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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 $$ R\widehat{x} + \widehat{x}R^{\top} = (Bx)^{\wedge}? $$

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  • $\begingroup$ What about writing down what that hat operator explicitly does, and then just do the actual computation with $3\times3$-matrices? $\endgroup$ – Torsten Schoeneberg Apr 17 at 19:24
  • $\begingroup$ 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 $R\in\mathbb{R}^{3\times 3}$. $\endgroup$ – a-deel Apr 18 at 3:50

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