The question is:
Show that if $V$ is an inner product space and {$v_1, . . . , v_n$} is an orthonormal basis, then
$v =\sum\limits_{i=1}^{n} \langle v, v_i\rangle v_i$, for all $v ∈ V$ .
To be perfectly honest, I'm unsure of where to begin with this question. Any help would be appreciated, thank you.
Best Answer
Every vector $v \in V$ can be written uniquely as: $$v = \sum_{j=1}^{n}\alpha_{j}v_{j},$$ once $\{v_{1},...,v_{n}\}$ is an orthonormal basis. Thus, using the linearity of the inner product, we get, for each $k=1,...,n$: $$\langle v, v_{k}\rangle = \langle \sum_{j=1}^{n}\alpha_{j}v_{j}, v_{k}\rangle = \sum_{j=1}^{n}\alpha_{j}\langle v_{j},v_{k}\rangle = \alpha_{k},$$ because $\langle v_{j},v_{k}\rangle = \delta_{jk}$. Thus, for each $k=1,...,n$, $\alpha_{k}=\langle v,v_{k}\rangle$ and: $$v = \sum_{j=1}^{n}\alpha_{j}v_{j} = \sum_{j=1}^{n}\langle v, v_{j}\rangle v_{j}$$