Let \(M\) be a member of the complex orthogonal group, \(O(n, C)\). Show that \(M^t\), \(\overline{M}\), and \(M^* = \overline{M}^t\) also belong to \(O(n, C)\).
Solution
For \(M\in O(n,\mathbb C)\), \[ M^tM=I. \]
1. The transpose \(M^t\) is orthogonal.
The equation \(M^tM=I\) shows that \(M\) is invertible and \[ M^{-1}=M^t. \] Hence also \[ MM^t=I. \] Therefore \[ (M^t)^tM^t=MM^t=I, \] so \[ M^t\in O(n,\mathbb C). \]
2. The entrywise conjugate \(\overline M\) is orthogonal.
Take complex conjugates of the defining equation: \[ \overline{M^tM}=\overline I. \] Because conjugation commutes with transpose and multiplication, \[ (\overline M)^t\overline M=I. \] Thus \[ \overline M\in O(n,\mathbb C). \]
3. The adjoint \(M^*=\overline M^t\) is orthogonal.
By step 2, \(\overline M\) belongs to the orthogonal group. Applying step 1 to \(\overline M\) shows that \[ \overline M^t=M^* \] also lies in \(O(n,\mathbb C)\).
Hence \[ \boxed{M^t,\ \overline M,\ M^*\in O(n,\mathbb C).} \]