The Hadamard is a two by two matrix:

$\begin{equation} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \end{equation} $

The Kronecker product of the Hadamard with itself is $H \otimes H$ or $H^{\otimes 2}$

Let X be a $2 \times 4$ complex valued matrix, and let $\overline{X}$ be its conjugate transpose.

Is it possible to write $H^{\otimes 2}$ as $ \overline{X} H X$ ?

More generally, is there a way to find a $ 4 \times 8$ matrix $Y$ such that $ H^{\otimes 3} = \overline{Y} \space\overline{X} H X \space Y$

and so on for higher order Kronecker products of the Hadamard?


$H\otimes H$ has rank $4$, but $\operatorname{rank}(PHQ)\le\operatorname{rank}(H)=2$.


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.