Hi there, those any of you have some ideas about this one?

O

Orizom

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I need to proof that given [imath]( u, v, w)[/imath] [imath]\in \mathbb{R}[/imath] basis of [imath]\mathbb{R}^3[/imath]
Then [imath](u \wedge v, u \wedge w, v \wedge w)[/imath] is basis in [imath]\mathbb{R}^3[/imath]
Thank you and sorry for the poor English
 
Yes, sorry! the [math]\wedge[/math] stands for vector product, so if I know that (u, v, w) is a basis of R^3 then that their vector product is a another basis of R^3. I think I have to work with the fact that the vector product is orthogonal to both the vectors, and so I'll have another triplet linearly indipendent vectors... I think I'm missing something
 
I think it can be proved by contradiction: assume that vector products don't form a basis and thus are linearly dependent.
Can you use the fact that vectors [imath]x,y,z[/imath] are linearly independent if and only if [imath]x\cdot y\wedge z \neq 0[/imath], where [imath]\cdot[/imath] is used for dot product ?
 
Yes I was thinking about the mixed product too, yet is still difficut to make a relationship between the two
 
Next hint: if the vector products do not form a basis then for some [imath]a,b,c[/imath] we have [imath]a u\wedge v + b u\wedge w + c v\wedge w =[/imath], where at least one member of [imath](a,b,c)[/imath] is non-zero.
Is this helpful?
 
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