4. (18 pts) Suppose that $\phi$ is an $n$-permutation, and that $P_{\phi}$ is its corresponding permutation matrix.\
Let $e_n = (e_1, e_2, \dots, e_n)$ be the standard basis for $\mathbb{R}^n$. Show that $P_{\phi}e_{\phi(i)} = e_i$.\
Given a vector space $V$, we can define the $k^{th}$ exterior power of $V$, denoted $\bigwedge^k V$, as the vector space spanned by expressions of the form
$\vec{v}_1 \wedge \vec{v}_2 \wedge \dots \wedge \vec{v}_k$
where $\vec{v}_i \in V$. Such expressions are sometimes called multivectors. This wedge product, "$\wedge$", satisfies the following axioms:
$\bullet$ Associativity: $(\vec{v}_1 \wedge \vec{v}_2) \wedge \vec{v}_3 = \vec{v}_1 \wedge (\vec{v}_2 \wedge \vec{v}_3)$.\
$\bullet$ Distrbutivity: $\vec{v} \wedge (\vec{u}_1 + \vec{u}_2) = (\vec{v} \wedge \vec{u}_1) + (\vec{v} \wedge \vec{u}_2)$.\
$\bullet$ Anticommutivity: $\vec{v} \wedge \vec{u} = -\vec{u} \wedge \vec{v}$.\
$\bullet$ Compatibility with scalar product: $(k\vec{v}) \wedge \vec{u} = \vec{v} \wedge (k\vec{u})$ where $k \in \mathbb{R}$.\
Because of the third property, $\vec{v} \wedge \vec{v} = 0$ for any vector $\vec{v}$. Because of the fourth property, we can write both sides of the equation as $k(\vec{v} \wedge \vec{u})$.
