Banach space adjoint as a matrix

Question

Question

Let $1< p,q < \infty$ be holder conjugates.

Let $A \colon \ell_n^p(\mathbb{C}) \to \ell_n^p(\mathbb{C})$ be a $n\times n$ matrix.

Let $A^* \colon \ell_n^q(\mathbb{C}) \to \ell_n^q(\mathbb{C}) $ be the Banach space adjoint of $A$.

Note $\ell_n^p(\mathbb{C})^* = \ell_n^q(\mathbb{C})$.

Show $A^*$ is the transpose of $A$.

My Understanding

Let $$x= (x_1. \cdots, x_n), \ y = (x_1, \cdots, y_n) \in \ell_n^p(\mathbb{C})$$

Let $$f_x(y) = x \cdot y = \sum_{k=1}^n x_k y_k, \text{where } f_x = \sum_{k=1}^n x_k e_k$$

Then we want to show that $$A^*(f_x)(y)= A^T(f_x)(y)$$

Note that $$x \cdot y = x^T y = f_x(y)$$

Also, by definition of adjoint we have $$A^*(f_x)= f_x \circ A$$

We have

$$A^*(f_x)(y)= f_x(Ay)= \sum_{k=1}^n x_k e_k=x^T \cdot (Ay) = (Ay)\cdot x^T = (Ay)^Tx^T= y^TA^Tx^T$$

which became a mess. But what I really want is to have the last equalities be $$=A^T(x^T y)= A^T(f_x)(y)$$

Any help is appreciated!

Answer 1

By the Banach space definition of an adjoint, $$(A^*f_x)(y)=(f_x\circ A)y=x\cdot(Ay)=x^\top Ay=(A^\top x)^\top y=(A^\top x)\cdot y$$ Since this is true for all $y$, it follows that $A^*f_x=f_{A^\top x}$.
This is commonly written as $A^*x=A^\top x$ if one abuses the notation and writes $x$ for $f_x$.