How to differentiate a tensor Frobenius norm?

Question

I am wondering how to calculate

$$\nabla_{\mathcal{T}} \|\mathcal{T}-\mathcal{C}\|_F^2$$

where $\mathcal{T}, \mathcal{C} \in \mathbb{R}^{n\times n\times n\times n}$ are $4$^th order tensors. Any help would be appreciated. Thank you.

Answer 1

Can't you just go by components? If $$ \Delta= \frac{d}{d\mathcal{T}} ||\mathcal{T}-C||_F^2 $$ where $\Delta\in\mathbb{R}^{n\times n\times n\times n}$ with \begin{align} \Delta_{ijk\ell} &= \frac{\partial}{\partial\mathcal{T}_{ijk\ell}} \sum_{\alpha,\beta,\gamma,\delta} (\mathcal{T}_{\alpha\beta\gamma\delta} - C_{\alpha\beta\gamma\delta})^2 \\ &= 2(\mathcal{T}_{ijk\ell} - C_{ijk\ell}) \end{align} So: $\Delta=2(T-C)$.

Answer 2

$ \def\p{\partial} \def\T{{\cal T}}\def\C{{\cal C}} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} $Write the function in terms of a quad-dot product, then calculate its differential and gradient. $$\eqalign{ \phi &= \|\T-\C\|^2_F \\ &= (\T-\C)::(\T-\C) \\ d\phi &= d\T::(\T-\C) \;+\; (\T-\C)::d\T \\ &= 2(\T-\C)::d\T \\ \grad{\phi}{\T} &= 2(\T-\C) \\ }$$