Difference between revisions of "Inner product"

Revision as of 18:23, 10 July 2015

Definition

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

$\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}$ (or sometimes $\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{C}$ )

Such that:

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

Properties

Notice that $\langle\cdot,\cdot\rangle$ is also linear (ish) in its second argument as:

[Expand]

$\langle x,\lambda y+\mu z\rangle =\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle$

$\langle x,\lambda y+\mu z\rangle$

$=\overline{\langle \lambda y+\mu z, x\rangle}$

$=\overline{\lambda\langle y,x\rangle + \mu\langle z,x\rangle}$

$=\bar{\lambda}\overline{\langle y,x\rangle}+\bar{\mu}\overline{\langle z,x\rangle}$

$=\bar{\lambda}\langle x,y\rangle+\bar{\mu}\langle x,z\rangle$

As required.

From this we may conclude the following:

$\langle x,\lambda y\rangle = \bar{\lambda}\langle x,y\rangle$ and
$\langle x,y+z\rangle = \langle x,y\rangle + \langle x,z\rangle$

This leads to the most general form:

[Expand]

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

Proof:

$\langle au+bv,cx+dy\rangle$

$=a\langle u,cx+dy\rangle+b\langle v,cx+dy\rangle$

$=a\overline{\langle cx+dy,u\rangle}+b\overline{\langle cx+dy,v\rangle}$

$=a(\overline{c\langle x,u\rangle} + \overline{d\langle y,u\rangle})+b(\overline{c\langle x,v\rangle}+\overline{d\langle y,v\rangle})$

$=a\overline{c}\langle u,x\rangle+a\overline{d}\langle u,y\rangle+b\overline{c}\langle v,x\rangle+b\overline{d}\langle v,y\rangle$

As required

Examples

Vector dot product

References

Jump up ↑ http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885
Jump up ↑ Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014
Jump up ↑ Functional Analysis - George Bachman and Lawrence Narici

[1] Jump up ↑ http://en.wikipedia.org/w/index.php?title=Inner_product_space&oldid=651022885

[2] Jump up ↑ Functional Analysis I - Lecture Notes - Richard Sharp - Sep 2014

[FA-3] Jump up ↑ Functional Analysis - George Bachman and Lawrence Narici

[1]

[2]

[3]

@@ Line 13: / Line 13: @@
 ==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>
+{{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:
 * <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:
-* {{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 Inline 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}}
+* {{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}}
+:'''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}}

Difference between revisions of "Inner product"

Revision as of 18:23, 10 July 2015

Contents

Definition

Properties

Examples

See also

References

Navigation menu

Views

Personal tools

Navigation

Search

Tools