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I encounted an least square problem involves hadamard product of two $100\times 1$ matrix: X, Y: $$ A_{N\times100}*(X_{100\times 1}∘Y_{100\times 1})_{100\times 1} = B_{N\times 1} $$ In above equation, * is a traditional matrix multiplication, ∘is the hadamard product. A and B are observed known matrix, X is also known . I known this is non-convex problem. My question: is there any method to get an approximate anser of Y?

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  • $\begingroup$ The pseudoinverse $A^+$ and Hadamard multiplication/division $\{\odot/\oslash\}$ can solve the problem $$\eqalign{ &A(x\odot y) = b \\ &(x\odot y) = A^+b \\ &y = (A^+b)\oslash x \\ }$$ $\endgroup$
    – greg
    Jan 4, 2022 at 13:05

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Let $\mathbf{y}=\mathbf{x} \circ \mathbf{x}$, a 100-by-1 vector with positive entries. The least-squares problem you are facing is now recast into a non-negative least-squares (NNLS) , i.e. minimize $$ \phi(\mathbf{y})=\| \mathbf{Ay-b} \|^2, s.t. \mathbf{y} \ge 0 $$ Once $\mathbf{y}$ is found, take the (elementwise) square root to obtain $\mathbf{x}$


Let $\mathbf{z}=\mathbf{x} \circ \mathbf{y}$, Since $\mathbf{x}$ is known, the method consists in 1) solving for $\mathbf{z}$ using standard LS and then 2) retrieve $\mathbf{y}$ by elementwise division : $$ \mathbf{y}=\frac{\mathbf{z}_{LS}}{\mathbf{x}} $$

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  • $\begingroup$ Thanks Steph. You means the X∘Y transfered to X∘X? $\endgroup$
    – jack tsang
    Dec 23, 2021 at 10:06
  • $\begingroup$ See my edit. I had not correctly read your question at first . Now it hould be ok. $\endgroup$
    – Steph
    Dec 23, 2021 at 12:32
  • $\begingroup$ I implemented my code with the same process as your described in the edited comments. While, the result did not OK. Because, in the original version of my problem, the X, Y are both unknown. I give a initial value for X to get an make sense result. My new question is if the X, Y are both need to be solved? $\endgroup$
    – jack tsang
    Dec 23, 2021 at 13:28
  • $\begingroup$ In your problem formulation , the vector x was said to be known. Without further hypothesis, it is easy to see that you cannot recover in an unique manner both vectors... $\endgroup$
    – Steph
    Dec 29, 2021 at 9:21

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