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Linear Algebra: Vectors/Proofs

  • Thread starter trulyfalse
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  • #1
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Hello PF!

Homework Statement


Prove the following: if u and v are two vectors in Rn such that u[itex]\cdot[/itex]w = v[itex]\cdot[/itex]w for all wεRn , then we have u = v

Homework Equations





The Attempt at a Solution


u[itex]\cdot[/itex]w - v[itex]\cdot[/itex]w = 0
w[itex]\cdot[/itex](u - v) = 0

I'm not sure what to do after applying the distributive property (in reverse). How do I go about proving that the vectors u and v are equal? I considered establishing two cases in which w = 0 and u-v = 0 but that doesn't help me out. Are there any properties that I can use to construct this proof?
 
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Answers and Replies

  • #2
Dick
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Hello PF!

Homework Statement


Prove the following: if u and v are two vectors in Rn such that u[itex]\cdot[/itex]w = u[itex]\cdot[/itex]v for all wεRn , then we have u = v

Homework Equations





The Attempt at a Solution


u[itex]\cdot[/itex]w - v[itex]\cdot[/itex]w = 0
w(u[itex]\cdot[/itex]v) = 0

I'm not sure what to do after applying the distributive property (in reverse). How do I go about proving that the vectors u and v are equal? I considered establishing two cases in which w = 0 and u-v = 0 but that doesn't help me out. Are there any properties that I can use to construct this proof?
You've got some typos in there. I'd fix them. But pick w=(u-v). If (u-v).(u-v)=0 then what can you say about (u-v). Look at the properties of the dot product.
 
  • #3
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Another way to think of it is this: [tex] w \cdot (u-v) = 0 [/tex] for every choice of w. What can you say about a vector t whose inner product with any vector whatever is 0? Given a particular vector t with non-zero components, you can always find a vector w such that [tex] w \cdot t \ne 0 [/tex] For example w = t.

So if u-v has any non-zero components, what can you conclude?

That word "every" is very powerful.
 
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  • #4
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If the vectors are non-zero, then we can conclude that u-v must be equal to zero, and therefore u=v. We can also conclude that vectors w and (u-v) must be orthogonal because their dot product is equal to zero. However, I am not sure how to prove that the vectors are nonzero. Am I supposed to assume that they are and work from there?

EDIT: Never mind, I've just realized (after reading bmath's post more carefully) that this represents EVERY choice of w. So if w does not equal zero, then u-v must be equal to zero to produce a dot product of zero. Therefore, u = v. If there are any gaps in my understanding please let me know. Thanks for your help guys.
 
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  • #5
Dick
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If the vectors are non-zero, then we can conclude that u-v must be equal to zero, and therefore u=v. We can also conclude that vectors w and (u-v) must be orthogonal because their dot product is equal to zero. However, I am not sure how to prove that the vectors are nonzero. Am I supposed to assume that they are and work from there?

EDIT: Never mind, I've just realized (after reading bmath's post more carefully) that this represents EVERY choice of w. So if w does not equal zero, then u-v must be equal to zero to produce a dot product of zero. Therefore, u = v. If there are any gaps in my understanding please let me know. Thanks for your help guys.
That's a little confused. You've got w.(u-v)=0 for EVERY w. So you can pick w=(u-v). So now you've got (u-v).(u-v)=0. You should have a property that tells you that for a vector V, V.V=0 if and only if V=0. So?
 
  • #6
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see answer below
 
  • #7
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it is perfectly possible for x and y to be non-zero vectors with [tex]x \cdot y = 0[/tex] Such vectors are called "orthogonal" . You can see this for yourself by looking at [tex] (1,0) \cdot (0,1) [/tex] The important point here is not that there is some vector w such that [tex] w \cdot (u-v) = 0 [/tex] but that you get 0 for every possible w.

There is in fact a property that tells you if [tex] v \cdot\ v = 0 [/tex] then v = 0. The dot product of v with itself is the square of the length of the vector. If it is 0 you have a vector of length 0 -- i.e. the zero vector.
 

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