Calculating the derivative $\frac{{\partial {{\bf{X}}^{{\rm{ - }}1}}}}{{\partial {\bf{X}}}}$

Question

How to calculate the derivation $\frac{{\partial {{\bf{X}}^{{\rm{ - }}1}}}}{{\partial {\bf{X}}}}$,where ${\bf{X}}$ is square matrix.Thanks a lot for your help!

Answer 1

If $F(X) = X^{-1}$, the differential of $f$ in $X$ is the linear function given by $$DF(X)H = - X^{-1}HX^{-1}$$ Proof: $$F(X+H) - F(X) - DF(X)H = (X+H)^{-1} - X^{-1} + X^{-1}HX^{-1}$$ $$= -(X+H)^{-1}HX^{-1} + X^{-1}HX^{-1} = (X^{-1}-(X+H)^{-1})HX^{-1},$$ so $$\|F(X+H) - F(X) - DF(X)H\|\le\|X^{-1}-(X+H)^{-1}\|\|H\|\|X^{-1}\| = \|X^{-1}-(X+H)^{-1}\|O(\|H\|).$$ And $$\lim_{H\to 0}\frac{\|F(X+H) - F(X) - DF(X)H\|}{\|H\|} = 0$$ by the continuity of $F$.

Answer 2

Let $\mathrm F (\mathrm X) := \mathrm X^{-1}$. Hence,

$$\mathrm F (\mathrm X + h \mathrm V) = (\mathrm X + h \mathrm V)^{-1} = \left( \mathrm X ( \mathrm I + h \mathrm X^{-1} \mathrm V) \right)^{-1} \approx ( \mathrm I - h \mathrm X^{-1} \mathrm V) \mathrm X^{-1} = \mathrm F (\mathrm X) - h \mathrm X^{-1} \mathrm V \mathrm X^{-1}$$

Thus, the directional derivative of $\mathrm F$ in the direction of $\mathrm V$ at $\mathrm X$ is $\color{blue}{- \mathrm X^{-1} \mathrm V \mathrm X^{-1}}$. Vectorizing,

$$\mbox{vec} (- \mathrm X^{-1} \mathrm V \mathrm X^{-1}) = - \left( \mathrm X^{-\top} \otimes \mathrm X^{-1} \right) \mbox{vec} (\mathrm V)$$

Answer 3

First note that $\frac{\partial F}{\partial X}$ is a 4th order tensor.

Let $F=X^{-1},\,$ then working out the problem in index notation yields $$\eqalign{ dF_{ij} &= -F_{ik}\,dX_{kl}\,F_{lj} \cr\cr \frac{\partial F_{ij}}{\partial X_{kl}} &= -F_{ik}\,F_{lj} \cr\cr }$$