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I have a constraint optimization problem formulated in a diagonal matrix form:

$ P_3:~ min_{x} \quad \|A X(t) - Y(t)\|^2 \\ \text{subject to} \quad X^*(t) \cdot X(t) = \mathbb{I} $

I need to take the partial derivative of the following function wrt x of this vectorization.

$ f(X,\Lambda) = \left[ (I \otimes A)vec X - vec Y \right]^ * \cdot \left[ (I \otimes A)vec X - vec Y \right] - \Lambda( (\mathbb{I} \otimes X^* vec X) - vec \mathbb(I))$

X,Y,A, $\Lambda$ are diagonal matrices, * is the conjugate transpose operation.

Any thoughts?

user3284182
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  • It looks like $L(x,\lambda)$ is supposed to be a Lagrangian. In that case it should be a scalar, not a matrix. – greg Mar 08 '23 at 04:06
  • It is a condensed version of a multiconstraint problem. But, regardless of this new format, please , could you help me with the matrix partial derivative ?? – user3284182 Mar 08 '23 at 14:10
  • @user3284182 The problem is that in order for $\frac{\partial L}{\partial X}$ to be representable as a matrix (as opposed to an order-3 or order-4 tensor), $L$ needs to be scalar valued. The formula that you've presented cannot possibly be correct even if all the matrices involved had real-number entries. At least, this is the case assuming that you mean what people usually are referring to when they write $\frac{\partial L}{\partial X}$. – Ben Grossmann Mar 08 '23 at 14:20
  • Okay. I understood now.. Let me rephrase it .. Please look at the edited question @BenGrossmann – user3284182 Mar 08 '23 at 16:49
  • Since the matrices are diagonal, the problem is equivalent to $$\eqalign{ x = {\rm diag}(X),\quad y = {\rm diag}(Y),\quad f = |Ax-y|^2_F }$$ You can find lots of posts (like this one) which calculate the gradient of $f.;$ – greg Mar 09 '23 at 00:37
  • @greg Thanks for your answer. But I need to find the formulation in this tensor form. Because actually, I have a constraint witch is given in a matrix form. Please see the edited question – user3284182 Mar 09 '23 at 22:09
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    So the matrix $X$ is diagonal and orthogonal. In terms of vector elements this means that $$\large{x_k = {\rm sign}(y_k/a_k) = \pm\tt1}$$ – greg Mar 09 '23 at 23:37
  • @greg, Yes.. this means that the constraint makes my x entries all have absolute values equal to 1. However, I need to find the partial derivatives anyway. Because I seek to rewrite the system with a new equation. – user3284182 Mar 10 '23 at 00:22

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