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1

I thought it might be worthwhile to add an explicit example. Consider $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ with $$[T] = \left[ \begin{array}{ccc} a & d & g \\ b & e & h \\ c & f & i \end{array}\right]$$ In particular, using the usual $e_1 = [ 1,0,0 ]^T=(1,0,0)$, $e_2 = (0,1,0)$, $e_3=(0,0,1)$ we have $$T(e_1) = (a,b,c), \ \ \ ... 4 Let T(v_i) = \sum_j a_{ji} \cdot v_j, so that (a_{ji}) is the matrix of T w.r.t. to (v_1,\dotsc,v_n). The wedges v_{i_1} \wedge \dotsc \wedge v_{i_k} with i_1<\dotsc<i_k form a basis of \Lambda^k(V). We have:$$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} ...

2

As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane $\Sigma$ is orthogonally projected onto another plane $\Pi$ then the area is multiplied by $|\cos\phi|$, where $\phi$ is the angle between the planes, or equivalently: the angle between the ...

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Let $\mathbf{n}$ be a unit vector in $\mathbb{R}^3$ be normal to the plane $P$. The projection $p$ of a vector $v\in\mathbb{R}^3$ onto the plane $P$ is given by $p(v)=v-(v\cdot\mathbf{n})\mathbf{n}$. The projection of the unit cube onto $P$ along the vector $\mathbf{n}$ is a hexagon whose interior has preimage which intersects the boundary of the unit cube ...

2

The norm is the same as the absolute value of the cross product followed by the dot product (or the absolute value of the determinant of a matrix with columns $a,b, c$): $$\| a \wedge b \wedge c \| = \lvert a \cdot (b \times c) \rvert = \lvert\det[ a\ b\ c]\rvert$$ Its geometric interpretation is quite different however. That is $a \wedge b \wedge c$ is ...

0

Antisymmetrization is an alternating function on tensors, and thus factors through the wedge product. The map $$a \wedge b \mapsto \frac{1}{2} \left(a \otimes b - b \otimes a\right)$$ has a one-sided inverse $$a \otimes b \mapsto a \wedge b$$ and thus the space of wedges is isomorphic to the space of anti-symmetric tensors. Some will take advantage ...

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