I know that unitary matrix A has |detA|=1. Previously, I thought, it means only 2 options: +1 and -1. But googling makes me think that, actually, det may be equal any number on unit circle. My problem is that I don't understand, why it is so.

Below, my reasoning: As solution to characteristic equation, complex eigenvalues should be in conjugate pair. By rule of multiplication of conjugate numbers, the product of such pair would be real number. Hence, determinant (equals the product of all eigenvalues) cannot be complex number.

Am I right? If not, could you provide an example of the matrix with complex determinant?

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    $\begingroup$ How about a $1\times1$ matrix $(u)$ for a non-real complex number $u$ on the unit circle (for instance $u=i$)? It is unitary. $\endgroup$ – Marc van Leeuwen May 3 '13 at 11:34

Characteristic equation will have roots occurring in conjugate pairs only if the co-efficient are real. In general, the co-efficient can be complex also. Thus, roots which do not come in conjugate pairs will come, thus determinant will be complex. Yet another one \begin{align} \begin{bmatrix}1 & 0 \\ 0 & i\end{bmatrix} \end{align}

  • $\begingroup$ Thank you very much for clarification and simple example. $\endgroup$ – DetDet May 3 '13 at 9:37

A simple example would be $$ \boldsymbol{A}= \begin{bmatrix} e^{i\pi/4} & 0 \\ 0 & e^{i\pi/4} \end{bmatrix}. $$ This matrix satisfies $\boldsymbol{A}^*\boldsymbol{A}=\boldsymbol{I}$ and has $\det\boldsymbol{A}=i$.


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