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Let $A$ and $B$ be $m \times n$ matrices with low-rank structures: $$ A = U_{A}\Sigma_{A}V_{A}^{T},\quad B= U_{B}\Sigma_{B}V_{B}^{T}, $$

Prove that Hadamard product $A\circ B$ admits the following representation $$ A\circ B = (U_{A}^T\odot U_{B}^T)^T (\Sigma_{A}\otimes\Sigma_{B})(V_{A}^{T}\odot V_{B}^{T}), $$ where $\odot$ represents the Khatri-Rao product, and $\otimes$ the Kronecker product.

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    $\begingroup$ The Khatri-Rao product, for those unfamiliar $\endgroup$
    – Ben Grossmann
    Commented May 26, 2019 at 16:26
  • $\begingroup$ This is an interesting problem, where did you encounter it? Also, could you clarify how exactly $U_A$ is meant to be partitioned for the Khatri-Rao product? $\endgroup$
    – Ben Grossmann
    Commented May 26, 2019 at 16:29
  • $\begingroup$ @Omnomnomnom $U_{a}$ is the matrix of left singular vectors of A. $\endgroup$
    – Georgiy
    Commented May 26, 2019 at 16:43
  • $\begingroup$ see math.stackexchange.com/q/1916069 $\endgroup$
    – Jean Marie
    Commented May 26, 2019 at 18:54

1 Answer 1

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We will use the following properties and definitions: \begin{align} (A\odot^T B)(C\odot D) = (AC)\circ(BD),\label{eq:khr-had}\\ (A\otimes B)(C\odot D) = (AC)\odot(BD),\label{eq:khr-kro} \end{align} It is easy to prove that Hadamard product of A and B admits the following representation: \begin{align} A \circ B &= (U_{A}\Sigma_{A}V_{A}^{T})\circ(U_{B}\Sigma_{B}V_{B}^{T})\nonumber\\ &= (U_{A}^T\odot U_{B}^T)^T (\Sigma_{A}V_{A}^{T}\odot \Sigma_{B}V_{B}^{T})\nonumber\\ &=(U_{A}^T\odot U_{B}^T)^T (\Sigma_{A}\otimes\Sigma_{B})(V_{A}^{T}\odot V_{B}^{T})\label{eq:repres} \end{align} where $\odot$ represents the Khatri-Rao product, and $\otimes$ the Kronecker product.

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