Calculating the map "induced" by the exterior product? I have an exam coming up (for differential topology) and my professor said the exam would be mostly computation oriented (particular emphasis on $\mathbb{R}^3$). I have proven some results theoretically about tensors in Guillemen and Pollack and but am not really confident in computation. I believe he said we should be able to "compute" the map "induced" by the exterior product. To (hopefully) be more precise he had put on the board, given a matrix:
$$A : V \to V$$
compute the corresponding:
$$\wedge^kA : \wedge^k V \to \wedge^k V$$
This was my best guess at a direct computation, but I am unsure if I am doing it correctly:
For example, say we have the matrix:
\begin{bmatrix}
3 & 1 & 1 \\
1 & 0 & 3 \\
2 & 1 & 0 \\
\end{bmatrix}
This is a map from $R^3 \to R^3$. Now I want to compute to compute:
$$\wedge^2 A : \wedge^2 R^3 \to \wedge^2R^3$$
I choose would then choose an ordered basis to be:
$$ \{ e^1\wedge e^2, e^1 \wedge e^3, e^2 \wedge e^3 \} $$
(I guess the order could change, but I assume this would be given in a question?)
I then apply $A^*$ to each of the basis to obtain a representation.
$$A^*e^1\wedge e^2$$
$$A^*e^1\wedge e^3$$
$$A^*e^2\wedge e^3$$
This becomes:
$$Ae^1\wedge Ae^2$$
$$Ae^1\wedge Ae^3$$
$$Ae^2\wedge Ae^3$$
This then implies:
$$ \begin{bmatrix} 3 \\ 1 \\ 2 \\ \end{bmatrix} \wedge 
   \begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix}$$
$$ \begin{bmatrix} 3 \\ 1 \\ 2 \\ \end{bmatrix} \wedge 
   \begin{bmatrix} 1 \\ 3 \\ 0 \\ \end{bmatrix}$$
$$ \begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix} \wedge 
   \begin{bmatrix} 1 \\ 3 \\ 0 \\ \end{bmatrix}$$
In the case that the dimension is equal, I believe the wedge should be just the dot product, and that this reduces down to the numbers $5$, $6$, and $1$. Then this becomes:
$$\wedge^2 A = \begin{bmatrix} 5 & 6 & 1 \end{bmatrix}$$
Is this correct? Is this how you compute the "map induced" by the exterior product on a matrix? (Sorry if this has been answered, I have had trouble finding concrete computations of exterior algebras...)
 A: Let us denote by $(e_1,e_2,e_3)$ the standard basis of $\mathbb{R}^3$. The space $\Lambda^2(\mathbb{R}^3)$ is a three dimensional vector space and $\Lambda^2(A) \colon \Lambda^2(\mathbb{R}^3) \rightarrow \Lambda^2(\mathbb{R}^3)$ is a linear map so it should be represented by a $3 \times 3$ matrix with respect to some choice of basis of $\Lambda^2(\mathbb{R}^3)$. Let us choose the basis $\mathcal{B} = (e_1 \wedge e_2, e_1 \wedge e_3, e_2 \wedge e_3)$. The order indeed matters and if would reorder our basis, we would get a different matrix (it would have the same columns, but in different order). To find $[\Lambda^2(A)]_{\mathcal{B}}$, we need to compute $\Lambda^2(A)(e_1 \wedge e_2), \Lambda^2(A)(e_1 \wedge e_3), \Lambda^2(A)(e_2 \wedge e_3)$ and express the result in terms of the basis $\mathcal{B}$.
For example,
\begin{align*}
\Lambda^2(A)(e_1 \wedge e_2) &= (Ae_1 \wedge Ae_2) = 
(3e_1 + e_2 + 2e_3) \wedge (e_1 + e_3) \\
&=  3 (e_1 \wedge e_1) + 3 (e_1 \wedge e_3) + e_2 \wedge e_1 + e_2 \wedge e_3 + 2(e_3 \wedge e_1) + 2(e_3 \wedge e_3) \\
&=3(e_1 \wedge e_3) + e_2 \wedge e_1 + e_2 \wedge e_3 + 2(e_3 \wedge e_1) \\&=
(-1)(e_1 \wedge e_2) + 3(e_1 \wedge e_3) - 2(e_1 \wedge e_3) + e_2 \wedge e_3 \\&= 
(-1) \cdot (e_1 \wedge e_2) + 1 \cdot (e_1 \wedge e_3) + 1 \cdot (e_2 \wedge e_3)
\end{align*}
Thus, the matrix $[\Lambda^2(A)]_{\mathcal{B}}$ will look like
$$ [\Lambda^2(A)]_{\mathcal{B}} = \begin{pmatrix} -1 & ? & ? \\ 1 & ? & ? \\
1 & ? & ? \end{pmatrix}. $$
The two other columns will be obtained by computing $\Lambda^2(A)(e_1 \wedge e_3)$ and $\Lambda^2(A)(e_2 \wedge e_3)$ similarly. Note that I have used the properties of the wedge product to cancel terms and reorder them (introducing a minus sign) in order to express $\Lambda^2(A)(e_1 \wedge e_2)$ in terms of the basis elements.
