Let $X,Y$ be vector space over $\mathbb C$
Let $B ,B_1$ be an orthonormal basis of $X,Y$ respectively And $B_1^*,B_2^*$ be basis of dual space $X^*,Y^*$ $$\begin{array}{c} Y & \xrightarrow{A^\dagger} & X \\ I_Y \downarrow ~~~~~&& ~~~~\downarrow I_X \\ Y^* & \xrightarrow{A^*} & X^* \end{array}$$ Such that the diagram commutes Now, matrix of dual map is Transpose of matrix of $A$. But matrix of adjoint matrix is conjugate transpose .
Could someone explain me why this happens. Why one matrix is transpose and another is conjugate transpose?In such cases how diagram commutes
Let $X$ have inner product $\langle -, - \rangle: X \times X \to \mathbb{C}$. I will suppose that the inner product is antilinear in its first argument and linear in its second argument, so $\langle \lambda x_1, \mu x_2 \rangle = \overline{\lambda} \mu \langle x_1, x_2 \rangle$.
Then the map $\phi_X: X \to X^*$ given by $x \mapsto \langle x, - \rangle$ cannot be represented by a matrix as it is not a linear map, which we can see by looking at $(\lambda x) \mapsto \langle \lambda x, - \rangle = \overline{\lambda} \langle x, - \rangle$. If $\phi_X(x) = f$, then $\phi_x(\lambda f) = \overline{\lambda} \phi_x(f)$.
If you did want to write down a "matrix" for $\phi_X$, then you would need to be sure to complex-conjugate all of the inputs before sending them through the matrix. (You should work out why this fixes the problem!). It also restores the commutativity of the diagram.