2
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Are you using the Frobenius inner product $\langle A\odot X, Y\rangle_{\mathbb{R}^{n\times m}} = \langle A\odot X, Y \rangle_F = \text{vec}(A\odot X)^\top \text{vec}(Y)$ ? $\endgroup$ Nov 11 '21 at 17:42
  • 1
    $\begingroup$ @Physics_Student Yes. Are you using a different one? $\endgroup$
    – user1551
    Nov 11 '21 at 17:42
  • $\begingroup$ Well actually I was super confused so didn't think about it, but yes I think Frobenius inner product is perfect! $\endgroup$ Nov 11 '21 at 17:43
  • $\begingroup$ Oh lovely I understand your proof! Thank you :) $\endgroup$ Nov 11 '21 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.