Consider $U \in R^{n*r} $, is there a way I can represent Khatri Rao product of $U$ with itself repeated $M$ times in any form of $U$?
$$(U \odot U \odot ....\odot U) = f(U)$$ where $(U \odot U)$ represents the Khatri Rao product. What would be this $f(U) \in R^{n*r} \to R^{n^M*r} $ ?
If the columns of the matrix are given by $$U = \left[\matrix{u_1&u_2\ldots&u_n}\right]$$ Then the columns of the iterated Khatri-Rao product are given $$U^{\odot M} = \left[\matrix{u_1^{\otimes M}&u_2^{\otimes M}\ldots&u_n^{\otimes M}}\right]$$ where the iterated Kronecker product of a vector with itself has been denoted by $$a^{\otimes M} = \operatorname*{\bigotimes}\limits_{k=1}^M a \;=\; a\otimes a\otimes\ldots\otimes a $$ If you like you can use the standard basis vectors $\{e_k\}$ to rearrange things. $$\eqalign{ u_k &= Ue_k \\ u_k^{\otimes M} &= U^{\otimes M}\,e_k^{\otimes M} \\ U^{\odot M} &= U^{\otimes M} \left[\matrix{e_1^{\otimes M}&e_2^{\otimes M}\ldots&e_n^{\otimes M}}\right] \\ &= U^{\otimes M}\,E_M \\ }$$ This reformulation swaps the iterated Khatri product for an iterated Kronecker product, multiplied by a very large sparse binary matrix.
