0I have a linear map $f:\mathbb{R}^{n\times m}\to \mathbb{R}^{n\times m}$ defined as
$$
f(X) = A \odot X \qquad \qquad \text{ for some } A\in\mathbb{R}^{n\times m}
$$
where $\odot$ is the Hadamard Product. What is the adjoint of $f$?
Attempted Solution
Let $e_i\in\mathbb{R}^{nm\times 1}$ the $i^{\text{th}}$ basis vector of $\mathbb{R}^{n\times m}$. One can reshape this into a $n\times m$ matrix. The first $n$ elements go in the first column, the second $n$ elements in the second and so on. The result of this is the matrix $E_i$ which has $0$ everywhere except for the entry in the $((i – 1) \text{ mod } n) + 1$ row and in the $(i – 1) \div n$ column, where $\div$ represents integer division.
$$
E_i = \begin{pmatrix}
0 & 0 & \cdots & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\\
0 & 0 & \cdots & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots &0 &\cdots & 0
\end{pmatrix}
$$
Then the matrix $X$ can be written in terms of these bases matrices as
$$
X = x_{11} E_1 + \cdots + x_{nm} E_{nm}
$$
Therefore the output of the linear map can be written as
$$
f(X) = x_{11} f(E_1) + \cdots + x_{nm} f(E_{nm})
$$
Similarly, denoting by $r(i) = ((i-1) \text{ mod } n) + 1$ and $c(i) = (i – 1) \div n$ the application of $f$ on one of these bases gives
$$
f(E_i) = A \odot E_i = a_{r(i), c(i)} E_i.
$$
Then the vector $(0, 0, \ldots, 0, a_{r(i), c(i)}, 0, \ldots, 0)^\top\in\mathbb{R}^{nm\times 1}$ will be the column vector of the matrix representing this linear operator. In other words
$$
M_f = \begin{pmatrix}
a_{1, 1} & \cdots & 0 \\
\vdots & \ddots & \vdots \\
0 & \cdots & a_{nm}
\end{pmatrix} = \text{Diagonal}(a_{11}, \ldots, a_{nm})
$$
However this can't be right since $M\in\mathbb{R}^{nm\times nm}$ can't multiply $X\in\mathbb{R}^{n\times m}$.
Best Answer
It is self-adjoint. Let $\mathbf a=\operatorname{vec}(A)=(a_1,\ldots,a_{mn})^T$ and define $\mathbf x,\mathbf y$ analogously. Then \begin{aligned} \langle A\odot X,\,Y\rangle_{\mathbb R^{n\times m}} &=\langle\operatorname{vec}(A\odot X),\,\operatorname{vec}(Y)\rangle_{\mathbb R^{nm}}\\ &=\langle\operatorname{vec}(A)\odot\operatorname{vec}(X),\,\operatorname{vec}(Y)\rangle_{\mathbb R^{nm}}\\ &=\sum_{k=1}^{mn}a_kx_ky_k\\ &=\langle\operatorname{vec}(X),\,\operatorname{vec}(A)\odot\operatorname{vec}(Y)\rangle_{\mathbb R^{nm}}\\ &=\langle\operatorname{vec}(X),\,\operatorname{vec}(A\odot Y)\rangle_{\mathbb R^{nm}}\\ &=\langle X,\,A\odot Y\rangle_{\mathbb R^{n\times m}}. \end{aligned}
However, note that on $\mathbb C^n$, a similar argument yields \begin{aligned} \langle A\odot X,\,Y\rangle_{\mathbb C^{n\times m}} =\langle X,\,\overline{A}\odot Y\rangle_{\mathbb C^{n\times m}}. \end{aligned} Therefore $f$ is not self-adjoint but $f^\ast(X)=\overline{A}\odot X$ in this case.