Can one compute the dot product knowing only the basis expansions of vectors v and w?

Question

Scenario: Suppose that $V$ is a finite dimensional vector space and that $\{e_1,...,e_n\}$ is a basis for $V$. Let $v,w\in V$ and suppose that $v=\sum_{i=1}^n \alpha _i e_i$ and $w=\sum_{i=1}^n \beta _i e_j$ where $\alpha_i$, $i=1,..,n$ and $\beta_j$, $j=1,...,n$. Can we compute $\langle v,w\rangle $ knowing only the basis expansions of $v$ and $w$ and the values of $\{\langle e_i,e_j\rangle :i,j=1,...,n\}$?

What I understand:

I know that $\langle v,w\rangle =\langle \sum_{i=1}^n \alpha _i e_i, \sum_{i=1}^n \beta _i e_j\rangle$ and that $\langle e_i,e_j\rangle=0$. The basis expansion would be any $\alpha_1 e_1 +\alpha_2 e_2+...+ \alpha_n e_n$ and similarly for $\beta_j e_j$. Is the problem asking whether the dot product is simply $\sum_{i=1}^n \alpha_i \beta_j$?

Answer 1

No, you have not been told that that the basis is orthonormal (not even orthogonal).

What you are being asked is to show that, because the inner product is bilinear, $$ \langle v,w\rangle=\sum_{k,j}\alpha_k\beta_j\langle e_k,e_j\rangle. $$ And the last expression depends precisely on the numbers you were given.

Answer 2

You can compute it given $e_i\cdot e_j$; just expand the product:

$$v\cdot w=(\sum_i a_ie_i)\cdot (\sum_j b_je_j)=\sum_{ij} a_ib_j(e_i\cdot e_j)$$