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Lineare Algebra und Analytische Geometrie ii: Noten zu Einer Vorlesung mit Historischen Anmerkungen von Erhard Scholz (in German)
Egbert Brieskorn
Synopsis "Lineare Algebra und Analytische Geometrie ii: Noten zu Einer Vorlesung mit Historischen Anmerkungen von Erhard Scholz (in German)"
Die Jordanzerlegung in halbeinfachen und nilpotenten Anteil lieferte uns die charakteristische Abbildung n M{n x n,K) ~ K , x die jeder Matrix A die Koeffizienten (a , ••• ,a ) des charakteristischen 1 n Polynoms von A zuordnet. Mit Hilfe dieser Abbildung hatten wir das Klassi fikationsproblem in zwei Teilprobleme A und B aufgespalten. Problem A Hier bestand das Problem in der Klassifikation der halbeinfachen Matrizen bis auf Konjugation. Das Hauptresultat war der Satz 11.45*. Die Konjugations klassen halbeinfacher Matrizen entsprechen bijektiv den Punkten des affinen Raumes ~. Eine Einteilung der halbeinfachen Konjugationsklassen in Typen ergibt sich in naturlicher Weise durch die algebraischen Multiplizitaten der Eigenwerte Ai • Dabei entsprechen die regularen Elemente, d.h. die n jenigen mit m = 1 , gerade den Punkten von K 1m Komplement der Disk- i n minantenmenge D cK , und den verschiedenen Typen von singul4ren Elementen entsprechen, wie wir an Beispielen gesehen haben, verschiedene Strata (d.h. Schichten) von D, welche man analytisch-geometrisch charakterisieren kann. 1m Fall K = Roder K = ~ sehen wir also, daB die Konjugationsklassen der halbeinfachen Anteile eine kontinuierliche Mannigfaltigkeit bilden, namlich einen affinen Raum Kn, und daB die weitere Typeneinteilung dieser Konju gationsklassen mit der analytischen Geometrie der Diskriminantenmengen n D c. K zusammenhangt. Leer más Leer menos P.when('DynamicIframe').execute(function(DynamicIframe){ var BookDescriptionIframe = null, bookDescEncodedData = "Die%20Jordanzerlegung%20in%20halbeinfachen%20und%20nilpotenten%20Anteil%20lieferte%20uns%20die%20charakteristische%20Abbildung%20n%20M%7Bn%20x%20n%2CK%29%20%7E%20K%20%2C%20x%20die%20jeder%20Matrix%20A%20die%20Koeffizienten%20%28a%20%2C%20%E2%80%A2%E2%80%A2%E2%80%A2%20%2Ca%20%29%20des%20charakteristischen%201%20n%20Polynoms%20von%20A%20zuordnet.%20Mit%20Hilfe%20dieser%20Abbildung%20hatten%20wir%20das%20Klassi%C2%AD%20fikationsproblem%20in%20zwei%20Teilprobleme%20A%20und%20B%20aufgespalten.%20Problem%20A%20Hier%20bestand%20das%20Problem%20in%20der%20Klassifikation%20der%20halbeinfachen%20Matrizen%20bis%20auf%20Konjugation.%20Das%20Hauptresultat%20war%20der%20Satz%2011.45*.%20Die%20Konjugations%C2%AD%20klassen%20halbeinfacher%20Matrizen%20entsprechen%20bijektiv%20den%20Punkten%20des%20affinen%20Raumes%20%7E.%20Eine%20Einteilung%20der%20halbeinfachen%20Konjugationsklassen%20in%20Typen%20ergibt%20sich%20in%20naturlicher%20Weise%20durch%20die%20algebraischen%20Multiplizitaten%20der%20Eigenwerte%20Ai%20%E2%80%A2%20Dabei%20entsprechen%20die%20regularen%20Elemente%2C%20d.h.%20die%C2%AD%20n%20jenigen%20mit%20m%20%3D%201%20%2C%20gerade%20den%20Punkten%20von%20K%201m%20Komplement%20der%20Disk-%20i%20n%20minantenmenge%20D%20cK%20%2C%20und%20den%20verschiedenen%20Typen%20von%20singul4ren%20Elementen%20entsprechen%2C%20wie%20wir%20an%20Beispielen%20gesehen%20haben%2C%20verschiedene%20Strata%20%28d.h.%20Schichten%29%20von%20D%2C%20welche%20man%20analytisch-geometrisch%20charakterisieren%20kann.