# relation between trace and hat operator (skew-symmetric matrices)

To avoid confusion, let me first introduce the notation (although pretty standard) which is required for the question that I want to ask. Let $$\mathsf{GL}(3,\mathbb{R})$$ be the set of $$3\times 3$$ real invertible matrices, $$\mathsf{SO}(3)$$ be the set of $$3\times 3$$ rotation matrices, and $$\mathfrak{so}(3)$$ be the set of $$3\times 3$$ skew symmetric matices. The isomorphism between $$\mathbb{R}^3$$ and $$\mathfrak{so}(3)$$ is given by the "hat'' operator, i.e., $$\widehat{\cdot}: \mathbb{R}^3 \rightarrow \mathfrak{so}(3)$$, sometimes also denoted by $$(\cdot)^{\wedge}: \mathbb{R}^3 \rightarrow \mathfrak{so}(3)$$ for notational conveniance. The $$3\times 3$$ identity matrix is represeted by $$I_{3\times 3}$$, trace of a matrix is represented by $$trace(\cdot)$$, and transpose of a matrix is represeted by $$\cdot^{\top}$$. For $$x\in\mathbb{R}^{3}$$ and $$A\in\mathbb{R}^{3\times 3}$$, how can we prove the following result? $$\widehat{x}A + A^{\top}\widehat{x} = \left(\left(trace(A)I_{3\times 3}-A\right)x\right)^{\wedge}.$$ Moreover, for $$x\in\mathbb{R}^3$$, $$R\in\mathsf{SO}(3)$$, and $$B\in\mathsf{GL}(3,\mathbb{R})$$, can we write $$R\widehat{x} + \widehat{x}R^{\top} = (Bx)^{\wedge}?$$

• What about writing down what that hat operator explicitly does, and then just do the actual computation with $3\times3$-matrices? – Torsten Schoeneberg Apr 17 at 19:24
• Thanks for the hint. It's pretty straight forward to prove both equalities. In fact the second equality is equal to $$R\widehat{x} + \widehat{x}R^{\top} = \left(\left(trace(R)I_{3\times 3}-R^{\top}\right)x\right)^{\wedge},$$ for each $R\in\mathbb{R}^{3\times 3}$. – a-deel Apr 18 at 3:50