An orientation on $\Lambda^k (V^\ast)$ defines an orientation on $V$

Below is an exercise from Guillemin and Pollack:

(a) Let $T$ be a nonzero element of $\Lambda^k (V^\ast)$, where $\dim⁡ > V=k$. Prove that two ordered bases $(v_1,\dots,v_k)$ and $(v_1',\dots,v_k')$ for $V$ are equivalently oriented if and only if $T(v_1,\dots,v_k )$ and $T(v_1',\dots,v_k' )$ have the same sign.

(b) Suppose that $V$ is oriented. Show that the one-dimensional vector space $\Lambda^k (V^\ast)$ acquires a natural orientation, by defining the sign of a nonzero element $T\in\Lambda^k (V^\ast)$ to be the sign of $T(v_1,\dots,v_k)$ for any positively oriented ordered basis $(v_1,\dots,v_k)$ for $V$.

(c) Conversely, show that an orientation of $\Lambda^k (V^\ast)$ naturally defines an orientation on $V$ by reserving the above.

I understand how to do (a) and (b), but I'm not sure whether I understand what exactly I have to check in (c).

So in (c), $\Lambda^k(V^\ast)-0$ is split into two equivalence classes, the equivalence relation being $T\simeq T'\iff T=aT'$ for $a> 0$. If $T$ is a positively oriented element of $\Lambda^k(V^\ast)$, then a basis $(v_1,\dots,v_k)$ is defined to be positively oriented provided $T(v_1,\dots,v_k)> 0$. If $T'$ is another positively oriented $k$-tensor, then it must differs from $T$ by a positive factor, and thus $T'(v_1,\dots,v_k)>0$ as well. Is that is what I'm supposed to check? Or is there something else to verify?

• Nothing else to verify, you got it right. – Amitai Yuval Apr 17 '18 at 4:56