%201m%20Fall%20K%20%3D%20Roder%20K%20%3D%20%7E%20sehen%20wir%20also%2C%20daB%20die%20Konjugationsklassen%20der%20halbeinfachen%20Anteile%20eine%20kontinuierliche%20Mannigfaltigkeit%20bilden%2C%20namlich%20einen%20affinen%20Raum%20Kn%2C%20und%20daB%20die%20weitere%20Typeneinteilung%20dieser%20Konju%C2%AD%20gationsklassen%20mit%20der%20analytischen%20Geometrie%20der%20Diskriminantenmengen%20n%20D%20c.%20K%20zusammenhangt.", bookDescriptionAvailableHeight, minBookDescriptionInitialHeight = 112, options = {}, iframeId = "bookDesc_iframe"; 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P.when('jQuery').execute(function(jQuery) { window.sitbWeblab = ''; var postReftags = function(result, isReaderOpening) { if (result === undefined || result === null) { return; } var urls = isReaderOpening ? result.onReaderOpen : result.onDetailPage; if (urls === undefined || urls === null) { return; } for (var i = 0; urls.length !== undefined && urls.length !== null && i Descripción del producto Reseña del editor Die Jordanzerlegung in halbeinfachen und nilpotenten Anteil lieferte uns die charakteristische Abbildung n M{n x n,K) ~ K , x die jeder Matrix A die Koeffizienten (a , ••• ,a ) des charakteristischen 1 n Polynoms von A zuordnet. Mit Hilfe dieser Abbildung hatten wir das Klassi fikationsproblem in zwei Teilprobleme A und B aufgespalten. Problem A Hier bestand das Problem in der Klassifikation der halbeinfachen Matrizen bis auf Konjugation. Das Hauptresultat war der Satz 11.45*. Die Konjugations klassen halbeinfacher Matrizen entsprechen bijektiv den Punkten des affinen Raumes ~. Eine Einteilung der halbeinfachen Konjugationsklassen in Typen ergibt sich in naturlicher Weise durch die algebraischen Multiplizitaten der Eigenwerte Ai • Dabei entsprechen die regularen Elemente, d.h. die n jenigen mit m = 1 , gerade den Punkten von K 1m Komplement der Disk- i n minantenmenge D cK , und den verschiedenen Typen von singul4ren Elementen entsprechen, wie wir an Beispielen gesehen haben, verschiedene Strata (d.h. Schichten) von D, welche man analytisch-geometrisch charakterisieren kann. 1m Fall K = Roder K = ~ sehen wir also, daB die Konjugationsklassen der halbeinfachen Anteile eine kontinuierliche Mannigfaltigkeit bilden, namlich einen affinen Raum Kn, und daB die weitere Typeneinteilung dieser Konju gationsklassen mit der analytischen Geometrie der Diskriminantenmengen n D c. K zusammenhangt. #productDescription { color: #333333; word-wrap: break-word; font-size: small; line-height: initial; margin: 0.5em 0px 0em 25px; } #productDescription_feature_div > h2.default { color: #CC6600; font-size: medium; margin: 0 0 0.25em; } #productDescription_feature_div > h2.books { color:#333 !important; font-size:21px !important; line-height: 1.3; padding-bottom: 4px; font-weight: normal; margin: 0px; } #productDescription_feature_div > h2.softlines { color:#333 !important; font-size:21px !important; line-height: 1.3; padding-bottom: 4px; font-weight: bold; margin: 0px; } #productDescription > p, #productDescription > div, #productDescription > table { margin: 0 0 1em 0; } #productDescription p { margin: 0em 0 1em 1em; } #productDescription h3 { font-weight: normal; color: #333333; font-size: 1.23em; clear: left; margin: 0.75em 0px 0.375em -15px; } #productDescription table { border-collapse: inherit !important; margin-bottom: 0; } #productDescription table img { max-width: inherit !important; } #productDescription table td { font-size: small; vertical-align: inherit !important; } #productDescription ul li { margin: 0 0 0 20px; } #productDescription ul li ul { list-style-type: disc !important; margin-left: 20px !important; } #productDescription ul ul li { list-style-type: