en>K9re11 at 19:41, 25 October 2014

2014-10-25T19:41:14Z

← Older revision		Revision as of 21:41, 25 October 2014
Line 1:		Line 1:
	~~The '''Steinitz exchange lemma'~~'' is a basic ~~theorem in [[linear algebra]] used, for example,~~ to ~~show that any two [[Basis (linear algebra)\|bases]] for a finite-[[Dimension (vector space)\|dimensional]] [[vector space]] have the same number of elements~~. ~~The result is named after~~ the ~~German mathematician [[Ernst Steinitz]]. The result~~ is ~~often called the '''Steinitz–Mac Lane exchange lemma''', also recognizing~~ the ~~generalization<ref>~~		Let's have an appearance on world's web hosting companies: Thousands and thousands of web hosts spread around the world and their business aim is basic - to get recognized and make earnings. With so much noise around the internet hosting market, it is a challenging job for us to pick up the right hosting.<br><br>
	{{citation\|last=Mac Lane\|first=Saunders\|authorlink=Saunders Mac Lane\|year=1936\|title=Some interpretations of abstract linear dependence in terms of projective geometry\|journal=American Journal of Mathematics\|volume=58\|pages=236–240\|doi=10.2307/2371070\| jstor=2371070 \| issue=1\|publisher=The Johns Hopkins University Press}}.</ref>
	~~by [[Saunders Mac Lane\|Saunders Mac Lane]]~~
	~~of Steinitz's lemma to [[matroid]]s~~.<~~ref~~>
	~~{{citation\|editor-last=Kung\|editor-first=Joseph P. S.\|title=A Source Book in Matroid Theory\|publisher=Birkhäuser\|mr=0890330\|isbn=0-8176-3173-9\|location=Boston\|year=1986}}.~~
	<~~/ref~~>

	~~== Statement ==~~		In any internet business, we depend on excellent webhosting to offer stable and dependable hosting services. Hostgator provides timely and promising qualities. Should you have virtually any [http://Browse.deviantart.com/?q=inquiries inquiries] regarding where along with how you can work with [http://www.alleyboy.com/forum/how-addon-domain-hostgator-695376 Cheap Hosting from Hostgator], it is possible to e-mail us with our own page. They invested a lot of cash for its servers which each of them using Dual Xeon servers with 4GB memory and IDE hard disks. These high speed servers offer high responsive uptime of up to 99.9 % without failure.<br><br>Hostgator provides a bunch of bundles to clients. Virtually all the bundles offered are cost effective. Hostgator shared hosting are charged $4.95 monthly for the basic shared hosting packaged. At charges which start from $9.95 yearly, you can delight in much more compact plans which truly come with a great deal of stuffs. There still have a lot packages which offering more affordable rates, however if you begin to research and try Hostgator, you will feel that every dollar and cent you invested is simply not wasted. Do your own research on the functions consisted of in every plans Hostgator offering, then you can differentiate the differences with others packages provided by various other internet hosts.<br><br>Hostgator also provides a lot of other features. For exaple, Hostgator assists you to back up the website promptly and safely. You have the option to host unlimited domains, sub domains and FTP accounts. Of cause, the attribute which pointed out just now will not be available if you are the basic hosting plan customer. For basic package, Hostgator still gives you single domain which packaged together with twenty sub domains. In my viewpoint, you can start your trial with Hostgator's basic bundle. When company gets mature and thinking for development, you can anytime upgrade your bundle to more advance one.<br><br>Hostgator provides control panel. This feature assists individuals to handle their domains and sub domains in even more versatile way. The control panel demo which comes in standard offers the valuable aid devices to new users. Out of many host around the globe, Hostgator's cpanel can be said succeeding the top position amongst them.<br><br>In term of client supports, Hostgator has a really strong client care team which working 24/7 to assist you take on all sorts of issues relevant to their [http://www.imdb.com/find?q=products&s=all products]. This is why till now, Hostgator still the finest choice for customers.<br><br>A last words for those who are getting began - do not be scare by what you are reading as building a great website is not that difficult. The work is in truth pretty simple and direct as long as you're getting the right information. To obtain begun, you'll first require a domain name and a trustworthy hosting service.
	~~If {''v''<sub>1</sub>~~, ..~~., ''v''<sub>''m''<~~/~~sub>} is a set of ''m'' [[Linear independence\|linearly independent]] vectors in a vector space ''V'', and {''w''<sub>1<~~/~~sub>, .~~..~~, ''w''<sub>''n''<~~/~~sub>}~~ [~~[Linear span\|span]] ''V'' then ''m'' ≤ ''n'' and, possibly after reordering the ''w''<sub>''i''<~~/~~sub>, the set {''v''<sub>1<~~/~~sub>, .~~..~~, ''v''<sub>''m''<~~/~~sub>, ''w''<sub>''m'' + 1<~~/~~sub>~~, ...~~, ''w''~~<~~sub~~>~~''n''~~<~~/sub~~>~~} spans ''V''~~.

	~~==Proof==~~

	We are ~~going to show that~~ for ~~any integer <math>k</math> satisfying <math>0\leq k\leq m</math>,~~ the ~~following assertion is valid~~. ~~Choosing <math>k=m</math> gives the result~~.

	~~(A) The set <math>\{ v_1~~,~~\ldots~~, ~~v_k~~,~~w_{k+1}~~,~~\ldots,w_n\}~~<~~/math~~> ~~spans~~ <~~math~~>~~V</math> (where~~ the ~~<math>w_j</math>~~ have ~~possibly been reordered~~, and ~~the reordering depends on <math>k</math>)~~.

	~~We will prove (A) by induction over <math>k</math>: Being clear for <math>k=0</math>~~, the ~~only thing that needs to~~ be ~~done is~~ the ~~inductive step~~.

	~~Assume that (A) holds for some <math>k</math> satisfying <math>0\leq k<m</math>~~. ~~Since <math>v_{k+1}\in V</math>~~, and ~~<math>\{ v_1~~,~~\ldots, v_k,w_{k+1},\ldots,w_n\}~~<~~/math~~> ~~spans~~ <~~math~~>~~V</math> (by~~ the ~~induction hypothesis)~~, ~~there exist~~ <~~math~~>~~\mu_1,\ldots,\mu_n~~<~~/math~~> ~~such that~~
	~~:<math>v_{k+1}=\sum_{j=1}^k \mu_j v_j+\sum_{j=k+1}^n \mu_j w_j.</math>~~
	~~At least one~~ of ~~<math>\{\mu_{k+1}~~,~~\ldots,\mu_n\}<~~/~~math> must be non-zero, otherwise this equality would contradict the linear independence~~ of ~~<math>\{ v_1,\ldots,v_m \}<~~/~~math>; note that this additionally implies that <math>k<n<~~/~~math>~~. ~~By reordering the <math>w_{k+1},\ldots,w_n</math>, we may assume that <math>\mu_{k+1}<~~/~~math> is not zero. Therefore, we have~~
	~~: <math>w_{k+1}~~= ~~\frac{1}{\mu_{k+1}}\left(v_{k+1} - \sum_{j~~=~~1}^k \mu_j v_j - \sum_{j=k+2}^n \mu_j w_j\right)</math>~~
	~~In other words, <math>w_{k+1}</math>~~ is ~~in the span of <math>\{ v_1~~,~~\ldots, v_{k+1},w_{k+2},\ldots,w_n\}</math> and so~~ the ~~latter must be the whole of~~ <~~math~~>V<~~/math~~>~~. We have thus shown that (~~A~~) holds~~ for ~~<math>k+1</math>, completing the inductive step~~.

	~~==Applications==~~
	The ~~Steinitz exchange lemma~~ is ~~a basic result in [[computational mathematics]], especially~~ in ~~[[numerical linear algebra\|linear algebra]]~~ and ~~in [[Matroid#Greedy_algorithm\|combinatorial algorithms]].<ref>Page v in Stiefel:~~
	~~{{cite book\|last=Stiefel\|first=Eduard L.\|authorlink=Eduard Stiefel\|title=An introduction to numerical mathematics\|edition=Translated by Werner C. Rheinboldt & Cornelie J. Rheinboldt from~~ the ~~second German\|publisher=Academic Press\|location=New York\|year=1963\|pages=x+286\|mr=181077}}~~
	~~</ref>~~

	~~== References ==~~

	~~<references/>~~

	* Julio R. ~~Bastida~~, '~~'Field extensions~~ and ~~Galois Theory'', [[Addison–Wesley\|Addison–Wesley Publishing Company]] (1984)~~.

