Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\det A = 1$?

share|cite|improve this question
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
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 $-1$.

share|cite|improve this answer
It may be worth noting that, however, $|\det A|=1$, that is, $A$ is a unitary matrix. – joriki Oct 3 '11 at 9:00
Indeed. Basically, the matrices Homan defines are the unitary matrices, i.e. $U(n)$ . – Raskolnikov Oct 3 '11 at 9:01
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.