Partial derivatives of matrix exponential

Question

I am trying to find an expression for

$$\frac{\partial \text{tr}\, Ae^{K}}{\partial v_i}$$

Given that $K$ is skew-symmetric

$$K = \begin{bmatrix} 0 & v_1 & v_2 \\ -v_1 & 0 & v_3 \\ -v_2 & -v_3 & 0 \end{bmatrix}$$

and $A$ is an arbitrary $3 \times 3$ matrix. It's easy to solve when $A = I_3$, but I'm having difficulties otherwise.

Answer 1

Given an an arbitrary $3\times 3$ skew symmetric matrix, $K$, calculate the quantity $$\eqalign{ \beta &= \frac{\|K\|_F}{\sqrt 2} = \sqrt\frac{K:K}{2} \cr }$$ The differential of this quantity can be expressed in terms of the differential of $K$ like so $$\eqalign{ \beta^2 &= \frac{K:K}{2} \cr 2\,\beta\,d\beta &= K:dK \cr\cr d\beta &= \frac{K:dK}{2\,\beta} \cr }$$

Then messing around with Rodrigues' rotation formula you'll find that $$\eqalign{ \exp(K) &= I + K\frac{\sin\beta}{\beta} + K^2\frac{1-\cos\beta}{\beta^2} \cr }$$ The function in question can be written as $$\eqalign{ f &= A^T:\exp(K) \cr &= A^T:\Big(I + K\frac{\sin\beta}{\beta} + K^2\frac{1-\cos\beta}{\beta^2}\Big) \cr }$$ Expand the differential $df$ in terms of $\{dK,d\beta\}$, then substitute the above expression for $d\beta$ to arrive at an expression entirely in terms of $dK$.

From that differential, the gradient with respect to $K$ is found to be $$\eqalign{ \frac{\partial f}{\partial K} &= A^T\frac{\sin\beta}{\beta}+(A^TK+KA^T)\frac{\cos\beta-1}{\beta^2} \cr &+ \,K\,\Big[(A^T:K)\frac{\beta\cos\beta-\sin\beta}{2\,\beta^3}+(A^T:K^2)\frac{\beta\sin\beta+2\cos\beta-2}{2\,\beta^4}\Big] \cr }$$ The derivative with respect to the components of $v$ can be found by taking the Frobenius product with various single-entry matrices $J_{ik}$ $$\eqalign{ \frac{\partial f}{\partial v_1}&=J_{12}:\frac{\partial f}{\partial K}\cr\cr \frac{\partial f}{\partial v_2}&=J_{13}:\frac{\partial f}{\partial K}\cr\cr \frac{\partial f}{\partial v_3}&=J_{23}:\frac{\partial f}{\partial K}\cr }$$

Answer 2

That follows is the Maple procedure for the calculation of $\frac{\partial \text{tr}\, Ae^{K}}{\partial v_1}$ where $v_1:=x,v_2:=y,v_3:=z$. Time of calculation: $<1"$. The result meets 5 screens...

restart: with(LinearAlgebra): K := Matrix(3, 3, [[0, x, y], [-x, 0, z], [-y, -z, 0]]): A := Matrix(3, 3, symbol = a): B := MatrixExponential(K): u := simplify(Trace(A.B)): f := unapply(u, x): simplify(diff(f(x), x));