If $v = \sum_{i} \xi_{i}v_{i}$ then $\sum \xi_{i}^2 = 1$?

Question

If $v$ is a unit vector and $v = \sum_{i} \xi_{i}v_{i}$ where $v_{i}$ are orthonormal vectors. Then how do we prove $\sum \xi_{i}^2 = 1$ ?

I thought that we have $<v,v_{i}> = \xi_{i}$ by taking inner products on both wrt to $v_{i}$, but how do we proceed next. Maybe this is an elementary result but I am stuck somewhere.

Answer 1

Since the vectors $v_1,\dots,v_n$ are orthonormal, it holds that $\langle v_i,v_j\rangle =0$ whenever $i\neq j$ and $\langle v_i,v_i\rangle =1$ for all $i$. Thus, $$ 1 = \lVert v \rVert^2 =\langle v,v\rangle = \sum_{i,j=1}^n \xi_i\xi_j\langle v_i,v_j\rangle = \sum_{i=1}^n \xi_i^2. $$

Answer 2

$\langle v,v \rangle = \sum_i \sum_j \xi_i \xi_j \langle v_i, v_j \rangle = ...$

Answer 3

$$1=\langle v,v \rangle=\langle \sum_i \xi_i v_i,\sum_j \xi_j v_j \rangle=\sum_{i,j}\xi_i \xi_j \langle v_i,v_j \rangle=\sum_{i=j} \xi_i \xi_j \langle v_i,v_j \rangle =\sum_k \xi_k^2 \|v_k \|^2 =\sum_k \xi_k^2.$$

Answer 4

Hint: $$1=\langle v, v\rangle=\langle v,\sum\eta_jv_j\rangle=\sum\eta_i\langle v,v_i\rangle=\dots$$