First in $\mathbb{R}^3$ for $u=(u_1,u_2,u_3), \ v=(v_1,v_2,v_3), \ w=(w_1,w_2,w_3)$ the triple product gives :

$\det(u,v,w)=(u \times v).w$, where $\times$ is the vectorial product.

Now for $\mathbb{R}^n$ and $u_1=(u_{1_1},...,u_{1_n}),\ ...\ ,u_n=(u_{n_1},...,u_{n_n})$ do we have a formula of the form :

$\det(u_1,...,u_n)=(u_1\times...\times u_{n-1}).u_n$ ?

It seems to only work for an orthonormal basis.

Thanks in advance !

  • $\begingroup$ Look up wedge product and volume element. $\endgroup$ Oct 26 '16 at 23:52
  • $\begingroup$ Related: What is the $3$-volume of the $3$-parallelepiped defined by $\{v_1, v_2, v_3\}$? (the answer contains a short account of the "cross product" in $\mathbf{R}^{n}$, which does not require an orthonormal basis). $\endgroup$ Oct 27 '16 at 0:01
  • $\begingroup$ @AndrewD.Hwang I think one of the problem is how we write the cross product. Because for instance $u\times v \times w$ is different from $(u\times v) \times w$ ... $\endgroup$
    – Maman
    Oct 27 '16 at 9:48
  • $\begingroup$ Assuming I understand your comment (and question): For vectors in $\mathbf{R}^{n}$ with $n > 3$, a "cross product" $u \times v$ isn't defined at all; what's defined is an ordered $(n - 1)$-fold product $u_1 \times \dots \times u_{n-1}$. This product is non-associative for a blunt reason: It makes no sense with a number of factors other than $n - 1$. If that doesn't address your point, it might help to add to your question how you define, e.g., $(1, 0, 0, 0) \times (0, 1, 0, 0)$ (a "cross product" of two standard basis vectors in $\mathbf{R}^{4}$). $\endgroup$ Oct 27 '16 at 10:21
  • $\begingroup$ @AndrewD.Hwang Ok I start to understand so I cannot write the formula with parenthesis in that case. $\endgroup$
    – Maman
    Oct 27 '16 at 10:32

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