Prove that $\phi_1 \wedge \cdots \wedge \phi_k (v_1, \cdots, v_k) = \frac{1}{k!}\det[\phi_i(v_j)].$

I have proved these two exercises:

(1) Suppose that $T \in \Lambda^p(V^*)$ and $v_1, \ldots, v_p \in V$ are linearly dependent. Prove that $T(v_1, \ldots, v_p) = 0$ for all $T \in \Lambda^p(V^*)$.

And

(2) Suppose that $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, where $k$ = dim$V$. Prove that $$\phi_1 \wedge \cdots \wedge \phi_k (v_1, \cdots, v_k) = \frac{1}{k!}\det[\phi_i(v_j)].$$

But now I am asked to prove

Show that whenever $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, then $$\phi_1 \wedge \cdots \wedge \phi_p (v_1, \cdots, v_p) = \frac{1}{k!}\det[\phi_i(v_j)].$$

But my concern is that, if $\phi_1, \ldots, \phi_k \in V^*$ are linearly dependent, it follows directly from (1), otherwise, (2) can be extend to any sub-dimension. So, that's it?

• That's the idea, but you should probably expand on why (2) can be extended. – anon Jun 25 '13 at 0:05
• @anon Is that because they are linearly independent, so also spans the $p$-dimension subspace..? – WishingFish Jun 25 '13 at 0:15
• This answer can be useful math.stackexchange.com/questions/426349/… – Avitus Nov 14 '13 at 17:53