disc !important; margin-left: 20px !important; } #productDescription > ul ul li { list-style-type: disc !important; } #productDescription ul li ul li { margin: 0 0 0 20px; } #productDescription .aplus p { margin: 0 0 1em 0; } #productDescription small { font-size: smaller; } #productDescription.prodDescWidth { max-width: 1000px } P.when('jQuery').execute(function($){ $("#productDescription table").each(function() { var width = $(this).attr('width'); if (width) width += 'px'; else width = 'auto'; $(this).css('width', width); var padding = $(this).attr('cellpadding'); if (padding) padding += 'px'; else padding = '0px'; $(this).css('padding', padding); }); }); ¡Díselo a la editorial! Me gustaría leer este libro en Kindle ¿No tienes un Kindle? Consigue un Kindle aquí o descarga una aplicación de lectura Kindle GRATUITA. Tapa blanda: 552 páginas Editor: Vieweg+Teubner Verlag; Edición: Softcover reprint of the original 1st ed. 1985 (1 de enero de 1985) Idioma: Alemán ISBN-10: 3322831779 ISBN-13: 978-3322831774 Calificación promedio de nuestros clientes acerca del producto: Sé el primero en calificar este artículo Si vendes este producto, ¿te gustaría sugerir actualizaciones mediante el soporte al vendedor? if (typeof uet === 'function' && typeof ues === 'function') {var scope = 'Detail_dp-ads-center-promo_Desktop';var placementId = 'd5e5e614-eb00-48e7-9be3-4047cfbef514';ues('wb', 'adplacements:' + scope.replace(/\_/g, ':'), {wb:1}); uet('bb', 'adplacements:' + scope.replace(/\_/g, ':'), {wb:1}); if (placementId) {ues('wb', 'adplacements:' + placementId, {wb:1});uet('bb', 'adplacements:' + placementId, {wb:1});}}(function(){function a(d,e){if(window.addEventListener){window.addEventListener(d,e,false);}else{if(window.attachEvent){window.attachEvent("on"+d,e);}}}function c(d,e){if(window.removeEventListener){window.removeEventListener(d,e,false);}else{if(window.detachEvent){window.detachEvent("on"+d,e);}}}var b=function(){(function(){(function(j,n){j.sfLogErrors=j.sfLogErrors||false;var o=o||function(s,r){r=r||new Error(s);if(j.ue&&typeof ue.count=="function"){ue.count("adplacements:safeFrameError",1);}if(!j.sfLogErrors){return;}if(j.ueLogError){j.ueLogError(r,{logLevel:"ERROR",attribution:"APE-safeframe",message:s+" "});}else{if(typeof console!=="undefined"&&console.error){console.error(s,r);}}};j.aanParams=j.aanParams||{};j.aanParams["dp-ads-center-promo"]="site=amazon.es;pt=Detail;slot=dp-ads-center-promo;pid=3322831779;bn=599367031;arid=c385e64e944b493ba56ba666d815f940";j["dp-ads-center-promo"]={};j["dp-ads-center-promo"].adStartTime=(new Date()).getTime();function d(){return j.innerHeight||n.documentElement.clientHeight;}function g(){return j.innerWidth||n.documentElement.clientWidth;}function e(t,r,s){if(t>0){return(s>t);}else{return(r>0);}}var f=function(){return(Date.now?Date.now():new Date().getTime());};throttle=function(s,u,y){var r,w,z;var x=null;var v=0;if(!y){y={};}var t=function(){v=y.leading===false?0:f();x=null;z=s.apply(r,w);if(!x){r=w=null;}};return function(){var B=f();if(!v&&y.leading===false){v=B;}var A=u-(B-v);r=this;w=arguments;if(Au){if(x){clearTimeout(x);x=null;}v=B;z=s.apply(r,w);if(!x){r=w=null;}}else{if(!x&&y.trailing!