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I'm trying to expand this Frobenius form $||C \circ (A-XB)||_F^2$ (here $\circ$ is the Hadamard point-wise multiplication). I want to find the minimum value with respect to X.

$$ \frac{\partial}{\partial X}||C \circ (A-XB)||_F^2 = 0$$

I've being trying to develop using the fact that the Frobenius form is $||A||_F^2=trace(AA^*)$ but the Hadamard product is always on my way.

How would you approach this?

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You can put $g(X)=C\circ (A-XB)$ where $\circ$ is the Hadamard product, and use the chain rule. – Davide Giraudo Jan 30 '12 at 16:41

You need to use the chain rule. $$\partial (\mathbf{X} \circ\mathbf{Y}) = \partial \mathbf{X} \circ \mathbf{Y} + \mathbf{X} \circ \partial \mathbf{Y} $$

Also, the fact that $$||\mathbf{A}||_2^2 = trace(\mathbf{A}^T\mathbf{A})$$

Look the following reference for matrix calculus it has all the stuff you would need.

Matrix Cookbook: [][1]

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Given $f = \|M\|_F^2 = M:M$, the differential is $\,\,df=2M:dM$

Substituting $M=C\circ(A-XB)$, we have $$ \eqalign { df &= 2(C\circ(A-XB)) : d(C\circ (A-XB)) \cr &= -2(C\circ(A-XB)) : C\circ dX\,B \cr &= -2(C\circ(A-XB)\circ C) : dX\,B \cr &= -2(C\circ(A-XB)\circ C)B^T : dX \cr &= \bigg(\frac {\partial f} {\partial X}\bigg) : dX \cr } $$ Setting the derivative to zero yields $$ \eqalign { \frac {\partial f} {\partial X} &= 0 = -2\,(C\circ(A-XB)\circ C)B^T \cr (XB\circ C\circ C)B^T&= (A\circ C\circ C)B^T \cr } $$ Now what is the best way to solve for $X$?

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Hi greg, Did you known how the solve this for X? I have the same question about this.… Thanks. – BioChemoinformatics Feb 28 at 23:02

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