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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)

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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

1 Answer 1

up vote 2 down vote accepted

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.

The permutation tensor represents a directed unit volume. As such volumes have different expressions in other coordinate systems, the Levi-Civita tensor will depend on the coordinate system used. However, it represents a fundamental, coordinate-system invariant object called the pseudoscalar of the space. And there are ways to use the pesudoscalar without resorting to the tensor approach, avoiding the problem of how the pseudoscalar has different expressions in different coordinates.

The pseudoscalar enforces the notion of duality. In 3d space, vectors are dual (orthogonal to, or normal to) planes. In some circles, we call directed planes bivectors. Using the pseudoscalar in this way converts back and forth between vectors and their dual bivectors. The skew-symmetric rank-2 tensor (matrix) you have here is the direct representation of such a bivector.

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Interesting. Thank you. I edited my question to provide more insight. –  pluton Dec 14 '12 at 23:20
    
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

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