I need help with the below proof,
$$A = \underset{A}{\text{argmin}}(\frac{1}{2} ||X_{(1)} - A(C \odot B)^T||^2_F + ||\Lambda \boxdot (A - \tilde{A})||_F^2 + \frac{\rho}{2} ||A - \tilde{A}||_F^2) \\ = (X_{(1)}(C \odot B) + \rho \tilde{A} - \Lambda ) ((C \odot B)^T(C \odot B) + \rho I_R)^{-1}$$
Specifications
$A, \Lambda$ and $ \tilde{A} \in \mathbb{R} ^{I \times R}$ $B \in \mathbb{R} ^{J\times R}, C \in \mathbb{R} ^{K \times R}, $ $ X_{(1)} \in \mathbb{R} ^{I \times JK}$
$A*B = trace(A^T B)$, $||A||_F^2 = trace(A^TA)$
$\odot:$ Kronecker product
$\boxdot:$ element wise product
I found a mistake copying from the paper$||\Lambda \boxdot(A - \tilde{A})||^2_F = (\Lambda * (A - \tilde{A}))$ (misunderstanding the relation between element wise product and trace)
$\dfrac{d}{dA} (\dfrac{1}{2} ||X_{(1)} - A(C \odot B)^T||_F^2) + (\Lambda * (A - \tilde{A})) + \dfrac{\rho}{2}||A - \tilde{A}||_F^2 = 0$
$= (1/2 * 2) (X_{(1)} - A(C \odot B)^T) \times -(C \odot B) + \Lambda + \rho(A-\tilde{A})$
$= -X_{(1)} (C \odot B) + A(C \odot B)^T(C \odot B)+\Lambda + \rho A - \rho \tilde{A}$
$ A ((C \odot B)^T (C \odot B) + \rho I_R) = X_{(1)} (C \odot B) - \Lambda+ \rho \tilde{A} $
$A = (X_{(1)} (C \odot B) - \Lambda+ \rho \tilde{A})((C \odot B)^T (C \odot B) + \rho I_R)^{-1}$