==false){x=setTimeout(t,A);}}return z;};};function l(u,w,v,r){try{var t=n.getElementById(u).getBoundingClientRect();if(e(t.top,t.bottom,d())&&e(t.left,t.right,g())){if(typeof uet=="function"){uet("bb","adplacements:viewablelatency:"+w,{wb:1});if(v){uet("bb","adplacements:viewablelatency:"+v,{wb:1});}}if(typeof uex=="function"&&j.ue&&typeof ue.count=="function"){if(j.apeViewableLatencyTrackers[r].loaded){uex("ld","adplacements:viewablelatency:"+w,{wb:1});if(v){uex("ld","adplacements:viewablelatency:"+v,{wb:1});}ue.count("adplacements:htmlviewed:loaded:"+w,1);if(v){ue.count("adplacements:htmlviewed:loaded:"+v,1);}}ue.count("adplacements:htmlviewed:"+w,1);if(v){ue.count("adplacements:htmlviewed:"+v,1);}}j.apeViewableLatencyTrackers[r].viewed=true;if(j.apeViewableLatencyTrackers[r].tracker){c("scroll",j.apeViewableLatencyTrackers[r].tracker);c("resize",j.apeViewableLatencyTrackers[r].tracker);}}}catch(s){j.apeViewableLatencyTrackers[r].valid=false;}}try{j.apeViewableLatencyTrackers=j.apeViewableLatencyTrackers||{};var q="ape_Detail_dp-ads-center-promo_Desktop_placement";var p="Detail_dp-ads-center-promo_Desktop".replace(/\_/g,":");var h="d5e5e614-eb00-48e7-9be3-4047cfbef514";var i="c385e64e944b493ba56ba666d815f940";j.apeViewableLatencyTrackers[i]=j.apeViewableLatencyTrackers[i]||{};j.apeViewableLatencyTrackers[i].valid=true;l(q,p,h,i);if(j.apeViewableLatencyTrackers[i].valid&&!j.apeViewableLatencyTrackers[i].viewed){j.apeViewableLatencyTrackers[i].tracker=throttle(function(){l(q,p,h,i);},20);a("scroll",j.apeViewableLatencyTrackers[i].tracker);a("resize",j.apeViewableLatencyTrackers[i].tracker);}}catch(k){if(j.apeViewableLatencyTrackers&&j.apeViewableLatencyTrackers.c385e64e944b493ba56ba666d815f940){j.apeViewableLatencyTrackers.c385e64e944b493ba56ba666d815f940.valid=false;}o("Error initializing viewable latency instrumentation",k);}try{if(j.DAsf){j.DAsf.loadAds();}else{var m=n.createElement("script");m.type="text/javascript";m.async=true;m.setAttribute("crossorigin","anonymous");m.charset="utf-8";m.src="https://images-eu.ssl-images-amazon.com/images/G/30/ape/sf/desktop/DAsf-1.50.b6f5f66._V479604094_.js?csm_attribution=APE-SafeFrame";m.onerror=function(){o("Error loading SafeFrame library");};(n.getElementsByTagName("head")[0]||n.getElementsByTagName("body")[0]).appendChild(m);}}catch(k){o("Error appending DAsf library",k);}}(window,document));})();};b();})(); (window.AmazonUIPageJS ? 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i var ocInitTimestamp = 1544002529; if (!window.$SearchJS && window.$Nav) { window.$SearchJS = $Nav.make('sx'); } if (window.$SearchJS) { } (function() { var precacheMaterials = ["https://images-eu.ssl-images-amazon.com/images/G/30/nav2/images/gui/searchSprite._CB272921296_.png"]; if (window.amznJQ) { amznJQ.addPL(precacheMaterials); } else if (window.P) { P.when('A').execute(function(A){ A.preload(precacheMaterials); }); } })(); #nav-prime-tooltip{ padding: 0 20px 2px 20px; background-color: white; font-family: arial,sans-serif; } .nav-npt-text-title{ font-family: arial,sans-serif; font-size: 18px; font-weight: bold; line-height: 21px; color: #E47923; } .nav-npt-text-detail, a.nav-npt-a{ font-family: arial,sans-serif; font-size: 12px; line-height: 14px; color: #333333; margin: 2px 0px; } a.nav-npt-a { text-decoration: underline; } Amazon Prime: Envíos rápidos, GRATIS e ilimitados y mucho más Los clientes de Amazon Prime disfrutan de Envío en 1 día GRATIS en dos millones de productos y Envío en 2 o 3 días en millones de productos más, Acceso a series y películas en Prime Video, incluyendo las series Prime Originals, 40 horas de música al mes con Prime Music, cientos de eBooks en Prime Reading, Acceso Prioritario a las Ofertas flash y Almacenamiento de fotos gratis e ilimitado en Amazon Drive. > Más información Hay un problema para cargar este menú. 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