1 problem found
A set of \(n\) distinct vectors \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n},\) where \(n\geqslant2\), is called regular if it satisfies the following two conditions:
Solution: \begin{align*} && \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\ \Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\ &&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\ &&&= (n-1)\beta + \alpha^2 \tag{i} \\ \Rightarrow && (n-1)\beta &= -\alpha^2 \end{align*} Suppose we have \(\mathbf{a}_j = \binom{x}{y}\), \(j \neq 1\) then \(x = \beta\). We also must have \(\beta^2 + y^2 = 1\), so there are at most two values for \(y\), ie two extra vectors. ie \(n = 2, 3\). If \(n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1\). If \(n = 3\) \begin{align*} && \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\ \Rightarrow && \beta = -1/2 \\ \Rightarrow && y &= \pm \frac{\sqrt{3}}{2} \end{align*} Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\(, \)\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$ (since we need \(\mathbf{a}_2 \cdot \mathbf{a}_2 = 1\)). \(\beta = \cos \theta\)). We can have \(\cos \theta = - 1\). Suppose we have \(\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}\), so \(x = \cos \theta\), and \(y^2 + z^2 = \sin^2 \theta\), so we can write it as: \(\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix}\). We must also have \(\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta\), so \(\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}\). Therefore there is one value for \(\cos \phi\), so at most two values for \(\sin \phi\), Therefore we can have either \(2, 3,4\) or \(5\) different values in the set. \(n = 2\), we've already handled. If \(n = 3\), then \(\beta = -\frac12\), \(\cos \phi = -1\), so we can only have two different values for \(\sin \theta\), ie: \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix} \right \}\) Finally, if \(n = 4\), we have \(\beta = -\frac13\), \(\cos \phi = \frac{-1/3}{2/3} = -\frac12\). \(\sin \theta = \pm \frac{\sqrt{3}}{2}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) If \(n = 5\), then \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)