# Prove that $A^t$ complex matrix is unitary.

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?

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.
• @PedroGomes That's because a matrix is unitary if and only if its columns have norm $1$ and any two distinct columns are orthogonal. – José Carlos Santos Aug 11 '17 at 13:01
Hint: It is useful to note that with matrix multiplication (just as with scalar multiplication), we have $\overline{AB}= \bar A \bar B$.