2
$\begingroup$

I know this question has been asked many times before, but I want to narrow it down into some more specific questions I have regarding this topic.

  1. Many descriptions online say that an inner product is the dot product for a real vector space, but then why is something like $\langle u, v \rangle = 3u_1v_1 + 4u_2v_2$ considered an inner product as well? What would be the purpose of such an inner product?

  2. I think geometrically it’s a bit difficult to understand why the kind of inner product I mentioned earlier is using the same concept of vector projections as a standard dot product.

When I learned about dot products, I always thought of $u \cdot v = |u||v|\cos(\theta)$ as the main definition and the fact that this is also equal to $u_1v_1 + u_2v_2$ was just a nice bonus to make calculations easier. With inner products, though, it seems we only learn about them in terms of the second definition but not the first, and so I’m struggling to understand them geometrically.

$\endgroup$
2
  • $\begingroup$ Same reason as: "Why do we use vector spaces other than $\mathbb R^n$?" Perhaps in the future you will study a course called "Linear Algebra" to understand this. $\endgroup$
    – GEdgar
    Nov 13, 2023 at 21:43
  • $\begingroup$ Are you familiar with polar coordinates? If I gave you $r$ and $\theta$ for two vectors what do you think the inner product would be? $\endgroup$
    – Jeff
    Nov 14, 2023 at 2:17

1 Answer 1

0
$\begingroup$

It is standard to establish the connection between the dot product and the cosine in Euclidean space.

I will suggest that the dot product is also connected to the covariance in a statistical setting.

That is $Cov(X,Y) = \frac {1}{N} \sum_\limits{i = 1}^N (x_i - \bar x)(y_i - \bar y) = \frac {1}{N}\left(\sum_\limits{i = 1}^N x_iy_i\right) - \bar x\bar y$

If we re-center our data sets, we exactly have $\frac 1N X\cdot Y$ which is one of our "non-Euclidean" inner products. All this inner product does is add in a scaling factor vs. the Euclidean inner product. This scaling factor makes data sets of different sizes more comparable.

But suppose, for some reason, we think that some measurements are more important than others. We can create a set of weighting factors $(\omega_1, \cdots, \omega_N)$ such that $\sum_\limits{i = 1}^N \omega_i = 1$ we then define a weighted average instead of the standard mean calculation, and a weighted covariance $\sum_\limits{i = 1}^N \omega_ix_iy_i$, instead of the Euclidean inner product.

This is one example, I hope it helps.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .