Let V be an inner product space
Let u,v belong to V
Is it true that if (u,v).(v,u) = (u,u).(v,v), then u, v are linearly dependent? If yes then why?
Thanks,
Aman
I'm confused now - by the brackets are we meaning the vector spaces generated by u and v or the inner product operation thingy? Or neither of these things? My first thought was the latter, but your question, and the reply, makes me wonder if the latter is what was meant...
Inner product.
@aman: I tend to think the statement is false.
For example, take $\displaystyle V = R^2, x,y \in V, <x, y> \ = \ <(x_1, x_2),(y_1, y_2)> \ = \ x_1y_1 + x_2y_2$ (This is indeed an inner product).
Then, you can take $\displaystyle x = (0,1)$ and $\displaystyle y = (1, 0)$ which are clearly not linearly dependent, yet you still get:
$\displaystyle <(0,1) \ , \ (1,0)> \cdot <(1,0) \ , \ (0,1)> = (0,0) = <(0,1) \ , \ (0,1)> \cdot <(1,0) \ , \ (1,0)>$
Which contradicts the statement.
This doesn't really make sense. It should be: $\displaystyle <(1,0),(0,1)><(0,1),(1,0)>=0$ and $\displaystyle <(0,1),(0,1)>=<(1,0),(1,0)>=1$
I think it's true: If the underlying field of $\displaystyle V$ is $\displaystyle \mathbb{C}$ then $\displaystyle <u,v><v,u>= \vert <u,v> \vert ^2 $(because $\displaystyle <u,v>= \overline{<v,u>}$ ) and $\displaystyle <u,u>=\Vert u \Vert ^2$ (where this last means the norm induced by the scalar product) doing the same for $\displaystyle v$, by the Cauchy-Schwarz inequality we have $\displaystyle \vert <u,v> \vert ^2 \leq \Vert u \Vert ^2 \Vert v \Vert ^2$ with equality iff $\displaystyle u,v$ are l.d. which proves the statement. The real case is easier
Thanks Jose27. Yes - I get your point when you say it is a form of Cauchy-Schwarz inequality. But I would appreciate if someone can help me prove that
"the equality holds IFF u,v are linearly dependent"
So basically I want to start from <u, v> <v,u> = <u,u> <v,v> and prove u,v and lineraly dependent
Thanks