Difference between revisions of "Inner product"
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==Properties== | ==Properties== | ||
Notice that <math>\langle\cdot,\cdot\rangle</math> is also linear (ish) 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> | + | {{Begin Inline Theorem}} |
| − | + | * <math>\langle x,\lambda y+\mu z\rangle =\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle</math> | |
| + | {{Begin Inline Proof}} | ||
| + | :<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. | ||
| + | {{End Proof}}{{End Theorem}} | ||
From this we may conclude the following: | From this we may conclude the following: | ||
* <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: | This leads to the most general form: | ||
| − | |||
{{Begin Inline Theorem}} | {{Begin Inline Theorem}} | ||
| − | + | * {{M|1=\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}} - which isn't worth remembering! | |
{{Begin Inline Proof}} | {{Begin Inline Proof}} | ||
| − | + | :'''Proof:''' | |
| + | :{{M|1=\langle au+bv,cx+dy\rangle}} | ||
| + | ::{{M|1= =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}} | ||
| + | : As required | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
Revision as of 18:23, 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 =\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
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
