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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 $e_i=Af_i,\;i=1,2,\dots,n$? –  anon Oct 3 '11 at 8:34
    
No. $e_i = \sum_{j=1}^n a_{ij} f_j$ and $A= (a_{ij})$. –  shaye Oct 3 '11 at 11:36
    
Homan: how are they each an orthonormal basis for $V$ (which I assume is $n$-dimensional) if they're each only a single vector? –  anon Oct 3 '11 at 11:40
    
For real vector spaces and two bases of the same orientation this will be true. –  Mark Oct 3 '11 at 14:38

1 Answer 1

up vote 5 down vote accepted

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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8  
It may be worth noting that, however, $|\det A|=1$, that is, $A$ is a unitary matrix. –  joriki Oct 3 '11 at 9:00
2  
Indeed. Basically, the matrices Homan defines are the unitary matrices, i.e. $U(n)$ . –  Raskolnikov Oct 3 '11 at 9:01
4  
So in general the determinant of $A$ is a complex number of modulus $1$, right? –  shaye Oct 3 '11 at 11:18
    
That's indeed right. –  Raskolnikov Oct 3 '11 at 11:28

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