Show that for a square matrix $A$ the equality $\det(A^*) = \overline{\det(A)}$ holds

Question

My idea is to start with the definition of the adjoint:
$$\det(A^*) = \det(\overline{A^T}) = \overline{\det(A^T)} = \overline{\det(A)}. $$ However, I do not have a proper reasoning for this step: $\det(\overline{A^T}) = \overline{\det(A^T)}$, so I doubt if I am on the right track. Can anyone provide a way to prove this?

Answer 1

You are on the right track. About the step you are asking for, by the properties of complex conjugation: $\overline{a+b}=\overline{a}+\overline{b}$ and $\overline{ab}=\overline{a}\overline{b}$. Therefore, recalling the definition of determinant of a $n\times n$ matrix $A$, we have that $$\overline{\det(A)} =\overline{\sum_{\sigma \in S_n} \text{sgn}(\sigma)\prod_{k=1}^n a_{k,\sigma(k)}}= \sum_{\sigma \in S_n} \text{sgn}(\sigma)\prod_{k=1}^n \overline{a_{k,\sigma(k)}}=\det(\overline{A}).$$ Notice that $\overline{\text{sgn}(\sigma)}=\text{sgn}(\sigma)$ because $\text{sgn}(\sigma)$ is $\pm 1$ (a real number).

Answer 2

Here is another way to think about this. The determinant of a ring $R$ is the unique map $\det\colon M_n(R)\to R$ which is multilinear alternating when viewed as a function of the column vectors $\mathbf v_1,\dots,\mathbf v_n$, and such that $D(I_n)=1$. (See Lang's Algebra XIII.4)

Given this characterization, it is enough to show that the map $A\mapsto \overline{\det(\overline A)}$ also satisfies the necessary properties. This shows, by uniqueness, that $\overline{\det(\overline A)}=\det(A)$, or $\det(\overline A)=\overline{\det(A)}$.

An advantage of this viewpoint is that one doesn't explicitly have to work with long formulas, and can instead just check that all the simple properties hold.

P.S.: This is a much more general phenomenon, that for any ring homomorphism $\sigma\colon R\to R'$ and any matrix $A\in M_n(R)$, we have $\sigma(\det A)=\det(\sigma(A))$.