Let $\mathcal{B}$ be an ordered basis for a vector space $V$ of dimension $n$. Denote the basis for the set of anti symmetric tensors ($\Lambda^{p}{V}$) by $\mathcal{B}^{p}$. For $n=3$, let $A=\left( a^{i}_{j} \right )$ be the matrix of the linear map $f: V \to V$ with respect to the basis $\mathcal{B}$. Describe the matrices of $\Lambda^{p}{f}$ with respect to $\mathcal{B}^{p}$ for $p=0,1,2,3$.
So far I have gotten $\Lambda^{0}f = \mathbb{1}_{\mathbb{F}}$.
And for $p=3$:
$\mathcal{B}=\{ e^{123} \} = \{ e^{1} \wedge e^{2} \wedge e^{3} \}$ \begin{align*} (\Lambda^{3}f(e^{123})) (e_{1}, e_{2}, e_{3}) := e^{123} (f(e_{1}), f(e_{2}), f(e_{3})) = e^{123} (a^{\alpha}_{1}e_{\alpha} , a^{\beta}_{2}e_{\beta}, a^{\gamma}_{3}e_{\gamma}) \\ = a^{\alpha}_{1} a^{\beta}_{2} a^{\gamma}_{3} e^{123} (e_{\alpha}, e_{\beta}, e_{\gamma}) \\ = a^{\alpha}_{1} a^{\beta}_{2} a^{\gamma}_{3} \delta_{\alpha}^{1} \delta_{\beta}^{2} \delta_{\gamma}^{3} = a^{1}_{1} a^{2}_{2} a^{3}_{3} \end{align*}
I'm not sure how to construct the matrices for p=1,2. As I have not found to get a matrix that is square for the transformation.
For p=2, I have tried:
$\mathcal{B}^{2} = \{e^{12}, e^{13}, e^{23}\}$ \begin{align*} (\Lambda^{2}f(e^{12}))(v^{\alpha} e_{\alpha}, w^{\beta} e_{\beta}) = e^{12} (v^{\alpha} f(e_{\alpha}), w^{\beta} f(e_{\beta})) \\ = v^{\alpha}w^{\beta} e^{12}(a_{\gamma}^{\alpha} e_{\gamma}, a_{\zeta}^{\beta} e_{\zeta}) \\ = v^{\alpha}w^{\beta} a_{\gamma}^{\alpha} a_{\zeta}^{\beta} e^{12}(e_{\gamma}, e_{\zeta}) \\ = v^{\alpha}w^{\beta} a_{\gamma}^{\alpha} a_{\zeta}^{\beta} \delta^{1}_{\gamma} \delta^{2}_{\zeta} \\ = v^{\alpha}w^{\beta} a_{1}^{\alpha} a_{2}^{\beta} \\ \end{align*}
But this is only 1 dimensional and I'm guessing a need a 3 dimensional vector to fill in the 3x3 matrix required as the basis is 3 dimensional?
Any help as to where to go from here would be much appreciated.
It seems you are actually referring to $\Lambda^pV^*$ and $\Lambda^pf^*$; these constructions are perfectly valid over $V$ rather than $V^*$, so I was confused at first.
Once we fix the basis $\mathcal B$, we can treat the elements of $V$ as column vectors and those of $V^*$ as row vectors with $e^1 = (1,0,0,\dotsc)$, $e^2 = (0,1,0,\dotsc)$, etc. Then $f(v) = Av$ and the dual $f^* : V^* \to V^*$ can be written $f^*(\omega) = \omega A$. Now, $\Lambda^pf^*$ is defined such that it is a homomorphism over $\wedge$, so $$\begin{aligned} {[}\Lambda^2f^*(e^i\wedge e^j)](e_k, e_l) & = [f^*(e^i)\wedge f^*(e^j)](e_k, e_l) = (e^iAe_k)(e^jAe_l) - (e^iAe_l)(e^jAe_k) \\& = a^i_ka^j_l - a^i_la^j_k. \end{aligned}$$ To get a matrix out of this, we have to decide on an order for $\mathcal B^p$ and assign a linear index accordingly. Writing this out in full generality is painful, but it will look something like the following: ordering $\mathcal B^2$ via $$ e^1\wedge e^2,\quad e^1\wedge e^3,\quad\dotsc,\quad e^1\wedge e^n,\quad e^2\wedge e^3,\quad\dotsc,\quad e^{n-1}\wedge e^n, $$ the matrix of $\Lambda^2f^*$ looks something like $$\begin{pmatrix} a^1_1a^2_2 - a^1_2a^2_1 & a^1_1a^3_2 - a^1_2a^3_1 & \cdots & a^1_1a^n_2 - a^1_2a^n_1 & a^2_1a^3_2 - a^2_2a^3_1 & \cdots & a^{n-1}_1a^n_2 - a^{n-1}_2a^n_1 \\ a^1_1a^2_3 - a^1_3a^2_1 & a^1_1a^3_3 - a^1_3a^3_1 & \cdots & a^1_1a^n_3 - a^1_3a^n_1 & a^2_1a^3_3 - a^2_3a^3_1 & \cdots & a^{n-1}_1a^n_3 - a^{n-1}_3a^n_1 \\ \vdots & \vdots && \vdots & \vdots && \vdots \\ a^1_1a^2_n - a^1_na^1_1 & a^1_1a^3_n - a^1_na^3_1 & \cdots & a^1_1a^n_n - a^1_na^n_1 & a^2_1a^3_n - a^2_na^3_1 & \cdots & a^{n-1}_1a^n_n - a^{n-1}_na^n_1 \\ a^1_2a^2_3 - a^1_3a^1_2 & a^1_2a^3_3 - a^1_3a^3_2 & \cdots & a^1_2a^n_3 - a^1_3a^n_2 & a^2_2a^3_3 - a^2_3a^3_2 & \cdots & a^{n-1}_2a^n_3 - a^{n-1}_3a^n_2 \\ \vdots & \vdots && \vdots & \vdots && \vdots \\ a^1_{n-1}a^2_n - a^1_na^1_{n-1} & a^1_{n-1}a^3_n - a^1_na^3_{n-1} & \cdots & a^1_{n-1}a^n_n - a^1_na^n_{n-1} & a^2_{n-1}a^3_n - a^2_na^3_{n-1} & \cdots & a^{n-1}_{n-1}a^n_n - a^{n-1}_na^n_{n-1} \end{pmatrix}$$