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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: $$\Lambda^k(T)(v_{i_1} \wedge \dotsc \wedge v_{i_k}) = T(v_{i_1}) \wedge \dotsc \wedge T(v_{i_k})=\sum_{j_1,\dotsc,j_k} ... 2 As remarked by Daniel Rust the area of the hexagon in question is the sum of the projected areas of the three "kinds" of cube facets. When a piece of a plane \Sigma is orthogonally projected onto another plane \Pi then the area is multiplied by |\cos\phi|, where \phi is the angle between the planes, or equivalently: the angle between the ... 1 Let \mathbf{n} be a unit vector in \mathbb{R}^3 be normal to the plane P. The projection p of a vector v\in\mathbb{R}^3 onto the plane P is given by p(v)=v-(v\cdot\mathbf{n})\mathbf{n}. The projection of the unit cube onto P along the vector \mathbf{n} is a hexagon whose interior has preimage which intersects the boundary of the unit cube ... 1 I thought it might be worthwhile to add an explicit example. Consider T:\mathbb{R}^3 \rightarrow \mathbb{R}^3 with$$ [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), \ \ \ ...

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