Let \[ y_j = \sum_{k=1}^{n} M_{jk}x_k \] where \(M\) is a member of \(O(n, C)\). Show that \[ \sum_j y_j^2 = \sum_j x_j^2. \]
Solution
Let \[ x=\begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix}, \qquad y=\begin{pmatrix}y_1\\ \vdots\\ y_n\end{pmatrix}. \] The defining equations for the \(y_j\) say exactly that \[ y=Mx. \] Since \(M\in O(n,\mathbb C)\), \[ M^tM=I. \] Therefore \[ \begin{aligned} \sum_{j=1}^n y_j^2 &=y^ty\\ &=(Mx)^t(Mx)\\ &=x^tM^tMx\\ &=x^tx\\ &=\sum_{j=1}^n x_j^2. \end{aligned} \] Hence \[ \boxed{\sum_j y_j^2=\sum_j x_j^2.} \]