Determinant of Adjoint matrix Suppose $A$ is an $n \times n$, $n > 0$ matrix with complex entries, and $A^*$ is the adjoint matrix of $A$. Show that $\det(A^*) = \overline{\det(A)}$.
My attempt:
By definition,
$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$
Similarly for $A^*$, by complete expansion,
$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{\sigma_1 1} \cdots A_{\sigma_n n}}$$
Consider the product $A_{\sigma_1 1} \cdots A_{\sigma_n n}$. If $j = \sigma_i$, then $A_{\sigma_i i} = A_{j \sigma^{-1}_{j}}$. Since $\sigma$ is a bijection, both $i, j$ range over $1, ..., n$. So,
$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma^{-1}) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$
However, $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\sigma^{-1})$. So:
$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma^{-1}_1} \cdots A_{n \sigma^{-1}_n}}$$
As before, $\sigma^{-1}$ ranges over all degree n permutations just as $\sigma$ does. Thus,
$$\det(A^*) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma_1} \cdots A_{n \sigma_n}}$$
$$\det(A^*) = \overline{\sum_{\sigma} (\operatorname{sgn} \sigma) A_{1 \sigma_1} \cdots A_{n \sigma_n}} = \overline{\det(A)}$$
QED.
Is this proof correct? I'm not sure because I derived it from the proof involving real entries. Any assistance is much appreciated.
 A: Why put everything together? It is easier to use some known results.
For $A\in \mathbb F^{n\times n}$, we have
$$\det A=\det A^T,$$
where $F$ is any field and $A^T$ is the transpose of $A$.
For $A\in \mathbb C^{n\times n}$, we have
$$\det \overline{A}=\overline{\det A}.$$
The reason is exactly as you wrote.
$$\begin{aligned}
\det \overline{A}&=\sum_{\sigma} \prod_{i} \overline{a_{i\sigma(i)}}\\
&=\overline{\sum_{\sigma} \prod_{i} a_{i\sigma(i)}}\\
&=\overline{\det A}.
\end{aligned}$$
The adjoint matrix you mean, is just $A^*=\overline{A^T}$.
It follows that
$$\det \overline{A^T}=\overline{\det A^T}=\overline{\det A}.$$
It is not necessary to proof $\det{A^T}=\det{A}$ again.
A: In your proof, when you apply the change of variables $\sigma^{-1}\mapsto\sigma$ in the sum, the coefficient $\operatorname{sgn}(\sigma)$ should change to $\operatorname{sgn}(\sigma^{-1})$. In fact, you don't need to artificially use $\operatorname{sgn}(\sigma^{-1})=\operatorname{sgn}(\sigma)$ before doing the change of variables.
A different proof would be: By definition $A^*:=(\overline A)^T$, where $\overline A$ denotes the componentwise conjugation. Then $\det(A^*)=\det((\overline A)^T)=\det(\overline A)$, and then, by linearity of conjugation,
$$\det(\overline A) = \sum_{\sigma} (\operatorname{sgn} \sigma) \overline {A_{1 \sigma_1} \cdots A_{n\sigma_n}} = \overline {\sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}}=\overline{\det A}$$
