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 $$X\in\mathbb{R}^{n\times m}$$.

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.
• Are you using the Frobenius inner product $\langle A\odot X, Y\rangle_{\mathbb{R}^{n\times m}} = \langle A\odot X, Y \rangle_F = \text{vec}(A\odot X)^\top \text{vec}(Y)$ ? Nov 11 '21 at 17:42