# From a vector to a skew symmetric matrix

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)

-
What you've defined in the question already is a linear mapping. What is the question then? – Dan Shved Dec 14 '12 at 17:00
yes, you are right. The question is to characterize the linear mapping $\Gamma$ such that $\mathbf{V}=\Gamma(\mathbf{v})$. – pluton Dec 14 '12 at 17:02
This is still not clear. What do you mean by "characterize"? You have already defined $\Gamma$ by saying how to find $\Gamma(v)$ for every $v \in \mathbb{R}^3$. What other characterization do you need? Or would you like to have an expression for components of this tensor, i.e. $\Gamma_{ijk}$ for all appropriate $i, j, k$? – Dan Shved Dec 14 '12 at 17:05
If so, then you can say that $V_{ji} = \sum e_{ijk} v^k$, where $e_{ijk}$ equals $0$ when at least two of $i,j,k$ are equal, $+1$ if $(i,j,k)$ is a cyclic permutation of $(1, 2, 3)$ and $-1$ otherwise. – Dan Shved Dec 14 '12 at 17:08
It is also possible that what you're really asking is if this definition is invariant, i.e. if it depends on the choice of basis. Well, the short answer is: it is not invariant. However, it is preserved under positive orthonormal changes of coordinates. – Dan Shved Dec 14 '12 at 17:11

The name of the tensor you're looking for is the Levi-Civita or permutation tensor. In cartesian coordinates, $\epsilon_{ijk}$ is equal to $+1$ for any even permutation of 123 and $-1$ for any odd permutation.
Is $A$ supposed to be some arbitrary matrix? – Muphrid Dec 15 '12 at 0:16
yes, $\mathbf{A}$ is a $3\times 3$ arbitrary matrix. – pluton Dec 15 '12 at 2:01
Seems like it'd just be $VAx$ then. – Muphrid Dec 15 '12 at 4:28
yes, but vector $\mathbf{x}$ should not be part of the final formula. It is just a vector onto which matrix $\mathbf{V}$ acts. – pluton Dec 15 '12 at 14:50