Difference between revisions of "Inner product"
From Maths
m |
m (Todo: add trivial properties, a bit more rejigging) |
||
| Line 1: | Line 1: | ||
==Definition== | ==Definition== | ||
| − | Given a {{Vector space}} (where {{M|F}} is either {{M|\mathbb{R} }} or {{M|\mathbb{C} }}), an ''inner product''<ref>http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885</ref><ref>Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014</ref> is a map: | + | Given a {{Vector space}} (where {{M|F}} is either {{M|\mathbb{R} }} or {{M|\mathbb{C} }}), an ''inner product''<ref>http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885</ref><ref>Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014</ref><ref name="FA">Functional Analysis - George Bachman and Lawrence Narici</ref> 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>) | * <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: | Such that: | ||
| Line 6: | Line 6: | ||
** Or just <math>\langle x,y\rangle = \langle y,x\rangle</math> if the inner product is into {{M|\mathbb{R} }} | ** Or just <math>\langle x,y\rangle = \langle y,x\rangle</math> if the inner product is into {{M|\mathbb{R} }} | ||
* <math>\langle\lambda x+\mu y,z\rangle = \lambda\langle y,z\rangle + \mu\langle x,z\rangle</math> ( [[Linear map|linearity in first argument]] ) | * <math>\langle\lambda x+\mu y,z\rangle = \lambda\langle y,z\rangle + \mu\langle x,z\rangle</math> ( [[Linear map|linearity in first argument]] ) | ||
| − | *: This may be | + | *: This may be alternatively stated as: |
| − | *:* <math>\langle\lambda x,y\rangle=\lambda\langle x,y\rangle</math> and | + | *:* <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 \ge 0</math> | + | ** <math>\langle x,x\rangle=0\iff x=0</math> |
==Properties== | ==Properties== | ||
| − | Notice that <math>\langle\cdot,\cdot\rangle</math> is also linear in its second argument as: | + | 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 = \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> | *<math>\langle x,\lambda y+\mu z\rangle = \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> | ||
| Line 18: | Line 18: | ||
* <math>\langle x,\lambda y\rangle = \bar{\lambda}\langle x,y\rangle</math> and | * <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> | * <math>\langle x,y+z\rangle = \langle x,y\rangle + \langle x,z\rangle</math> | ||
| + | This leads to the most general form: | ||
| + | * {{M|1=\langle au+bv,cx+dy\rangle=a\langle u,cx+dy\rangle+b\langle v,cx+dy\rangle}}{{M|1= =a\overline{\langle cx+dy,u\rangle}+b\overline{\langle cx+dy,v\rangle} }}{{M|1= =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})}}{{M|1= =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}} | ||
| + | {{Begin Theorem}} | ||
| + | Proof of claim: {{M|1=\langle x,\alpha y+\beta z\rangle=\overline{\alpha}\langle x, y\rangle +\overline{\beta}\langle x,z\rangle}} | ||
| + | {{Begin Proof}} | ||
| + | |||
| + | {{End Proof}}{{End Theorem}} | ||
==Examples== | ==Examples== | ||
Revision as of 11:38, 10 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]
- This may be alternatively stated as:
- [math]\langle x,x\rangle \ge 0[/math] but specifically:
- [math]\langle x,x\rangle=0\iff x=0[/math]
Properties
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 = \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]
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\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]
Proof of claim: [ilmath]\langle x,\alpha y+\beta z\rangle=\overline{\alpha}\langle x, y\rangle +\overline{\beta}\langle x,z\rangle[/ilmath]
Examples
See also
References
- ↑ http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885
- ↑ Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014
- ↑ Functional Analysis - George Bachman and Lawrence Narici
