0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Begin with: $v=\sum_i a_i v_i$ for some $a_i$ since $v_i$ form a basis. Now how would you compute $a_i$? $\endgroup$ – Michal Adamaszek Nov 25 '19 at 11:47
  • 2
    $\begingroup$ 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. $\endgroup$ – Ben Grossmann Nov 25 '19 at 11:49
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $w = \sum_{i=1}^n \langle v,v_i \rangle v_i$.

$\endgroup$
1
  • $\begingroup$ Okay my mistake, I've edited that out of the question now. Thank you for your help. $\endgroup$ – Nipster Nov 25 '19 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.