Multilinear algebra and matrices 
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 $$\wedge^k(T)(v_1\wedge\dots\wedge v_k) = T(v_1) \wedge \dots \wedge T(v_k)$$


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 $$\begin{bmatrix}
1 &-1  &-1 \\ 
 1&1  &-1 \\ 
1 &1  &1 
\end{bmatrix}$$
 A: 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
$$
\Lambda^2(T) = \begin{pmatrix}
1 & -1 & -1 \\
1 & 1 & -1 \\
1 & 1 & 1 \end{pmatrix}.
$$
