Derivative of matrix $\frac{\partial^2 \Pi}{\partial (A^{-1}X)^2}=A^T \frac{\partial ^2\Pi}{\partial X^2}A$, where $A$ is an orthogonal matrix.

Question

I dont understand why $\frac{\partial^2 \Pi}{\partial (A^{-1}X)^2}=A^T \frac{\partial ^2\Pi}{\partial X^2}A$, where $A$ is an orthogonal matrix.

Answer 1

Analyze the function in terms of the variables $w{\rm\;and\;}x$ $$\eqalign{ \def\o{{\tt1}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\vecc#1{\op{vec}\LR{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\qiq{\quad\implies\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} w &= A^T x \quad\; \qiq dw = A^T\,dx \\ \Pi &= \Pi(w) \quad\qiq g = \grad{\Pi}{w},\;\;P = \grad{g}{w} = \grad{^2\Pi}{w^2} \\ &= \Pi(A^Tx) \qiq h = \grad{\Pi}{x},\;\;Q = \grad{h}{x} = \grad{^2\Pi}{x^2} \\ \\ d\Pi &= g:\c{dw} \qquad\qquad\qquad \{ {\rm differential\;of\;}\Pi \} \\ &= g:\c{A^Tdx} \\ &= Ag:dx \\ \grad{\Pi}{x} &= Ag \;\equiv\; h \\ \\ dh &= A\:\c{dg} \\ &= A\c{P\;dw} \\ &= AP\:A^Tdx \\ \grad{h}{x} &= APA^T \;\equiv\; Q \qquad\quad \{ {\rm desired\;relationship} \} \\ \\ }$$

---

In the preceding, a colon denotes the Frobenius product $$\eqalign{ A:B &= \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; \trace{A^TB} \\ A:A &= \frob{A}^2 \qquad \{ {\rm Frobenius\;norm} \}\\ A:B &= B:A \;=\; B^T:A^T \\ C:\LR{AB} &= \LR{CB^T}:A \;=\; \LR{A^TC}:B \\ }$$