How to find the derivative of the commutator map?

Let $f : \mbox{GL}_n(\mathbb R) \to \mbox{GL}_n(\mathbb R)$ be defined by $$f(X) = K X K^{-1} X^{-1}$$ where $K$ is invertible matrix. How to find derivative of $f$?

Can I use here product rule of differentiation as follows?

$$f'(X)=KX'K^{-1}X^{-1}+KXK^{-1}(X^{-1})'$$

Am I correct? Any Help will be appreciated.

Solution 1:

The derivative is given by:

$$\frac{\partial}{\partial X} KXK^{-1}X^{-1} = \big((K^{-1}X^{-1})^T\otimes K\big) - \big((X^{-1})^T\otimes KXK^{-1}X^{-1}\big)$$

By using the rules $\partial(\mathbf{X} \mathbf{Y})=(\partial \mathbf{X}) \mathbf{Y}+\mathbf{X}(\partial \mathbf{Y})$, $\frac{\partial K X}{\partial X}=I\otimes K$ and $\frac{\partial X^{-1}}{\partial X}=-X^{-T}\otimes X^{-1}$

cf. matrix cook book and http://www.matrixcalculus.org/

On the other hand, using Gabriels approach gives:

$$\begin{aligned} f(X+H) &= K(X+H)K^{-1}(X+H)^{-1} \\ &= K(X+H)K^{-1}(I-X^{-1}H)^{-1}X^{-1} \\ &\sim f(X) + KHK^{-1} X^{-1} - KXK^{-1} X^{-1} H X^{-1} \end{aligned}$$

as the first order Taylor approximation. Note that both are equivalent because

$$ f(X+H)\sim f(X) + f'(X)\cdot H $$

and here

$$\begin{aligned} f'(X)\cdot H &= \big((K^{-1}X^{-1})^T\otimes K\big) - \big((X^{-1})^T\otimes KXK^{-1}X^{-1}\big)\cdot H \\ &= KHK^{-1} X^{-1} - KXK^{-1} X^{-1} H X^{-1} \end{aligned}$$

since in this case the tensor contraction "$\cdot$" works like $(U\otimes V) \cdot H = VHU^T$ where on the RHS it is regular matrix multiplication. This can be seen by looking at the index notation, cf. by answer here How to reconcile these two Jacobi formuli)