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Let $V_1, V_2, ... V_{n-1} \in R^n$ be independent vectors. I want to prove that their cross product $V_1 \times V_2 \times ...\times V_{n-1} \ne 0$.

I know that the cross product is equal to the determinant of: $$ \begin{pmatrix} e_1 & e_2 & \cdots & e_n \\ V_{1,1} & V_{1,2} & \cdots & V_{1,n} \\ \vdots & \vdots & \ddots \\ V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n} \\ \end{pmatrix} $$ which is then equal to $$e_1\begin{vmatrix} V_{1,2} & V_{1,3} & \cdots & V_{1,n} \\ \vdots & \vdots & \ddots \\ V_{n-1,2} & V_{n-1,3} & \cdots & V_{n-1,n} \\ \end{vmatrix} + \space... \space + e_n\begin{vmatrix} V_{1,1} & V_{1,2} & \cdots & V_{1,n-1} \\ \vdots & \vdots & \ddots \\ V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n-1} \\ \end{vmatrix}$$

So it is enough to prove that one of those sub-determinants is not zero.

However when I to assume they all zero, so the rows are linearly dependent and to get to a contradiction I got stuck.

Any help here?

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  • $\begingroup$ I might try the contrapositive. $\endgroup$ – David Peterson Apr 7 at 23:03
  • $\begingroup$ what do you mean? $\endgroup$ – Gabi G Apr 7 at 23:05
  • $\begingroup$ Assume that cross product is zero, then show that as a result, the vectors cannot be linearly independent. $\endgroup$ – D.B. Apr 7 at 23:09
  • $\begingroup$ That is what I tried, but it got messed up with all the equations $\endgroup$ – Gabi G Apr 7 at 23:13
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If $V_1, V_2, \dots, V_{n-1}$ are independent in $\mathbb R^n$, there is an $n^{\text{th}}$ vector $V_n \in \mathbb R^n$ forming a basis with them.

Then we have $$ (V_1 \times V_2 \times \dots \times V_{n-1}) \mathbin{\boldsymbol{\cdot}} V_n = \det\begin{bmatrix}V_{n1} & V_{n2} & \cdots & V_{nn} \\ V_{11} & V_{12} & \cdots & V_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ V_{n-1,1} & V_{n-1,2} & \cdots & V_{n-1,n}\end{bmatrix} $$ and this determinant is nonzero because $V_1, V_2, \dots, V_n$ are linearly independent. So $V_1 \times V_2 \times \cdots \times V_{n-1}$ can't be the zero vector, otherwise it could not have a nonzero dot product with $V_n$.

If you're not convinced that the dot product above is equal to the determinant, expand the cross product as you have done, then take the dot product with $V_n$. Since $$(a_1 e_1 + \dots + a_n e_n) \mathbin{\boldsymbol{\cdot}} b = (a_1 b_1 + \dots + a_n b_n),$$ you get the same expansion, but for the determinant with $V_n$ in it.

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  • $\begingroup$ It's a pretty smart answer, thank you! $\endgroup$ – Gabi G Apr 7 at 23:27
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The other answer covers it, but here is an approach that uses the Riesz Representation theorem: define the linear functional

$$\phi(x)=\det(V_1,\cdots, V_{n-1},x)$$

We apply Riesz to get a nonzero $z\in \mathbb R^n$ such that

$$\phi(x)=\langle x,z\rangle$$

Then,

$$\det(V_1,\cdots, V_{n-1},x)=\langle x,z\rangle$$

Now,

$$\langle e_i,z\rangle=\det(V_1,\cdots, V_{n-1},e_i)$$

is the $i^{\text{th}}$ subdeterminant in your post. If each of these were zero then,

$$\langle e_i,z\rangle =0;\ 1\le i\le n$$

and so $z=0$, a contradiction.

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