If $a$, $b$, $c$ are three linearly independent vectors show that the vectors $a \times b$, $b \times c$, $c \times a$ are also linearly independent.
I have tried proving with three linearly independent vectors and working out the determinants of the cross products, but I cannot seem to get them to equal 0.
Any help would be appreciated, thank you.
Assume that $$\lambda(a\times b)+\mu(b\times c)+\nu(c\times a)=\vec 0\ .$$ Taking the scalar product with $c$ gives $$\lambda\>(a\times b)\cdot c=0\ .$$ Since the triple product is assumed $\ne0$ it follows that $\lambda=0$, and similarly for $\mu$ and $\nu$.
You are given that $$a\cdot b\times c=\Delta\neq0$$ and you need to show that $$(a\times b)\cdot((b\times c)\times(c\times a))\neq0$$
Writing, for convenience, $n=c\times a$, we can evaluate the inner bracket $$(b\times c)\times n=c(n\cdot b)-b(n\cdot c)$$ using the BACCAB rule. Then simplifying further, we get $$(b\times c)\times n=c[(c\times a\cdot b)-b(0)]$$
Therefore, $$(a\times b)\cdot((b\times c)\times(c\times a))=(a\times b)\cdot c(c\times a\cdot b)=\Delta^2\neq 0$$
Hint It's enough (in fact, equivalent) to show that the triple product $$\newcommand{\bfa}{{\bf a}}\newcommand{\bfb}{{\bf b}}\newcommand{\bfc}{{\bf c}} \newcommand{\bfx}{{\bf x}}\newcommand{\bfy}{{\bf y}}\newcommand{\bfz}{{\bf z}}[(\bfa \times \bfb) \times (\bfb \times \bfc)] \cdot (\bfc \times \bfa)$$ of $\bfa \times \bfb, \bfb \times \bfc, \bfc \times \bfa$ is nonzero.
We can rewrite this quantity using the occasionally useful cross product identity $$ (\bfx \times \bfy) \times (\bfx \times \bfz) = (\bfx \cdot(\bfy \times \bfz)) \bfx . $$
Substituting gives that the triple product is $$(\bfa \cdot (\bfb \times \bfc))^2,$$ but this is just the square of the triple product of the original vectors $\bfa, \bfb, \bfc$; since the set of these vectors is linearly independent, this latter triple product is nonzero.