Let $V$ be a finite dimensional vector space over $\mathbf{C}$ with a hermitian inner product. Let $e=(e_1,\ldots,e_n)^t$ and $f=(f_1,\ldots,f_n)^t$ be orthonormal bases for $V$.
There is a matrix $A$ such that $e =A f$.
Is $\det A = 1$?
Let $V$ be a finite dimensional vector space over $\mathbf{C}$ with a hermitian inner product. Let $e=(e_1,\ldots,e_n)^t$ and $f=(f_1,\ldots,f_n)^t$ be orthonormal bases for $V$.
There is a matrix $A$ such that $e =A f$.
Is $\det A = 1$?
No, as a counterexample, take the matrix
$A = \left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right) \; .$
And take for $f$ the standard basis
$f_1=\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \; , \; f_2=\left(\begin{array}{c} 0 \\ 1 \end{array}\right) \; .$
Clearly, the determinant of $A$ is $-1$.