Trace of $p \wedge p$ Hello everyone here is a little problem I have some trouble with :
Let $V$ a vector space with $v_1,v_2,v_3$ as a basis and we define $p$ an endomorphism of $V$ with the following matrix :
$$A=\begin{pmatrix}
 a&b  &c \\ 
 d&e  &f \\ 
 g&h  &i 
\end{pmatrix}$$

I want to prove that the trace of $p\wedge p$ is the sum of the three minors of order $2$ of $A$

A hint that a one of my friend gave me is to compute the values of the diagonal elements of the matrix $p\wedge p$ in the basis $v_1\wedge v_2,v_1\wedge v_3,v_2\wedge v_3$ of $V\wedge V$ but I don't know how to do if anyone can help compute one or two coefficients of this matrix so I can do it for the others it would be a lot appreciated , thanks in advance
 A: Let $\alpha_1,\alpha_2,\alpha_3$ denote the dual basis associated with $v_1,v_2,v_3$. Note that $\alpha_1 \wedge \alpha_2, \alpha_1 \wedge \alpha_3, \alpha_2 \wedge \alpha_3$ is the dual basis associated with $v_1 \wedge v_2, v_1 \wedge v_3, v_2 \wedge v_3$, where we have
$$
(\alpha \wedge \beta)(v \wedge w) = \det \pmatrix{\alpha(v) & \alpha(v)\\ \beta(v) & \beta(w)}.
$$
Note that for a space $V$ with basis $B = (v_1,\dots,v_n)$ and dual basis $\alpha_1,\dots,\alpha_n$, the $i$th diagonal element of the matrix of $\phi:V \to V$ with respect to $B$ is $\alpha_i(\phi(v_i))$. With that established, note that
$$
(\alpha_i \wedge \alpha_j)(p\wedge p)(v_i \wedge v_j) = 
(\alpha_i \wedge \alpha_j)(pv_i \wedge pv_j) = \\
\det \pmatrix{\alpha_i \circ pv_i & \alpha_i \circ pv_j\\
\alpha_j \circ pv_i & \alpha_j \circ pv_j}.
$$

Proof of the determinant formula:
Let $m_{ij} = \alpha_i (pv_j)$. We have $pv_j = \sum_i m_{ij}v_i$. Thus, we have
$$
\begin{align}
(p \wedge p)(v_i \wedge v_j) &=
pv_i \wedge pv_j 
\\ & = 
\left(\sum_{k} m_{ki}v_k \right) \wedge
\left(\sum_{\ell} m_{\ell j}v_\ell \right) 
\\ & = 
\sum_{k,\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
\\ & = 
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k>\ell} -m_{ki}m_{\ell j} (v_\ell \wedge v_k)
\\ & = 
\sum_{k<\ell} m_{ki}m_{\ell j} (v_k \wedge v_\ell)
+ \sum_{k'<\ell'} -m_{\ell'i}m_{k' j} (v_{k'} \wedge v_{\ell'})
\\ & = 
\sum_{k<\ell} (m_{ki}m_{\ell j} - m_{kj}m_{\ell i}) (v_k \wedge v_\ell)
= 
\sum_{k<\ell} \det\pmatrix{m_{ki} & m_{\ell i}\\ m_{kj} & m_{\ell j}} \cdot (v_k \wedge v_\ell).
\end{align}
$$
