# Proof: How to start it?

• B
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.

member 587159
What is your definition of the dot product? In what space are you working? I assume ##\mathbb{R}^n##

fresh_42
Mentor
Use the distributive law step by step: ##(u+v)\cdot (u-v)= u\cdot (u-v) + v\cdot (u-v)##.

Yes I am
What is your definition of the dot product? In what space are you working? I assume ##\mathbb{R}^n##

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.

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 ?

fresh_42
Mentor
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.

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..

fresh_42
Mentor
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##.