Let $V$ be a $n$-dim vector space, $\{e_1,\cdots,e_n\}$ is a positive orthonormal basis, $\{v_1,\cdots,v_n\}$ is an arbitrary positive basis, how to show $$ v_1\wedge \cdots\wedge v_n =\sqrt{det(\langle v_i,v_j\rangle)}\ e_1\wedge \cdots \wedge e_n $$
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Note that $$v_1\wedge\cdots\wedge v_n={\rm det}\ [v_1\cdots v_n]\ e_1\wedge\cdots \wedge e_n $$
If $V=[v_1\cdots v_n]$, then $$ {\rm det}\ V =\sqrt{{\rm det}\ V^TV} $$ Here $V^TV$ has entry $(v_i,v_j)$ so that we complete the proof
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$\begingroup$ what is ${\rm det}\ [v_1\cdots v_n]$ ? Why $v_1\wedge\cdots\wedge v_n={\rm det}\ [v_1\cdots v_n]\ e_1\wedge\cdots \wedge e_n$ ? $\endgroup$ Dec 12, 2016 at 3:10
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$\begingroup$ ${\rm det}\ [v_1\cdots v_n]$ is determinant of a matrix $[v_1\cdots v_n]$ And recall the definition of determinant Wedge product $v_1\wedge\cdots \wedge v_n$ satisfies the definition of determinant $\endgroup$– HK LeeDec 12, 2016 at 3:15
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