Difference between revisions of "Inner product"

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* [[Inner product examples]]
 
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* [[Inequalities for inner products]]
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==References==
 
==References==

Revision as of 14:56, 12 July 2015

Definition

Given a vector space, [ilmath](V,F)[/ilmath] (where [ilmath]F[/ilmath] is either [ilmath]\mathbb{R} [/ilmath] or [ilmath]\mathbb{C} [/ilmath]), an inner product[1][2][3] is a map:

  • [math]\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}[/math] (or sometimes [math]\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{C}[/math])

Such that:

  • [math]\langle x,y\rangle = \overline{\langle y, x\rangle}[/math] (where the bar denotes Complex conjugate)
    • Or just [math]\langle x,y\rangle = \langle y,x\rangle[/math] if the inner product is into [ilmath]\mathbb{R} [/ilmath]
  • [math]\langle\lambda x+\mu y,z\rangle = \lambda\langle y,z\rangle + \mu\langle x,z\rangle[/math] ( linearity in first argument )
    This may be alternatively stated as:
    • [math]\langle\lambda x,y\rangle=\lambda\langle x,y\rangle[/math] and [math]\langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle[/math]
  • [math]\langle x,x\rangle \ge 0[/math] but specifically:
    • [math]\langle x,x\rangle=0\iff x=0[/math]

Terminology

Given a vector space [ilmath]X[/ilmath] over either [ilmath]\mathbb{R} [/ilmath] or [ilmath]\mathbb{C} [/ilmath], and an inner product [ilmath]\langle\cdot,\cdot\rangle:X\times X\rightarrow F[/ilmath] we call the space [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] an:

Properties

  • The most important property by far is that: [ilmath]\forall x\in X[\langle x,x\rangle\in\mathbb{R}_{\ge 0}][/ilmath] - that is [ilmath]\langle x,x\rangle[/ilmath] is real

Proof:

Notice that we (by definition) have [ilmath]\langle x,x\rangle=\overline{\langle x,x\rangle}[/ilmath], so we must have:
  • [ilmath]a+bj=a-bj[/ilmath] where [ilmath]a+bj:=\langle x,x\rangle[/ilmath], and by equating the real and imaginary parts we see immediately that we have:
    • [ilmath]b=-b[/ilmath] and conclude [ilmath]b=0[/ilmath], that is there is no imaginary component.

To complete the proof note that by definition [ilmath]\langle x,x\rangle\ge 0[/ilmath].

Thus [ilmath]\langle x,x\rangle\in\mathbb{R}_{\ge 0}[/ilmath] - as I claimed.

Notice that [math]\langle\cdot,\cdot\rangle[/math] is also linear (ish) in its second argument as:

  • [math]\langle x,\lambda y+\mu z\rangle =\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle[/math]

[math]\langle x,\lambda y+\mu z\rangle[/math]
[math]=\overline{\langle \lambda y+\mu z, x\rangle}[/math]
[math]=\overline{\lambda\langle y,x\rangle + \mu\langle z,x\rangle}[/math]
[math]=\bar{\lambda}\overline{\langle y,x\rangle}+\bar{\mu}\overline{\langle z,x\rangle}[/math]
[math]=\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle[/math]
As required.

From this we may conclude the following:

  • [math]\langle x,\lambda y\rangle = \bar{\lambda}\langle x,y\rangle[/math] and
  • [math]\langle x,y+z\rangle = \langle x,y\rangle + \langle x,z\rangle[/math]

This leads to the most general form:

  • [ilmath]\langle au+bv,cx+dy\rangle=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle[/ilmath] - which isn't worth remembering!

Proof:
[ilmath]\langle au+bv,cx+dy\rangle[/ilmath]
[ilmath]=a\langle u,cx+dy\rangle+b\langle v,cx+dy\rangle[/ilmath]
[ilmath]=a\overline{\langle cx+dy,u\rangle}+b\overline{\langle cx+dy,v\rangle}[/ilmath]
[ilmath]=a(\overline{c\langle x,u\rangle} + \overline{d\langle y,u\rangle})+b(\overline{c\langle x,v\rangle}+\overline{d\langle y,v\rangle})[/ilmath]
[ilmath]=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle[/ilmath]
As required


Notation

Typically, [ilmath]\langle\cdot,\cdot\rangle[/ilmath] is the notation for inner products, however I have seen some authors use [ilmath]\langle a,b\rangle[/ilmath] to denote the ordered pair containing [ilmath]a[/ilmath] and [ilmath]b[/ilmath]. Also, notably[3] use [ilmath](\cdot,\cdot)[/ilmath] for an inner product (and [ilmath]\langle\cdot,\cdot\rangle[/ilmath] for an ordered pair!)

Immediate theorems

Here [ilmath]\langle\cdot,\cdot\rangle:X\times X\rightarrow \mathbb{C} [/ilmath] is an inner product

Theorem: if [ilmath]\forall x\in X[\langle x,y\rangle=0][/ilmath] then [ilmath]y=0[/ilmath]

Suppose that [ilmath]y\ne 0[/ilmath], then by hypothesis:
  • [ilmath]\forall x\in X[\langle x,y\rangle =0][/ilmath]
Specifically that means for [ilmath]y\in X[/ilmath] we have [ilmath]\langle y,y\rangle=0[/ilmath]
  • Of course by definition, [ilmath]\langle y,y\rangle\ge 0[/ilmath] for [ilmath]\forall y\in X[/ilmath], and specifically
    • [ilmath]\langle x,x\rangle = 0\iff x=0[/ilmath]
So we have [ilmath]\langle y,y\rangle =0[/ilmath] contradicting that [ilmath]y\ne 0[/ilmath]
  • We conclude that if [ilmath]\forall x\in X[\langle x,y\rangle=0][/ilmath] then we must have [ilmath]y=0[/ilmath]
    (As required)

Norm induced by

  • Given an inner product space [ilmath](X,\langle\cdot,\cdot\rangle)[/ilmath] we can define a norm as follows[3]:
    • [ilmath]\forall x\in X[/ilmath] the inner product induces the norm [ilmath]\Vert x\Vert:=\sqrt{\langle x,x\rangle}[/ilmath]

TODO: Find out what this is called, eg compared to the metric induced by a norm

Prominent examples

See also

References

  1. ↑ http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885
  2. ↑ Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014
  3. ↑ 3.0 3.1 3.2 3.3 3.4 Functional Analysis - George Bachman and Lawrence Narici
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