crossproducts of linearly independent vectors

hello, i am asked to proof that, for 3 linearly independent vectors, the crossproducts of each of the pairs of vectors are also linealry independent. i have tried to start this off looking at various properties of these vectors but i feel i am short of a step or two. i have that the crossproducts cant equal 0, and looking at deriving some kind of proof from this. but i am stuck. it seems a simple question but i got stuck.

Suppose for some real numbers , , and . What happens when you take the dot product with, say, ?

- Hollywood

sorry, you take the dot product of x1 and what else? i interpret the form of your reply as a proof by contradiction? but i dont know exactly the instruction. thanks for reply. can you clarify please?

To show the three vectors are independent, my plan is to suppose that a linear combination of them is zero:



and prove that , , and must all be zero. I don't know if that would be contradiction - I suppose it depends on how your book defined linearly independent - but it's a common way of proving that a group of things is linearly independent.

So if we suppose that , we can take the dot product of both sides with to get:

.

The right-hand side is obviously zero, and we can distribute the dot product on the left to get:

.

Now look at . The vector is perpendicular to both and , so since it is perpendicular to , its dot product with must be zero: so . Do you see how to finish the proof now?

- Hollywood

thank you,
first, what do you mean by how 'your book' defined linearly independent? what do u mine by book in this context?

second, thank you for clarification. what i meant by contradiction is what you are saying but just the other way round, i.e. show that they cant equal 0 unless the scalars are all zero. so its the same concept really.

so.. as
.

and also then too, but then the a's can be anyhing if the vector parts are 0?

que? amirite? ... the final part is "EXpress any vector in terms of these latter vectors". which ip ressume means, i have to say express a in terms of axb, bxc, cxa? how can i do this? i have a lot of identities for dot product, cross product, etc, but none that can express an original vector in terms of cross

Sorry - I got busy in my "real life".

is zero and is zero, but is definitely not zero, since we are assuming , , and are linearly independent. So you can conclude .

By taking the dot product with you get , and by taking the dot product with , you get .

- Hollywood

thanks hollywood thanks, ok so it is a simple "this isnt zero, but the product is zero so that must be zero" type thing, ok i am used to that with solving functions

"EXPRess any vector in terms of these latter vectors" is the final part.
i think this measn you have to take say a

and write a = something to do with these crossproduct vectors
but i cant manage to isolate in this way? do ou have an idea what to do with this? i think this is just a small task but i cant figure it. i understand the rest of the question now
how can i do this?

You know from the previous result that it is possible to write . Now you need to find , , and . Here's a hint: you've already done this for a=0.

- Hollywood

hey, thanks, i am a little confused. a,b,c are my three vectors. which you wrote as x1,x2,x3; i realise i ddint specify a name for them. but when i say "a" i am talking one of the original vectors. i think i am supposed to express "any" vector (what is it mean by any? i think it means i can pick one of the original vectors) and express them as something to do with the crossproduct vectros

I read it as any possible vector, but it's pretty much the same process if you interpret it as just a, b, and c.

in lieu of that then what do i do? i dont understnad yuor last euqation a = a1 ... (x1 etc..

how does it relate to the question?

what am i looking to show?

are we just looking to show that because they are linearly independent they form a basis? i.e. you are just showing that a can be formed as a linear combination? or is there more to it?do i need to actually solve anything i.e. find out what the componenets of this vector are? i do nt have any numerics this is all just raw. is what you wrote the actualy answer and thats just the end of it?

You showed that , , and are linearly independent, so they're a basis. This means that any vector x can be written as a linear combination of those three: . You use the same method as before - taking the dot product with a, then b, then c - to determine , , and in terms of x, a, b, and c.

-Hollywood

im sory, i get that if you dot with either component of the cros, it elimainates the term, but how do we then get these "r1, r2, r3" from that when we have no number values? i do what you say. then x.a = r3 a.(bxc), x.b = r2 b.(axc), x.c = r1 c.(axb)

can we then cancel the dots on each side. i.e "de-dotify" so x=r3(bxc), x= r1(axb). but then we just have that these are equal. i dont know whats the "final expression" for it.

Quote:

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Originally Posted by learning

hello, i am asked to proof that, for 3 linearly independent vectors, the crossproducts of each of the pairs of vectors are also linealry independent. i have tried to start this off looking at various properties of these vectors but i feel i am short of a step or two. i have that the crossproducts cant equal 0, and looking at deriving some kind of proof from this. but i am stuck. it seems a simple question but i got stuck.

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a.bxc is the volume of the parallelipiped determined by vedtors a,b, and c.

(AxB)xC = (A.C)B - (A.B)C, from which you get:
(AxB)x(CxD) = (A.BxD)B – (A.BxC)D
Then
(axb)x(bxc) = (a.bxc)b – (a.bxb)c = (a.bxc)b
(axb)x(bxc).(cxa) = (a.bxc)(b.cxa) = |a.bxc|^2
|a.bxc|^2 is unequal to zero so the volume determined by the three cross products is unequal to zero and they are non-coplanar, ie, linear independent.