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Given $\wedge^k(V)$ an alternating multilinear space and $T : V \to W$ a linear map, then we have $$v_1 \wedge \dots \wedge v_k \in\wedge^k(V).$$ Define $$\wedge^k(T)(v_1\wedge\dots\wedge v_k) = T(v_1) \wedge \dots \wedge T(v_k)$$

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Can someone explain to me why $T(e_1) \wedge T(e_2) = (e_1 + e_2) \wedge (e_2 + e_3)$? Specifically why is $T(e_2) = e_2 + e_3$?

Because I was (for some reason) under the impression that $T(e_2) = e_1 \wedge e_3.$

EDIT: How do they get the matrix for $$\begin{bmatrix} 1 &-1 &-1 \\ 1&1 &-1 \\ 1 &1 &1 \end{bmatrix}$$

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The linear transformation $T$ takes vectors in $\mathbb{R}^3$ to vectors in $\mathbb{R}^3$, so $T(e_2)$ needs to be a vector in $\mathbb{R}^3$ and not an element of $\Lambda^2 (\mathbb{R}^3)$. Specifically, $$ T(e_2) = \begin{pmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} 0\\ 1 \\ 0 \end{pmatrix} =\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = e_2 + e_3. $$ Similarly, we get that $T(e_1) = e_1 + e_2$, so we must have that $$ T(e_1) \wedge T(e_2) = (e_1 + e_2) \wedge (e_2 + e_3). $$

Edit: To address your follows-up question, recall that to find the matrix of the linear operator $\Lambda^2(T)$ with respect to the basis $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$, it suffices to find the image of the basis vectors under $\Lambda^2(T)$. The first such computation is done above, giving that $$ \Lambda^2(T) (e_1 \wedge e_2) = e_1 \wedge e_2 + e_1 \wedge e_3 + e_2 \wedge e_3. $$ Let's compute the image of the next basic vector: \begin{align} \Lambda^2(T)(e_1 \wedge e_3) &= T(e_1) \wedge T(e_3)\\ &= (e_1 + e_2) \wedge (e_1 + e_3)\\ &= e_1 \wedge e_1 + e_1 \wedge e_3 + e_2 \wedge e_1 + e_2 \wedge e_3\\ &= -e_1 \wedge e_2 + e_1 \wedge e_3 + e_2 \wedge e_3. \end{align} Similarly, we find that $\Lambda^2(T)(e_2 \wedge e_3) = -e_1 \wedge e_2 - e_2 \wedge e_3 + e_2 \wedge e_3$. Recall that the first column of the matrix $\Lambda^2(T)$ will be the image of the first basis vector $e_1 \wedge e_2$, i.e. the vector $(1,1,1)$; the second column of the matrix will be the image of the second basis vector $e_1 \wedge e_3$, i.e. the vector $(-1,1,1)$; the third column of the matrix is the image of the third basis vector $e_2 \wedge e_3$, i.e. $(-1,-1,1)$. Therefore, the matrix of $\Lambda^2(T)$ with respect to the basis $\{ e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3 \}$ is given by $$ \Lambda^2(T) = \begin{pmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{pmatrix}. $$

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  • $\begingroup$ Oh so it is the standard basis. $\endgroup$ – Hawk Sep 26 '14 at 0:37
  • $\begingroup$ How did they get $$\begin{bmatrix} 1 &-1 &-1 \\ 1&1 &-1 \\ 1 &1 &1 \end{bmatrix}$$? $\endgroup$ – Hawk Sep 26 '14 at 2:15
  • $\begingroup$ Just edited my answer with a follow-up. $\endgroup$ – msteve Sep 26 '14 at 2:26
  • $\begingroup$ What does the symbol $V$ here mean? The upside down wedge? $\endgroup$ – Hawk Sep 26 '14 at 2:37
  • $\begingroup$ My mistake, should have been a wedge product. Fixed it now. $\endgroup$ – msteve Sep 26 '14 at 2:38

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