\omega\) is a primitive nth root of unity. Then \[\lambda_t\;=\;0\;\Longleftrightarrow\;\omega^t\;=\;\frac{-1}{2} \] which is an impossible equation for \(t\in \{0,...,n-1\}\). Hence, the eigenvalues are all nonzero and thus the deteminant of \(A\) is nonzero and the correspondence between the initial points and the new points constructed by trisection is a bijection.

3rd example

In this case, \(\alpha_{0}=1-p\) and \(\alpha_{1}=p\), for some \(p\) between 0 and 1, and \(\alpha_{j}=0\) for \(j>1\). Matrix \(A\) has exactly \(n\) distinct complex eigenvalues, given by \[\lambda_t\;=\;(1-p)+p\omega^t \] for every \(t\in \{0,...,n-1\}\), where \(\omega\) is a primitive nth root of unity. Then \[\lambda_t\;=\;0\;\Longleftrightarrow\;\omega^t\;=\; \frac{1-p}{p} \] which is an impossible equation for \(t\in \{0,...,n-1\}\) whenever \(\frac{1-p}{p}\neq 1\), that is, is \(p\neq \frac{1}{2}\) (if \(p=\frac{1}{2}\), then \(\alpha_0=\alpha_1=\frac{1}{2}\) and \(\alpha_{j}=0\) for \(j > 1\) already analysed in the first example). If \(p\neq \frac{1}{2}\), then the eigenvalues are all nonzero and the determinant of \(A\) is nonzero (indeed, one may show that the determinant of \(A\) is equal to \((1-p)^{n}-(-p)^{n}\)) and the correspondence between the initial points and the new points constructed by this procedure is a bijection. Note that for \(p=\frac{2}{3}\), then \(\alpha_0=\frac{1}{3}\), \(\alpha_1=\frac{2}{3}\), and \(\alpha_{j}=0\) for \(j> 1\), a case already analysed in the second example.

4th example

In this case, \(\alpha_1=\alpha_{n-1}=\frac{1}{2}\) and \(\alpha_{j}=0\) for \(j\neq 1,n-1\), and \(A\) has \(n\) complex eigenvalues (not necessarily distinct), given by \[\lambda_t\;=\;\frac{\omega^t+\omega^{(n-1)t}}{2} \;=\;\frac{\omega^t+\omega^{-t}}{2}\;=\;\cos\frac{2\pi t}{n} \] for every \(t\in \{0,...,n-1\}\), \(\omega\) denoting primitive nth root of unity. Then \[\lambda_t\;=\;0\;\Longleftrightarrow\;\omega^t\;=\;-\omega^{-t} \;\Longleftrightarrow\; \omega^{2t}\;=\;-1 \;\Longleftrightarrow\; \frac{4\pi t}{n}\;=\;\pi \;\Longleftrightarrow\; 4t\;=\;n \] Hence, if \(n\) is not a multiple of 4, the eigenvalues are all nonzero and the determinant of \(A\) is nonzero and there is a bijection between the initial points and the new points constructed by this procedure. Otherwise, if \(n\) is a multiple of 4, then \(\lambda_{n/4}=0\) and the determinant of \(A\) is 0, whereby the correspondence between the initial points and the new points constructed by this process is not bijective.