Which Subspaces do Antisymmetric Tensors Represent? So antisymmetric tensors represent volumetric subspaces (I've asked this here instead of on phys.stackexchange because it seems like more of a math question)?  How exactly would one know WHICH subspace a given antisymmetric tensor then represents?
For instance, let's take a simple case: a 2D subspace in $\Bbb R^3$.  If we had a tensor with components $$\begin{bmatrix} 0 & 4 & -16\\ -4 & 0 & 5\\ 16 & -5 & 0\\ \end{bmatrix}$$ with respect to the standard basis, which subspace would this represent?
Likewise, if I had a plane (subspace) specified by $3x + 5y +9z = 0$, which antisymmetric tensor would I use to represent it?
 A: Let's take an orthogonal basis of $\mathbb{R}^3$: $e_1, e_2, e_3$. We can write vectors $a = \sum_i a_i e_i$ and $b = \sum_i b_i e_i$. The oriented subspace spanned by these vectors is a 2-blade, denoted $a \wedge b$:
$$a \wedge b = \left(\sum_i a_i e_i\right) \wedge \left(\sum_j b_j e_j\right) = \sum_{ij} a_i b_j e_i \wedge e_j$$
where in the last line we used the linearity of the wedge product. The wedge product is also defined to be anti-symmetric, because it represents a directed subspace: $a \wedge b = -b \wedge a$, for all vectors $a,b$ including the basis vectors. So we can group the terms containing $e_i \wedge e_j$ and $e_j \wedge e_i$ in the sum:
$$a \wedge b = \sum_{ij} a_i b_j e_i \wedge e_j = \sum_{i < j} (a_i b_j - a_j b_i) e_i \wedge e_j$$
Hence, a 2-blade (and a sum of 2-blades which is called a bivector) is described by $n(n-1)/2$ coefficients (i.e. n choose 2). If we put these coefficients in the upper triangle of a matrix, and put their negation in the lower triangle, we can identify the bivectors with the anti-symmetric matrices. For blades we find:
$$
a \wedge b \sim a b^T - ba^T
$$
So given an anti-symmetric matrix $B$, we wish to find a basis for the subspace it represents as a blade, i.e. we wish to factor the blade $B = a \wedge b$, or in matrix language $B = a b^T - ba^T$. It is important to note that while this can be done for any anti-symmetric matrix in 3D, that is not true for higher dimensions. In n dimensions, an anti-symmetric matrix may be associated with a bivector, which may or may not be factorizable (i.e. may or may not be a blade that represents a subspace).
The linear map on vectors defined by the matrix $B$ is a contraction: $B a = a \rfloor B$ (on the left-hand side $B$ is a matrix, on the right hand side it is a blade). Geometrically, this corresponds to a projection of $a$ onto the subspace, followed by a 90 degree rotation and a uniform scaling (the amount of scaling depends on the norm of $B$). From this it is clear that the vector $n$ orthogonal to the subspace (i.e. its normal vector) is an eigenvector with eigenvalue $\lambda= 0$ (since the geometric operation just described acts on $n$ as: $Bn = 0 n$. Furthermore, $B$ has two complex-conjugate eigenvectors, the real and imaginary components of which span the invariant subspace of the operator $B$, which is the subspace it represents as a blade.
If you want to get a real-valued basis right away, it is easy to see that you can obtain it by diagonalizing the symmetric matrix $B^2$.
Going in the other direction, we have the normal vector $n = (3,5,9)$ that in 3D uniquely determines a 2D subspace. Algebraically, we can easily obtain a 2-blade by dualization (taking the orthogonal complement): $B = n^*$. This may be written using a contraction, which is a generalization of the dot product: $B = n^* = n \rfloor I$ where $I = e_1\wedge e_2 \wedge e_3$. Geometrically, we take the entire space $I$ and "take out" the normal vector $n$ leaving the subspace orthogonal to it: $B$. If you learn a bit of geometric algebra, you can easily obtain the coordinate formula for this operation. However, unless you are doing computational work, its usually much cleaner to leave things at the algebraic level.
