Definition 1. Let $A \in M_{m \times n}(F)$. We define the adjoint or conjugate transpose of $A$ to be the $n \times m$ matrix $A^*$ such that $(A^*)_{ij} = \overline{A_{ji}}$ for all $i, j$.
Theorem 1
Theorem 1.Let $A, B \in M_{m \times n}(F)$, and let $C \in M_{n \times p}$. Then (a)$(A+B)^* = A^* + B^*$ (b)$(cA)^* = \overline{c} A^*, \forall c \in F$. (c)$(AC)^* = C^*A^*.$ (d)$A^{**} = A$. (e)$I^* = I$.
Proof. It is immediate from Theorem 4. $\blacksquare$
Theorem 2
Theorem 5. Let $A \in M_{m \times n}(F)$. Then rank($A^*$) = rank($A^*A$) = rank($AA^*$) = rank($A$).
Proof. Clearly $N(L_{A^*A}) \supseteq N(L_A)$. If $A^*Ax = \mathbf{0}$, then $\langle A^*Ax, x \rangle = \langle Ax, Ax \rangle$. Thus $Ax = \mathbf{0}$. This means that $N(L_{A^*A}) = N(L_A) \iff $ rank($A^*A$) = rank($A$). Similarily, we can show that rank($AA^*$) = rank($A^*$). By Theorem 1, rank($A$) = rank($A^*A$) $\leq$ rank($A^*$) and rank($A^*$) = rank($AA^*$) $\leq$ rank($A$). Thus rank($A^*$) = rank($A$). $\blacksquare$
Theorem 3
Theorem 3. Let $A \in M_{n \times n}(F)$. Then $\det(A^*) = \overline{\det(A)}$.
Proof. Use the induction on $n$. If $n= 1$, the result is trivial. Suppose that the statement is true for $n - 1$ where $n -1 \geq 1$. Then we have $$\det(A^*) = \sum_{j=1}^n (-1)^{i+j} A^*_{ij} \det(\widetilde{A^*_{ij}}) = \sum_{j=1}^n (-1)^{i+j} \overline{A_{ji}} \det(\widetilde{A_{ij}}^*) \\ = \sum_{j=1}^n \overline{(-1)^{i+j} A_{ji} \det(\widetilde{A_{ij}})} = \overline{\sum_{j=1}^n (-1)^{i+j} A^t_{ij} \det(\widetilde{A^t_{ij}})} = \overline{\det(A^t)} = \overline{\det(A)}.$$ $\blacksquare$