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I'm solving the problem 4-7 from Spivak's "Calculus on Manifolds", which is the following: show that every non-zero $\omega \in \Lambda^n(V)$ (according to Spivak's notation it's the set of all alternating n-tensors on $n$-dimensional space $V$ over real numbers) is the volume element determined by some inner product $T$ and orientation $\mu$.

Thanks for any help.

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Since $\omega$ is non-zero, we can choose $e_1,\dots,e_n \in V$ such that $\omega(e_1,\dots,e_n) = 1$. Define an inner product on $V$ by declaring that $e_i$ are orthonormal. More formally, given $v = \sum_{i=1}^n a_i e_i$ and $w = \sum_{i=1}^n b_i e_i$, set

$$ \left< v, w \right> := \sum_{i=1}^n a_i b_i. $$

Define an orientation $\mu$ on $V$ by declaring $e_1,\dots,e_n$ to be positively oriented. Then $\omega$ is the volume form associated to $(\left< \cdot, \cdot \right>, \mu)$.

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    Thank you! Actually, I did something different in the end, I said $\omega = \lambda \cdot \det$ (since $\Lambda^n(V)$ is one dimensional), then we can define a basis consisting of $v_i=\frac{e_i}{\sqrt[n]{\lambda}}$, thus $\omega(v_1,\ldots,v_n) = 1$ and we define inner product as $T(x,y) = \sqrt[n]{\lambda^2}\langle x, y \rangle$, is it right? – 2017-01-28
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    I'm not sure how Spivak develops everything but in an abstract vector space there is no $\det$. If $V = \mathbb{R}^n$ and the $e_i$ are the standard basis vectors then your proof more or less works, except $\lambda$ can be negative and then the orientation comes into play. – 2017-01-28
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    Thanks! Indeed, I forgot, if $\lambda < 0$ we can just replace, say, $v_1$ by $-v_1$. By the way, if $V$ is not equal to $\mathbb{R}^n$, can we still use the same argument applied to the transformation matrices between bases? I'm asking because there was a theorem (4-6) about it, if you don't have a copy of his book it's OK, anyway thanks for your answers. – 2017-01-28