# Derivative of $tr(A \exp{\hat{\omega}})$ with respect to $\omega \in \mathbb{R}^3$

I want to calculate the partial derivative of $tr(A \exp{\hat{\omega}})$ with respect to $\omega = [\omega_1, \omega_2, \omega_3 ]^\top \in \mathbb{R}^3$, where $A \in \mathbb{R}^{3\times 3}$ is a constant matrix and $\exp$ is the matrix exponential map. Here $$\hat{\omega} = \left[\begin{array}{ccc} 0 & -\omega_{3} & \omega_{2}\\ \omega_{3} & 0 & -\omega_{1}\\ -\omega_{2} & \omega_{1} & 0 \end{array}\right]$$ Obviously the derivative should be in $\mathbb{R}^3$. $A$ could be any matrix with determinant non-zero and I would like the answer not to depend on the individual terms of $A$.

Any solution or suggestion of approach or reference is highly appreciated

Attempt so far : let $\delta$ be a real scalar and let $v \in \mathbb{R}^3$ be a variation of $\omega$.

then $$tr(A\exp(\hat{\omega}+\delta\hat{v}))=tr(A\exp(\hat{\omega})\exp(\delta\hat{v}))=tr(A\exp(\hat{\omega})(I+\delta\hat{v}+\frac{1}{2!}\delta^{2}\hat{v}^{2}+\cdots))$$ Now $$tr(A\exp(\hat{\omega}+\delta\hat{v}))-tr(A\exp(\hat{\omega}))= tr(A\exp(\hat{\omega})(\delta\hat{v}+\frac{1}{2!}\delta^{2}\hat{v}^{2}+\cdots))$$ removing higher order terms of $\delta$ and dividing by $\delta$, we get $tr(A\exp(\hat{\omega})\hat{v})$. However the expression is not of correct order. I am not understanding how to obtain a $3\times 1$ expression from this.

## 1 Answer

Note that in general, we don't have $\exp(A + B) = \exp(A)\exp(B)$!

If you want to calculate the derivative of your map $f$ at the origin, we have

$$df|_{0}(v) = \frac{d}{dt} \operatorname{tr}(A \exp(t\hat{v}))|_{t=0} = \operatorname{tr} \left( \frac{d}{dt} A \exp(t \hat{v})|_{t = 0} \right) = \operatorname{tr}(A \hat{v}).$$

For other points, you'll need to calculate

$$df|_{\omega}(v) = \frac{d}{dt} \operatorname{tr} (A \exp(\hat{\omega} + t\hat{v}))|_{t = 0} = \operatorname{tr} \left( A \frac{d}{dt} \exp(\hat{\omega} + t\hat{v})|_{t = 0}\right).$$

Set $X = \frac{\omega}{t} + v$. Using Rodrigues' formula, we have

$$\exp(\hat{\omega} + t\hat{v}) = \exp (t \hat{X}) = I + \sin(t) \hat{X} + (1 - \cos t) \hat{X}^2 = \\ I + \frac{\sin t}{t} \hat{\omega} + \sin(t) \hat{v} + \frac{1 - \cos t}{t^2} \hat{\omega}^2 + \frac{1 - \cos t}{t}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega}) + (1 - \cos t) \hat{v}^2 = \\ I + t \left( \hat{v} + \frac{1}{2}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega})\right) + \dots$$

where the $\dots$ denote higher order terms. Thus,

$$df|_{\omega}(v) = \operatorname{tr} \left( A \left( \hat{v} + \frac{1}{2}(\hat{\omega}\hat{v} + \hat{v}\hat{\omega})\right) \right).$$