# Show that questions involving orthonormal basis and inner product space

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.

• Begin with: $v=\sum_i a_i v_i$ for some $a_i$ since $v_i$ form a basis. Now how would you compute $a_i$? – Michal Adamaszek Nov 25 '19 at 11:47
• Regarding the original version of the question: the Gram-Schmidt process is for constructing an orthonormal basis. Since we already are given an orthonormal basis, I can't see why constructing one would be useful. – Ben Grossmann Nov 25 '19 at 11:49

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: $$v = \sum_{j=1}^{n}\alpha_{j}v_{j} = \sum_{j=1}^{n}\langle v, v_{j}\rangle v_{j}$$

One approach is as follows: it suffices to note that if $$\langle v,v_i\rangle = \langle w,v_i \rangle$$ for all $$i = 1,\dots,n$$, then $$v$$ and $$w$$ must be the same vector. Apply this fact to $$w = \sum_{i=1}^n \langle v,v_i \rangle v_i$$.

• Okay my mistake, I've edited that out of the question now. Thank you for your help. – Nipster Nov 25 '19 at 11:49