Let $h$ be the standard Hermitian product in $\mathbb{C}^n$, that is:
$$h(x, y) = \sum_{j=1}^n x^j\overline{y}^j$$
How can I prove the following property?
$$h(Ax, y) = h(x, \overline{A}^ty)$$
for every complex matrix A of order $n\times n$ whose and every $x, y\in\mathbb{C}^n$. I've tried the obvious, developing both sides of the previous equality, but I don't get the same result.
Let $A = (a_{ij})$. Then: $$ h(Ax,y) = \sum_{j=1}^n (Ax)_j \overline y_j = \sum_{j=1}^n \left(\sum_{i=1}^n a_{ji}x_i\right)\overline y_j = \sum_{i,j=1}^n a_{ji}x_i\overline y_j \\ = \sum_{i=1}^n x_i\left(\sum_{j=1}^n a_{ji}\overline{y_j}\right)= \sum_{i=1}^n x_i\left(\sum_{j=1}^n \overline{\overline{a_{ji}}y_j}\right)\\= \sum_{i=1}^n x_i\overline{\left(\sum_{j=1}^n \overline{a_{ji}}y_j\right)} = \sum_{i=1}^n x_i\overline{\left(\overline{A^t}y\right)}_i= h(x,\overline{A^t}y) $$