	~~[[Category:Linear algebra]]~~
	~~[[Category:Lemmas]]~~
	~~[[Category:Matroid theory]]~~

en>Rzh at 03:34, 19 May 2009

2009-05-19T03:34:05Z

New page

The '''Steinitz exchange lemma''' is a basic theorem in [[linear algebra]] used, for example, to show that any two [[Basis (linear algebra)|bases]] for a finite-[[Dimension (vector space)|dimensional]] [[vector space]] have the same number of elements. The result is named after the German mathematician [[Ernst Steinitz]]. The result is often called the '''Steinitz–Mac Lane exchange lemma''', also recognizing the generalization<ref>
{{citation|last=Mac Lane|first=Saunders|authorlink=Saunders Mac Lane|year=1936|title=Some interpretations of abstract linear dependence in terms of projective geometry|journal=American Journal of Mathematics|volume=58|pages=236–240|doi=10.2307/2371070| jstor=2371070 | issue=1|publisher=The Johns Hopkins University Press}}.</ref>
by [[Saunders Mac Lane|Saunders Mac Lane]]
of Steinitz's lemma to [[matroid]]s.<ref>
{{citation|editor-last=Kung|editor-first=Joseph P. S.|title=A Source Book in Matroid Theory|publisher=Birkhäuser|mr=0890330|isbn=0-8176-3173-9|location=Boston|year=1986}}.
</ref>

== Statement ==
If {''v''1, ..., ''v''''m''} is a set of ''m'' [[Linear independence|linearly independent]] vectors in a vector space ''V'', and {''w''1, ..., ''w''''n''} [[Linear span|span]] ''V'' then ''m'' ≤ ''n'' and, possibly after reordering the ''w''''i'', the set {''v''1, ..., ''v''''m'', ''w''''m'' + 1, ..., ''w''''n''} spans ''V''.

==Proof==

We are going to show that for any integer <math>k</math> satisfying <math>0\leq k\leq m</math>, the following assertion is valid. Choosing <math>k=m</math> gives the result.

(A) The set <math>\{ v_1,\ldots, v_k,w_{k+1},\ldots,w_n\}</math> spans <math>V</math> (where the <math>w_j</math> have possibly been reordered, and the reordering depends on <math>k</math>).

We will prove (A) by induction over <math>k</math>: Being clear for <math>k=0</math>, the only thing that needs to be done is the inductive step.

Assume that (A) holds for some <math>k</math> satisfying <math>0\leq k<m</math>. Since <math>v_{k+1}\in V</math>, and <math>\{ v_1,\ldots, v_k,w_{k+1},\ldots,w_n\}</math> spans <math>V</math> (by the induction hypothesis), there exist <math>\mu_1,\ldots,\mu_n</math> such that
:<math>v_{k+1}=\sum_{j=1}^k \mu_j v_j+\sum_{j=k+1}^n \mu_j w_j.</math>
At least one of <math>\{\mu_{k+1},\ldots,\mu_n\}</math> must be non-zero, otherwise this equality would contradict the linear independence of <math>\{ v_1,\ldots,v_m \}</math>; note that this additionally implies that <math>k<n</math>. By reordering the <math>w_{k+1},\ldots,w_n</math>, we may assume that <math>\mu_{k+1}</math> is not zero. Therefore, we have
: <math>w_{k+1}= \frac{1}{\mu_{k+1}}\left(v_{k+1} - \sum_{j=1}^k \mu_j v_j - \sum_{j=k+2}^n \mu_j w_j\right)</math>
In other words, <math>w_{k+1}</math> is in the span of <math>\{ v_1,\ldots, v_{k+1},w_{k+2},\ldots,w_n\}</math> and so the latter must be the whole of <math>V</math>. We have thus shown that (A) holds for <math>k+1</math>, completing the inductive step.

==Applications==
The Steinitz exchange lemma is a basic result in [[computational mathematics]], especially in [[numerical linear algebra|linear algebra]] and in [[Matroid#Greedy_algorithm|combinatorial algorithms]].<ref>Page v in Stiefel:
{{cite book|last=Stiefel|first=Eduard L.|authorlink=Eduard Stiefel|title=An introduction to numerical mathematics|edition=Translated by Werner C. Rheinboldt & Cornelie J. Rheinboldt from the second German|publisher=Academic Press|location=New York|year=1963|pages=x+286|mr=181077}}
</ref>

== References ==

<references/>

* Julio R. Bastida, ''Field extensions and Galois Theory'', [[Addison–Wesley|Addison–Wesley Publishing Company]] (1984).

[[Category:Linear algebra]]
[[Category:Lemmas]]
[[Category:Matroid theory]]

Petersson trace formula - Revision history

en>K9re11 at 19:41, 25 October 2014

en>Rzh at 03:34, 19 May 2009