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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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    Do you mean $e_i=Af_i,\;i=1,2,\dots,n$?2011-10-03
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    No. $e_i = \sum_{j=1}^n a_{ij} f_j$ and $A= (a_{ij})$.2011-10-03
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    Homan: how are they each an orthonormal basis for $V$ (which I assume is $n$-dimensional) if they're each only a single vector?2011-10-03
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    For real vector spaces and two bases of the same orientation this will be true.2011-10-03

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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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    It may be worth noting that, however, $|\det A|=1$, that is, $A$ is a unitary matrix.2011-10-03
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    Indeed. Basically, the matrices Homan defines are the unitary matrices, i.e. $U(n)$ .2011-10-03
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    So in general the determinant of $A$ is a complex number of modulus $1$, right?2011-10-03
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    That's indeed right.2011-10-03