# 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. Aug 11, 2017 at 13:01
Hint: It is useful to note that with matrix multiplication (just as with scalar multiplication), we have $\overline{AB}= \bar A \bar B$.