# crossproducts of linearly independent vectors

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• January 30th 2013, 08:32 PM
learning
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.
• January 30th 2013, 11:15 PM
hollywood
Re: crossproducts of linearly independent vectors
Suppose $a_1(x_1 \times x_2)+a_2(x_1 \times x_3)+a_3(x_2 \times x_3)=0$ for some real numbers $a_1$, $a_2$, and $a_3$. What happens when you take the dot product with, say, $x_1$?

- Hollywood
• January 31st 2013, 01:52 AM
learning
Re: crossproducts of linearly independent vectors
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?
• January 31st 2013, 07:20 AM
hollywood
Re: crossproducts of linearly independent vectors
To show the three vectors are independent, my plan is to suppose that a linear combination of them is zero:

$a_1(x_1 \times x_2)+a_2(x_1 \times x_3)+a_3(x_2 \times x_3)=0$

and prove that $a_1$, $a_2$, and $a_3$ 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 $a_1(x_1 \times x_2)+a_2(x_1 \times x_3)+a_3(x_2 \times x_3)=0$, we can take the dot product of both sides with $x_1$ to get:

$x_1 \cdot (a_1(x_1 \times x_2)+a_2(x_1 \times x_3)+a_3(x_2 \times x_3)) = x_1 \cdot 0$.

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

$a_1(x_1 \cdot (x_1 \times x_2))+a_2(x_1 \cdot (x_1 \times x_3))+a_3(x_1 \cdot (x_2 \times x_3))=0$.

Now look at $x_1 \cdot (x_1 \times x_2)$. The vector $x_1 \times x_2$ is perpendicular to both $x_1$ and $x_2$, so since it is perpendicular to $x_1$, its dot product with $x_1$ must be zero: so $x_1 \cdot (x_1 \times x_2)=0$. Do you see how to finish the proof now?

- Hollywood
• January 31st 2013, 10:06 AM
learning
Re: crossproducts of linearly independent vectors
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
$a_1(x_1 \cdot (x_1 \times x_2))+a_2(x_1 \cdot (x_1 \times x_3))+a_3(x_1 \cdot (x_2 \times x_3))=0$.

$(x_1 \cdot (x_1 \times x_2) = 0$ and also $(x_1 \cdot (x_1 \times x_3 = 0$ then $a_3(x_1 \cdot (x_1 \times x_2) = 0$too, but then the a's can be anyhing if the vector parts are 0?
• January 31st 2013, 11:16 PM
learning
Re: crossproducts of linearly independent vectors
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
• February 1st 2013, 11:35 PM
hollywood
Re: crossproducts of linearly independent vectors
Sorry - I got busy in my "real life".

$(x_1 \cdot (x_1 \times x_2))$ is zero and $(x_1 \cdot (x_1 \times x_3))$ is zero, but $(x_1 \cdot (x_2 \times x_3))$ is definitely not zero, since we are assuming $x_1$, $x_2$, and $x_3$ are linearly independent. So you can conclude $a_3 = 0$.

By taking the dot product with $x_2$ you get $a_2=0$, and by taking the dot product with $x_3$, you get $a_1=0$.

- Hollywood
• February 3rd 2013, 02:34 AM
learning
Re: crossproducts of linearly independent vectors
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?
• February 3rd 2013, 06:00 AM
hollywood
Re: crossproducts of linearly independent vectors
You know from the previous result that it is possible to write $a=a_1(x_1 \times x_2)+a_2(x_1 \times x_3)+a_3(x_2 \times x_3)$. Now you need to find $a_1$, $a_2$, and $a_3$. Here's a hint: you've already done this for a=0.

- Hollywood
• February 3rd 2013, 09:56 AM
learning
Re: crossproducts of linearly independent vectors
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
• February 3rd 2013, 10:18 PM
hollywood
Re: crossproducts of linearly independent vectors
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.
• February 4th 2013, 12:09 AM
learning
Re: crossproducts of linearly independent vectors
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?
• February 4th 2013, 06:44 AM
hollywood
Re: crossproducts of linearly independent vectors
You showed that $a \times b$, $a \times c$, and $b \times c$ are linearly independent, so they're a basis. This means that any vector x can be written as a linear combination of those three: $x=r_1(a \times b)+r_2(a \times c)+r_3(b \times c)$. You use the same method as before - taking the dot product with a, then b, then c - to determine $r_1$, $r_2$, and $r_3$ in terms of x, a, b, and c.

-Hollywood
• February 4th 2013, 10:42 AM
learning
Re: crossproducts of linearly independent vectors
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.
• February 4th 2013, 01:17 PM
Hartlw
Re: crossproducts of linearly independent vectors
Quote:

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.

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.
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