High School Vector Proof: Solving for IuI and IvI using Dot Product Properties

[stolencookie]

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]

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

 

[stolencookie]

Yes I am

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

 

[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.

 

[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 ?

 

[fresh_42]

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.

 

[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..

 

[fresh_42]

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 $|u|^2=u \cdot \overline{u}= i\cdot (-i) +2^2 = 5$.

 

[member 587159]

Do you still need help?

$(u+v).(u-v) = 0 \iff (u+v).u - (u+v).v = 0 \iff u.u + v.u - u.v - v.v = 0 \iff \dots$

 

[stolencookie]

I got it just got distracted ...