Find a matrix in \(O(3, C)\) whose first row is \((2i, 2i, 3)\).
Solution
The prescribed row already has transpose norm one: \[ (2i)^2+(2i)^2+3^2=-4-4+9=1. \] Choose \[ r_1=(2i,2i,3), \qquad r_2=\frac1{\sqrt2}(1,-1,0). \] Then \[ r_1r_2^t=0, \qquad r_2r_2^t=1. \]
To find a third row orthogonal to \(r_2\), take its first two entries equal. The condition of orthogonality to \(r_1\) then gives \[ 4ia+3z=0. \] Choosing \(a=3/\sqrt2\) gives \(z=-2\sqrt2\,i\), and hence \[ r_3=\left(\frac3{\sqrt2},\frac3{\sqrt2},-2\sqrt2\,i\right). \] Now \[ r_3r_3^t =\frac92+\frac92+(-2\sqrt2\,i)^2 =9-8 =1, \] and directly \[ r_1r_3^t=r_2r_3^t=0. \]
Therefore the matrix \[ \boxed{ M= \begin{pmatrix} 2i&2i&3\\ \frac1{\sqrt2}&-\frac1{\sqrt2}&0\\ \frac3{\sqrt2}&\frac3{\sqrt2}&-2\sqrt2\,i \end{pmatrix}} \] has mutually transpose-orthonormal rows, so \[ MM^t=I. \] Thus \(M\) is invertible with \(M^{-1}=M^t\), and consequently \[ M^tM=I. \] Hence \[ M\in O(3,\mathbb C), \] and its first row is the required \((2i,2i,3)\).