# Did I give enough justification when I extend to $p$-dimensional?

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 $$\phi_1 \wedge \cdots \wedge \phi_p (v_1, \dots, v_p) = \frac{1}{k!}\det[\phi_i(v_j)].$$

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, because they are linearly independent, so also spans the $p$-dimension subspace.

• Well, the $\frac{1}{k!}$ must of course be $\frac{1}{p!}$, otherwise choose $p = 1$ to find it wrong. – Daniel Fischer Jun 25 '13 at 18:55
• This looks like Guillemin and Pollack's cockamamy definition of wedge product, wherein the volume of the unit cube in $\mathbb R^k$ is $1/k!$. Personally, I would delete the factorials entirely here. You certainly mean to have $k=p$ throughout in your final set-up. Think about extending a set of linearly independent vectors to a basis. – Ted Shifrin Jun 25 '13 at 19:06
• Well, many people are really fond of it. I always find trouble with it. - yes, it is from Guillemin and Pallack. – WishingFish Jun 25 '13 at 19:09
• I'm extraordinarily fond of the book, having taken the course from Guillemin as he was writing it. However, I've taught out of the book close to a dozen times and I never use their crazy definition of wedge. – Ted Shifrin Jun 25 '13 at 19:11
• I enjoy reading it some time despite of finding trouble with it many time. Though, what would be a good alternative, wedge definition wise and book wise..? – WishingFish Jun 25 '13 at 19:12