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Vectors

 
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May6-11, 06:36 PM   #1
 

Vectors


1. The problem statement, all variables and given/known data
Show that is u ≠ 0 and au = bu, then a=b. Here u is a vector and a and b are real numbers


2. Relevant equations



3. The attempt at a solution
I don't see how this is possible. Maybe is it said If A=B and u ≠ 0 then au = bu. If a and b are different then they are just going to scale. The only way I can think of is just dividing by u on both sides but this seems to easy.
 
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May6-11, 07:54 PM   #2
 
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Quote by topgear View Post
1. The problem statement, all variables and given/known data
Show that is u ≠ 0 and au = bu, then a=b. Here u is a vector and a and b are real numbers


2. Relevant equations



3. The attempt at a solution
I don't see how this is possible. Maybe is it said If A=B and u ≠ 0 then au = bu. If a and b are different then they are just going to scale. The only way I can think of is just dividing by u on both sides but this seems to easy.
Division by vectors is not defined.

You're given that u0 and that au = bu.
That's equivalent to au - bu = 0. Can you think of something you can do to work with the left side of this equation?
 
May8-11, 08:56 PM   #3
 
take out u then divide by it?
 
May8-11, 09:40 PM   #4
 
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Vectors


Quote by topgear View Post
take out u then divide by it?
I suppose you mean "factor u out." That's reasonable to do. What isn't reasonable or even valid, as I said before, is dividing by a vector.
 
May8-11, 09:47 PM   #5
 
Then how should I make u go away
 
May8-11, 09:52 PM   #6
 
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You can't make u go away. What equation do you get when you do the factoring?
 
May8-11, 09:53 PM   #7
 
U(a-b)=0
 
May8-11, 09:57 PM   #8
 
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I would write it as (a - b)u = 0, since I usually write scalar multiples of vectors as kv rather than vk.

So (a - b)u = 0. That can happen if u = 0 (which can't happen in this problem). How else can this happen?
 
May8-11, 10:00 PM   #9
 
So that means a-b=0 therefore a=b
 
May9-11, 12:18 AM   #10
 
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Yes.
 
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linear algebra, proof, vectors
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