I'm having trouble getting this proof right, its somthing like this
Let $ \{e_1, e_2, ... e_n \}$ be an orthonormal basis, of a vector space with internal product $\mathbb V$, and $\theta_i$, the angle between $e_i$ and $v\neq 0 \in \mathbb V$, then:
$\sum_{i=0}^{n}\cos^{2}{\theta_i}=1$
In general I'm trying to proof it, for a dot product, in a space $\mathbb R^n$... i have try using the expansion theorem, point product definition, but i can't find it right!
I have this
$\cos{\theta_i}= \frac{\langle e_i,v\rangle}{||v||} $
writing $v$ in terms of the basis we get (using the expansion theorem)
$v=\sum_{i=0}^{n}=\frac{\langle v,e_i \rangle}{||e_i||^2}e_i$, where $||e_i||=1$
(not sure if the theorem its call like that, but at least in Hoffman's Linear algebra book spanish version, and Claudio Pita's that is how the call the theorem) i have checked in a lot of books, and I'm not entirely sure that I've approached the problem the right way, any suggestions?
You are very close to showing what you want. I think what you are missing is that $\|\mathbf{v}\|^{2}= \langle \mathbf{v}, \mathbf{v} \rangle$.
Given any orthonormal basis $\lbrace \mathbf{e}_{1}, \dots, \mathbf{e}_{n}\rbrace$, you can decompose $\mathbf{v}$ as follows: $$ \mathbf{v} = \sum_{i=1}^{n} \frac{\langle \mathbf{v}, \mathbf{e}_{i}\rangle}{\|\mathbf{e}_{i}\|} \mathbf{e}_{i} = \sum_{i=1}^{n} \langle \mathbf{v}, \mathbf{e}_{i}\rangle \mathbf{e}_{i}. \quad (1) $$ Now, you have $$ \|\mathbf{v}\|^{2}= \langle \mathbf{v}, \mathbf{v} \rangle = \langle \mathbf{v}, \sum_{i=1}^{n} \langle \mathbf{v}, \mathbf{e}_{i}\rangle \mathbf{e}_{i} \rangle = \sum_{i=1}^{n} \langle \mathbf{v}, \mathbf{e}_{i}\rangle^{2}. \quad (2) $$ Dividing both sides by $\|\mathbf{v}\|^{2}$, you get your desired result.