Linear Algebra: Vectors/Proofs

[trulyfalse]

Hello PF!

Homework Statement

Prove the following: if u and v are two vectors in Rⁿ such that u\cdotw = v\cdotw for all wεRⁿ , then we have u = v

Homework Equations

The Attempt at a Solution

u\cdotw - v\cdotw = 0
w\cdot(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?

 

[Dick]

Hello PF!

Homework Statement

Prove the following: if u and v are two vectors in Rⁿ such that u\cdotw = u\cdotv for all wεRⁿ , then we have u = v

Homework Equations

The Attempt at a Solution

u\cdotw - v\cdotw = 0
w(u\cdotv) = 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.

 

[brmath]

Another way to think of it is this: w \cdot (u-v) = 0 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 w \cdot t \ne 0 For example w = t.

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

That word "every" is very powerful.

 

[trulyfalse]

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.

 

[Dick]

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?

 

[brmath]

see answer below

 

[brmath]

it is perfectly possible for x and y to be non-zero vectors with x \cdot y = 0 Such vectors are called "orthogonal" . You can see this for yourself by looking at (1,0) \cdot (0,1) The important point here is not that there is some vector w such that w \cdot (u-v) = 0 but that you get 0 for every possible w.

There is in fact a property that tells you if v \cdot\ v = 0 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.