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Vectors |
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| May6-11, 06:36 PM | #1 |
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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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You're given that u ≠ 0 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 |
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take out u then divide by it?
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| May8-11, 09:40 PM | #4 |
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Vectors |
| May8-11, 09:47 PM | #5 |
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Then how should I make u go away
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| 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?
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| May8-11, 09:53 PM | #7 |
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U(a-b)=0
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| 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 |
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So that means a-b=0 therefore a=b
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| May9-11, 12:18 AM | #10 |
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Yes.
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| linear algebra, proof, vectors |
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