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Let $V$ be a vector space over field $\mathbb{C}$ with scalar product $\langle,\rangle$, in which $v\in V$. Let $A$ be an unitary complex matrix, so that $\langle Av,Av\rangle=\langle v,v\rangle$. Is it true $\bar A=A?$

I know $\langle Av,Av\rangle=\langle v,\bar A^tAv\rangle$, so I can infer $\bar A^tA=I$. Therefore $A^{-1}=\bar A^t\implies(A^{-1})^{-1}=(\bar A^t)^{-1}\implies$...The problem rises because I cannot eliminate $t$, since it is not true that only the transpose is the inverse.

However in exercises:

Let A be a complex unitary matrix.

(a) Show that $A^t$ is unitary.

Real case:

$(A^t)^t=A=(A^t)^{-1}$

I tried to solve it as for the real matrix:

Complex case:

$\overline{(A^t)^t}=\bar A=?$ I cannot complete it because of the conjugate

Question:

How can I prove complex matrix $A^t$ is unitary?

Thanks in advance!

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The matrix $A$ is unitary if and only if its inverse is $\overline A^t$. And if $\overline A^t$ is unitary, then its conjugate (which is $A^t$) is unitary too.

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  • $\begingroup$ I would like to prove that the conjugate of an unitary matrix is an unitary matrix. Thanks for the answer! $\endgroup$ – Pedro Gomes Aug 11 '17 at 12:36
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    $\begingroup$ @PedroGomes That's because a matrix is unitary if and only if its columns have norm $1$ and any two distinct columns are orthogonal. $\endgroup$ – José Carlos Santos Aug 11 '17 at 13:01
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Hint: It is useful to note that with matrix multiplication (just as with scalar multiplication), we have $\overline{AB}= \bar A \bar B$.

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  • $\begingroup$ How shall I use the multiplication to complete my proof? $\endgroup$ – Pedro Gomes Aug 11 '17 at 12:36
  • $\begingroup$ Using this fact, show that the conjugate of a unitary matrix is unitary. $\endgroup$ – Omnomnomnom Aug 11 '17 at 15:00

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