Let a complex matrix ${\bf A} \in \mathbb{C}^{m×n}$, where $m>n$, and its pseudoinverse ${\bf A}^{+} \in \mathbb{C}^{n×m}$, is there any explicit formula for the following derivative:
$d\frac{|| {\bf {\bf A}^{+}B - C}||_F^2}{d{\bf A}} $, where ${\bf B} \in \mathbb{C}^{m \times l}$ where $m>n>l$ and ${\bf C} \in \mathbb{C}^{n \times l}$ and $||.||_F$ is the Frobenius norm?
The differential of the pseudoinverse is well known but complicated $$\eqalign{ dA^+ &= A^+{A^+}^HdA^H(I-AA^+)+(I-A^+A)dA^H{A^+}^HA^+-A^+dAA^+ \cr }$$ Keep the $dA$ terms and ignore the $dA^H$ terms in accordance with the so-called Wirtinger or $\mathbb{CR}$-calculus.
For convenience, define the matrix $$X = C-A^+B$$ Write the function in terms of this variable. Then find its differential and gradient. $$\eqalign{ \phi &= \|X\|_F^2 \,\,\in {\mathbb R} \cr &= X^*:X \cr d\phi &= X^*:dX \cr &= X^*:(-dA^+B) \cr &= X^*B^T:A^+\,dA\,A^+ \cr &= {A^+}^TX^*B^T{A^+}^T:dA \cr &= {A^+}^T(C-A^+B)^*B^T{A^+}^T:dA \cr G=\frac{\partial\phi}{\partial A} &= {A^+}^T(C-A^+B)^*B^T{A^+}^T \cr }$$ Given the gradient wrt $A$, it's a simple matter to find the gradient wrt $A^H$ or $A^*$ $$\eqalign{ \frac{\partial\phi}{\partial A^H} &= G^H,\quad \frac{\partial\phi}{\partial A^*} &= G^* \cr }$$ In some intermediate steps, a colon was used to denote the trace/Frobenius product, i.e. $$A:B = {\rm Tr}(A^TB)$$