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What does inner product actually mean? So far most of the cases that I encounter seems to suggest that dot product is the only useful inner product. I mean most of the things that we discuss about defines inner product as dot product. For example, when we talk about norm or length of a vector. If that is the case, why do we need inner product? I don't really get the purpose of defining such thing.

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Inner product tells you how much of one vector is pointing in the direction of another one. If e is a unit vector then $<f, e>$ is the component of f in the direction of e and the vector component of f in the direction e is $<f, e>e$. The vectors f and e are orthogonal when $<f, e> = 0$, in which case f has zero component in the direction e.
In 2 dimensions and with dot product this is familiar. We can write any vector v as $<v, i>i + <v, j>j$ where i and j are unit vectors in the direction of the x and y axes.

More interestingly, consider the space of real square integrable continuous functions on the interval [0, 1] and inner product $<f, g>=\int_{0}^{1}f(x)g(x)dx$. With respect to this inner product the functions $u_n=sin(2n\pi x)$ and $v_n=cos(2n\pi x)$ are orthonormal for each n, just like the usual basis vectors in n dimensions. A function (vector) in our space can now be written as (modulo some technicalities) $$\sum_{0}^{\infty}(<f,u_n>)sin(2n\pi x)+<f,v_n>cos(2n\pi x))$$

the real fourier series for f. That is one use of the inner product - it lets you visualize function spaces geometrically.

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  • $\begingroup$ Strictly speaking, your example is not an inner product since it's not positive definite. $\endgroup$
    – John
    Oct 24, 2014 at 10:38
  • $\begingroup$ I'm only looking at real valued functions to illustrate the point. $\endgroup$
    – Paul
    Oct 24, 2014 at 10:41
  • $\begingroup$ Inner product requires $<f,g>=0 \implies f=g$. But $\int_0^1 f\,dx=0 \implies f=0$ almost everywhere. $\endgroup$
    – John
    Oct 24, 2014 at 10:43
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    $\begingroup$ Kind of. This is why $L^2$ is secretly a space of equivalence classes, rather than functions. Basically, given two functions $f$ and $g$, we call them equivalent if they are equal almost everywhere (look up). Then, we take all square integrable functions, and form the corresponding equivalence classes and define the vector space structure on the equivalence classes (the downside to this is you need to take a second to make sure these operations are well defined, but thats not too bad). Now, if you have $\int |f|^2 =0$ then $f=0$ a.e., so the equivalence class of $f$ is that of $0$, as needed $\endgroup$
    – msm
    Jan 14, 2021 at 17:56
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    $\begingroup$ There's a lot of content in my above comment, so if its confusing I'm sorry. The main point is that there are ways to get around needing $f=0$ at every point. I could get more in to the motivation for doing that, but I've already overcomplicated things. $\endgroup$
    – msm
    Jan 14, 2021 at 18:03
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The inner product is not necessarily the best way to think of it; I would argue that orthogonality is the more useful concept. It's nice to know when things are perpendicular, and the inner product provides us with a way of determining that in contexts more general than, say, $\mathbb{R}^N$.

Why is it useful? Well, it's nice to be able to decompose vector spaces (even infinite dimensional ones!) into collections of mutually orthogonal subspaces.

For example, given a subspace $V \subset W$ of some vector space, when do you know that you can write $W = V \oplus V'$ in a nice way? Well, if there is an inner product on $W$ then there is a natural way to do so; just define $V' = V^\perp$. Without that there is not necessarily a natural choice of complement.

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Inner product is a generalization for dot product on a space $X$. And with this generalization, we can define the angle between two elements in $X$ and we can talk about perpendicular, projection of one element on the other element's direction, minimum distance and so on.

For example, we can define an inner product on $C[0,1]$ by $<f,g>=\int_{0}^1 f(x)g(x)\, dx$.

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