I've learned that the dot product is just one of many possible inner product spaces. Can someone explain this concept? When is it useful to define it as something other than the dot product?


3 Answers 3


As for the utility of inner product spaces: They're vector spaces where notions like the length of a vector and the angle between two vectors are available. In this way, they generalize $\mathbb R^n$ but preserve some of its additional structure that comes on top of it being a vector space. Familiar friends like Cauchy-Schwarz, the parallelogram rule, and orthogonality all work in inner product spaces.

(Note that there is a more general class of spaces, normed spaces, where notions of length make sense always, but an inner product cannot necessarily be defined.)

The dot product is the standard inner product on $\mathbb R^n$. In general, any symmetric, positive definite matrix will give you an inner product on $\mathbb C^n$. And you can have inner products on infinite dimensional vector spaces, like

$$ \langle \, f, \, g \, \rangle = \int_a^b \ f(x)\overline{g(x)} \, dx$$

for $f, g$ square-integrable functions on $[a,b]$.

This becomes useful, for example, in applications like Fourier series where you want a basis of orthonormal functions for some function space (it's not just the trigonometric functions that work).

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    $\begingroup$ The notion of the length of the vector is defined in any normed vector space, independent of whether it is an inner product vector space. This is worth clarifying. I believe vector spaces where the inner product of a vector with itself equals the norm squared have a special name (Euclidian and Hilbert spaces are good examples). $\endgroup$
    – Noldorin
    Jul 21, 2010 at 7:28
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    $\begingroup$ Fair enough. Inner product spaces have a norm, but normed spaces do not necessarily have an inner product. Edited to reflect this. $\endgroup$ Jul 21, 2010 at 7:44
  • $\begingroup$ How do you get an inner product from a symmetric, positive definite matrix? I think I've heard this before. $\endgroup$
    – Casebash
    Jul 21, 2010 at 8:01
  • $\begingroup$ Thanks :) I'm liking the answer more now, +1 $\endgroup$
    – Noldorin
    Jul 21, 2010 at 8:09

An inner product space is a vector space for which the inner product is defined. The inner product is also known as the 'dot product' for 2D or 3D Euclidean space. An arbitrary number of inner products can be defined according to three rules, though most are a lot less intuitive/practical than the Euclidean (dot) product.

Side note:

It may seem slightly esoteric, but as a physicist the obvious application of inner product spaces are Hilbert spaces used in quantum mechanics. The inner product of an eigenfunction with a wavefunction in Hilbert space gives the corresponding eigenvalue.

  • $\begingroup$ Can you give an example of a vector space where the inner product is not defined? $\endgroup$ May 5, 2014 at 13:48
  • $\begingroup$ @Imray, math.stackexchange.com/questions/247425/… $\endgroup$
    – Noldorin
    May 5, 2014 at 20:36
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    $\begingroup$ Probably the thing to think about is not a space where an inner product is not defined. Usually in a given vector space many different inner products exist. An inner product space is a vector space together with a particular inner product specified as the one to be used. $\endgroup$
    – GEdgar
    Nov 30, 2015 at 14:05

An inner produce space is a juncture of a vector space with an operation (and is thus an ordered pair consisting of the pair for the vector space $(\mathbb V, \mathbb F)$, and the operation (usually denoted$ \langle,\rangle $). So this would look like $((V,F),\langle,\rangle)$. Obviously there's structure in $\mathbb V$, and $\mathbb F$, and$ (\mathbb V, \mathbb F)$ consistent with a vector space (vector multiplication; scalar multiplication; vector-scalar multiplication; and vector, and scalar addition). The inner product is then a function (with $\mathbb V$ the set of vectors, and $\mathbb F$ the set of scalars): $$ \langle,\rangle:\mathbb V \to \mathbb F $$ where the notation $\langle\vec a , \vec b\rangle = c$ means the product of a and b(in $\mathbb V$) reutrns the value C in $\mathbb F$.

there are then a set of properties for the product which must adhere (in order for the product space to have results that are consistent and useful):

I'm going to drop the vector notation above the elements since it should be clear that anything in the product is from the vector space, and any result is a scalar from the field. $$ \langle a,b\rangle = \overline {\langle b,a\rangle} $$ which is to say that the complex conjugate of $\langle a,b\rangle$ is $\langle b, a\rangle$. in the real numbers this then becomes simply $\langle a,b\rangle = \langle b,a\rangle$.

The first argument must also be linear, that is (with $x$ a scalar): $$ \langle xa+y,b\rangle = \langle xa,b\rangle + \langle y,b\rangle $$

And lastly a relation to some norm on the inner product space: $$ \langle x,x\rangle \geq 0 $$ That is, a norm can always be defined in terms of the inner product: $ ||x||$ = $\sqrt{\langle x,x\rangle}$

This norm existing from the definition of scalar multiplication, and the existence of an inner product space proves that there is always a norm in the inner product space (the one defined by the square root of the inner product.

The simplest example other than a dot product is probably the inner product of two functions using integration which returns a value in the scalar filed in which they are defined to be vectors (which has been mentioned previously).

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    $\begingroup$ I changed various occurrences of things like $<a,b>$ to $\langle a,b\rangle$. $\endgroup$ Aug 23, 2012 at 3:18
  • $\begingroup$ Complex conjugation only makes sense on subfields of the complex numbers, and positivity only makes sense in ordered fields. So $\mathbb{F}$ can't be just any field here. $\endgroup$
    – Zhen Lin
    Aug 23, 2012 at 3:35
  • $\begingroup$ @Zhen: True; I didn't want to over-complicate the answer with jargon that, while necessary in more advanced practice, doesn't inherently help in the development of the intuitions of the construction. $\endgroup$
    – Vilid
    Aug 30, 2012 at 5:33

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