An \(n \times n\) matrix \(A\) over the field of complex numbers is said to be unitary if \(AA^* = I\) (\(A^*\) denotes the conjugate transpose of \(A\)). If \(A\) is unitary, show that \(|\det A| = 1\).
Solution
From \(AA^*=I\), \[ 1=\det I=\det(A)\det(A^*). \] Since \(A^*=\overline{A}^{\,t}\), \[ \det(A^*)=\overline{\det A}. \] Therefore \[ 1=(\det A)\overline{\det A}=|\det A|^2, \] so \(|\det A|=1\).