Proof: How to start it?

  • #1
stolencookie
Member advised that homework must be posted in one of the homework sections
I am having trouble with this proof. I just need a step in the right direction. Let u and v be vectors.
(u+v)*(u-v)=0, then IuI=IvI I have to use properties of the dot product.
I started off by combining both using this property u*(v+w)=u*v+u*w (u,v,w are vectors)
I got lost in all of my mess, after I combined them. Was this a good place to start? I am just so lost right now.
 

Answers and Replies

  • #2
member 587159
What is your definition of the dot product? In what space are you working? I assume ##\mathbb{R}^n##
 
  • #3
14,846
12,337
Use the distributive law step by step: ##(u+v)\cdot (u-v)= u\cdot (u-v) + v\cdot (u-v)##.
 
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  • #4
stolencookie
Yes I am
What is your definition of the dot product? In what space are you working? I assume ##\mathbb{R}^n##
 
  • #5
stolencookie
Use the distributive law step by step: ##(u+v)\cdot (u-v)= u\cdot (u-v) + v\cdot (u-v)##.

That helps a lot actually I think I know how to do it now.
 
  • #6
stolencookie
What is your definition of the dot product? In what space are you working? I assume ##\mathbb{R}^n##
Does it make a difference if i am working in Rn ?
 
  • #7
14,846
12,337
Yes, it does for the definition of ##|u|## as ##\sqrt{u\cdot u}##. In ##\mathbb{C}^n## it would be ##\sqrt{u \cdot \overline{u}}## with complex conjugation in one factor.
 
  • #8
stolencookie
Yes, it does for the definition of ##|u|## as ##\sqrt{u\cdot u}##. In ##\mathbb{C}^n## it would be ##\sqrt{u \cdot \overline{u}}## with complex conjugation in one factor.
I am confused now..
 
  • #9
14,846
12,337
If we have a vector ##u=\begin{bmatrix}1\\2\end{bmatrix}## then ##|u|^2=u\cdot u = 1^2 +2^2 = 5## with the length ##\sqrt{5}##.
If we consider ##u=\begin{bmatrix}i\\2\end{bmatrix}## then the same formula would result in ##|u|^2=u\cdot u = i^2 +2^2 = -1 + 4 = 3## which is wrong, as I didn't actually change the length at all. So in this case the calculation goes ##|u|^2=u \cdot \overline{u}= i\cdot (-i) +2^2 = 5##.
 
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  • #10
member 587159
Do you still need help?

##(u+v).(u-v) = 0 \iff (u+v).u - (u+v).v = 0 \iff u.u + v.u - u.v - v.v = 0 \iff \dots##
 
  • #11
stolencookie
I got it just got distracted ....
 

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