0
$\begingroup$

I am studying Linear Algebra and my professor proposed an exercise:

Let $\mathbb{R^n}$ a vector space, $v=(v_1,...,v_n)$ and $u=(u_1,...,u_n)$, then there is a general form for Inner Product between $u$ and $v$?

My attempt:

I know that the inner product is bilinear, comutative, $\langle u , u \rangle \geq 0 $ and $\langle u,u \rangle = 0 \iff u=0.$ So I think that the general form of Inner Product is:

$ \langle u,v \rangle = \displaystyle \sum_{j=1}^n \alpha_jv_ju_j,\hspace{0.5cm}$ where $\alpha_j \geq0, \forall j=1,...,n.$

But, how I arrive in this expression? and every Inner Product has this form?

$\endgroup$
  • 1
    $\begingroup$ So you know that $u = \sum u_i e_i$ and $v = \sum v_j e_j$, where $\{ e_i \}$ is the standard basis. Now what's $\langle u, v \rangle$? $\endgroup$ – M. Vinay Mar 17 at 4:07
  • $\begingroup$ this is a notation for Inner product, like $v^t \cdot u$ $\endgroup$ – user638057 Mar 17 at 4:11
  • $\begingroup$ Oh, I know. I'm asking you, now that you've written $u$ and $v$ in terms of the standard basis, what will be their inner product $\langle u, v \rangle$, for any arbitrary inner product $\langle \cdot, \cdot \rangle$? Try writing it, and conclude something from it. $\endgroup$ – M. Vinay Mar 17 at 4:14
  • $\begingroup$ ok, $\langle u,v \rangle = \displaystyle \sum_{j=1}^n v_j u_j$, because $\langle e_i,e_j \rangle = 0$ for $i \neq j, $ or $1$ if $i=j.$ , it's over? $\endgroup$ – user638057 Mar 17 at 4:20
  • 1
    $\begingroup$ No, $\langle e_i, e_j \rangle = \delta_{ij}$ for a particular inner product. In general it could be something else, since we're considering an arbitrary inner product. Nevertheless it cannot be anything at all, since it's an inner product, which has to satisfy the properties you mentioned. So use that to draw your conclusions $\endgroup$ – M. Vinay Mar 17 at 4:23

Your Answer

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

Browse other questions tagged or ask your own question.