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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 $\det A = 1$?

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    For real vector spaces and two bases of the same orientation this will be true.2011-10-03

1 Answers 1

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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 $-1$.

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    That's indeed right.2011-10-03