Is there an existing linear mapping that maps a 3-dimensional vector: $\mathbf{v}=\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}$ to a corresponding skew-symmetric matrix: $\mathbf{V}=\begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0\end{pmatrix}$ A tensor of order 3 should probably be defined.
Edit The question is related to the following one: knowing that there exists a matrix $\mathbf{V}\in\mathbb{R}^{3,3}$ such that for a given vector $\mathbf{v}\in\mathbb{R}^3$: $\forall\mathbf{x}\in\mathbb{R}^3,\quad\mathbf{V}\mathbf{x}=\mathbf{v}\times \mathbf{x}\quad\Leftrightarrow\quad \mathbf{V}=\mathrm{CPM}(\mathbf{v})$ where CPM means cross-product matrix, can we express in a frame-invariant fashion the quantity: $ \mathbf{V}_\mathrm{A}=\mathrm{CPM}(\mathbf{Av})$ where $\mathbf{A}$ is any $3\times 3$ real matrix?
(the result is $\mathbf{V}_\mathrm{A}=(\mathbf{VA})^T-\mathbf{VA}+\mathrm{tr}(\mathbf{A})\mathbf{V}$ but was obtained by calculating each coordinate of the left-hand and right-hand side matrices and subsequently identifying each term)