## 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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Mentor
 Quote by topgear 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?

 take out u then divide by it?

Mentor

## Vectors

 Quote by topgear 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.

 Then how should I make u go away
 Mentor You can't make u go away. What equation do you get when you do the factoring?
 U(a-b)=0
 Mentor 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?
 So that means a-b=0 therefore a=b
 Mentor Yes.

 Tags linear algebra, proof, vectors