Matrix derivative of Frobenius norm involving Hadamard product

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Problem: Solve $\frac{\partial\left|\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})\right|_ {F}^{2}}{\partial\boldsymbol{W}}$ and $\frac{\partial\left|\boldsymbol{A}\circ ( \boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})\right|_{F}^{2}}{\partial\boldsymbol{X}}$, where $\circ$ denotes the Hadamard product, and all variables are matrices.

Solution

We define $\boldsymbol{Z}=\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})$, then we have

\[\begin{aligned} \partial\|\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})\|_F^2 &= \partial \boldsymbol{Z}:\boldsymbol{Z} \\ &= 2\boldsymbol{Z}:d\boldsymbol{Z} \\ &= 2\boldsymbol{Z}:d(\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})) \\ &= 2\boldsymbol{Z}:\boldsymbol{A}\circ d(-\boldsymbol{W}^\top\boldsymbol{X}) \\ &= -2(\boldsymbol{A}\circ\boldsymbol{Z}):(d\boldsymbol{W}^\top\cdot\boldsymbol{X} +\boldsymbol{W}^\top\cdot d\boldsymbol{X}) \\ &= -2(\boldsymbol{A}\circ\boldsymbol{Z}):(d\boldsymbol{W}^\top\cdot\boldsymbol{X}) \\ &\quad -2(\boldsymbol{A}\circ\boldsymbol{Z}):(\boldsymbol{W}^\top\cdot d\boldsymbol{X}) \\ &= -2(\boldsymbol{A}\circ\boldsymbol{Z})\boldsymbol{X}^\top:d\boldsymbol{W}^\top -2\boldsymbol{W}(\boldsymbol{A}\circ\boldsymbol{Z}):d\boldsymbol{X} \\ &= -2\boldsymbol{X}(\boldsymbol{A}\circ\boldsymbol{Z})^\top:d\boldsymbol{W} -2\boldsymbol{W}(\boldsymbol{A}\circ\boldsymbol{Z}):d\boldsymbol{X}. \end{aligned}\]

Therefore, we have

\[\begin{aligned} \frac{\partial\|\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})\|_F^2}{\partial\boldsymbol{W}} &= -2\boldsymbol{X}(\boldsymbol{A}\circ\boldsymbol{Z})^\top \\ &= -2\boldsymbol{X}(\boldsymbol{A}\circ\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X}))^\top, \end{aligned}\]

and

\[\begin{aligned} \frac{\partial\|\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})\|_F^2}{\partial\boldsymbol{X}} &= -2\boldsymbol{W}(\boldsymbol{A}\circ\boldsymbol{Z}) \\ &= -2\boldsymbol{W}(\boldsymbol{A}\circ\boldsymbol{A}\circ (\boldsymbol{Y}-\boldsymbol{W}^\top\boldsymbol{X})). \end{aligned}\]

Notes

  1. The Frobenuis product of two matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ is defined as $\boldsymbol{A}: \boldsymbol{B}=\text{tr}(\boldsymbol{A}^\top\boldsymbol{B})$.
  2. $\boldsymbol{A}:\boldsymbol{B}\circ\boldsymbol{C}=\boldsymbol{A}\circ\boldsymbol{B}:\boldsymbol{C}$.