# Prove that if vectors are independent then their cross product is not $0$

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?

• I might try the contrapositive. – David Peterson Apr 7 '19 at 23:03
• what do you mean? – Gabi G Apr 7 '19 at 23:05
• Assume that cross product is zero, then show that as a result, the vectors cannot be linearly independent. – D.B. Apr 7 '19 at 23:09
• That is what I tried, but it got messed up with all the equations – Gabi G Apr 7 '19 at 23:13

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.

• It's a pretty smart answer, thank you! – Gabi G Apr 7 '19 at 23:27

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.