5
$\begingroup$

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 $\sum_{i=1}^n \alpha_i \beta_j$?

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$

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.