{"id":457,"date":"2025-11-12T08:15:22","date_gmt":"2025-11-12T07:15:22","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=457"},"modified":"2025-11-13T06:54:47","modified_gmt":"2025-11-13T05:54:47","slug":"alg-interseccion-incidencia-y-paralelismo","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=457","title":{"rendered":"ALG: Intersecci\u00f3n, incidencia y paralelismo"},"content":{"rendered":"<p>Vamos a tratar la posici\u00f3n relativa de dos variedades afines: \\(L_1=P+C_1\\) y \\(L_2=Q+C_2\\). Diremos que se cortan si el conjunto \\(L_1\\cap L_2\\) no es vac\u00edo. Si \\(L_1\\cap L_2=\\phi\\); es decir, si no se cortan, puede ocurrir que \\(C_1\\subseteq C_2\\) (o \\(C_2\\subseteq C_1\\) ) en cuyo caso se dice que son paralelas; en caso contrario se dice que se cruzan.<\/p>\n<p>Si conocemos las ecuaciones impl\u00edcitas de las dos variedades, el conjunto \\(L_1\\cap L_2\\) viene dado por los puntos cuyas coordenadas, respecto del sistema de referencia considerado, son las soluciones del sistema que resulta de reunir todas las ecuaciones impl\u00edcitas. Si denotamos por \\(n=dim(E)\\), siendo \\(E\\) el espacio af\u00edn, \\(r=dim(L_1)\\) y \\(s=dim(L_2)\\), y suponiendo que \\(r\\leq s\\), el sistema formado por todas las ecuaciones es un sistema de \\(2n-r-s\\) ecuaciones, que podemos escribir en forma matricial: \\(AX=b\\). Seg\u00fan que el rango de la matriz de coeficientes coincida con el rango de la ampliada obtenemos la diferencia entre variedades que se cortan o que no se cortan. Si \\(rg(A)\\) es \\(n-r\\), entonces la dimensi\u00f3n de \\(C_1\\cap C_2\\) ser\u00e1 \\(r\\) y por tanto \\(C_1\\cap C_2=C_1\\), con lo que \\(C_1\\subseteq C_2\\).<\/p>\n<p>Podemos resumirlo en el siguiente cuadro:<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"aligncenter size-full wp-image-180\" title=\"incidencia\" src=\"http:\/\/uploads.jesussoto.es\/2011\/12\/incidencia2.png\" alt=\"\" width=\"501\" height=\"102\" \/><\/p>\n<p>Que es equivalente a:<\/p>\n<table id=\"yzpi\" style=\"font-size:1em\" border=\"1\" align=\"center\" width=\"70%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#D6EAF8\">\n<tbody  style=\"font-size:14px;\">\n<tr>\n<td>\\(\\textbf{rg}\\,A=\\textbf{max}\\{\\textbf{rg}\\,L_1,\\textbf{rg}\\,L_2\\}\\)<\/td>\n<td>\\(\\textbf{rg}\\,A=\\textbf{rg}\\,Ab\\)<\/td>\n<td><strong>Incidentes \\((\\mathbf{-})\\)<\/strong><\/td>\n<\/tr>\n<tr>\n<td>\\(\\textbf{rg}\\,A=\\textbf{max}\\{\\textbf{rg}\\,L_1,\\textbf{rg}\\,L_2\\}\\)<\/td>\n<td>\\(\\textbf{rg}\\,A\\neq\\textbf{rg}\\,Ab\\)<\/td>\n<td><strong>Paralelas \\((\\mathbf{=})\\)<\/strong><\/td>\n<\/tr>\n<tr>\n<td>\\(\\textbf{rg}\\,A&gt;\\textbf{max}\\{\\textbf{rg}\\,L_1,\\textbf{rg}\\,L_2\\}\\)<\/td>\n<td>\\(\\textbf{rg}\\,A= \\textbf{rg}\\,Ab\\)<\/td>\n<td><strong>Se cortan \\((\\mathbf{+})\\)<\/strong><\/td>\n<\/tr>\n<tr>\n<td>\\(\\textbf{rg}\\,A&gt;\\textbf{max}\\{\\textbf{rg}\\,L_1,\\textbf{rg}\\,L_2\\}\\)<\/td>\n<td>\\(\\textbf{rg}\\,A\\neq \\textbf{rg}\\,Ab\\)<\/td>\n<td><strong>Se cruzan \\((\\mathbf{\\times})\\)<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar la posici\u00f3n relativa de las variedades \\[\\begin{align*} r_1&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ 4 x-2 y-2 z=0,\\ -3 x+4 y-2 t+8=0\\}\\\\ r_2&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ 6 y-2 z+8=0,\\ -4 x+6 y-2 t+10=0\\}\\end{align*}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv31() {\n  var htmlShow31 = document.getElementById(\"html-show31\");\n  if (htmlShow31.style.display === \"none\") {\n    htmlShow31.style.display = \"block\";\n  } else {\n    htmlShow31.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv31()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show31\" style=\"display: none;\">\nConstruyamos las matrices del sistema: \\[\\begin{bmatrix}4 &#038; -2 &#038; -2 &#038; 0\\\\<br \/>\n-3 &#038; 4 &#038; 0 &#038; -2\\\\<br \/>\n0 &#038; 6 &#038; -2 &#038; 0\\\\<br \/>\n-4 &#038; 6 &#038; 0 &#038; -2\\end{bmatrix}\\begin{bmatrix}x\\\\ y\\\\ z\\\\ t\\end{bmatrix}=\\begin{bmatrix}0\\\\ -8\\\\ -8\\\\ -10\\end{bmatrix}\\]<br \/>\nLos rangos nos dicen que <\/p>\n<p>\\[\\mathbf{rank}\\begin{bmatrix}4 &#038; -2 &#038; -2 &#038; 0\\\\<br \/>\n-3 &#038; 4 &#038; 0 &#038; -2\\\\<br \/>\n0 &#038; 6 &#038; -2 &#038; 0\\\\<br \/>\n-4 &#038; 6 &#038; 0 &#038; -2\\end{bmatrix}=3\\neq \\mathbf{rank}\\begin{bmatrix}4 &#038; -2 &#038; -2 &#038; 0 &#038; 0\\\\<br \/>\n-3 &#038; 4 &#038; 0 &#038; -2 &#038; -8\\\\<br \/>\n0 &#038; 6 &#038; -2 &#038; 0 &#038; -8\\\\<br \/>\n-4 &#038; 6 &#038; 0 &#038; -2 &#038; -10\\end{bmatrix}\\]<\/p>\n<p>Y como \\(\\mathbf{rank}\\ A=3>\\mathbf{max}\\{\\mathbf{dim}\\ r_1,\\mathbf{dim}\\ r_2\\}=2\\), resulta que las variedades se cruzan.\n<\/p><\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar la posici\u00f3n relativa de las variedades \\[\\begin{align*} r_1&#038;:\\left\\{\\begin{bmatrix}x&#038; y\\\\ z&#038; t\\end{bmatrix}\\in\\mathcal{M}_2(\\mathbb{R});\\ x-y-z=0,\\ -3 x+2 y-t+4=0\\right\\}\\\\ r_2&#038;:\\left\\{\\begin{bmatrix}x&#038; y\\\\ z&#038; t\\end{bmatrix}\\in\\mathcal{M}_2(\\mathbb{R});\\ -y-2z+1=0,\\ -4x+7y-2t+11=0\\right\\}\\end{align*}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv31a() {\n  var htmlShow31a = document.getElementById(\"html-show31a\");\n  if (htmlShow31a.style.display === \"none\") {\n    htmlShow31a.style.display = \"block\";\n  } else {\n    htmlShow31a.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv31a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show31a\" style=\"display: none;\">\nPlanteemos el sistema y comprobemos los rangos de sus matrices:<br \/>\n\\[\\mathbf{rank}\\begin{bmatrix}1 &#038; -1 &#038; -1 &#038; 0\\\\<br \/>\n-3 &#038; 2 &#038; 0 &#038; -1\\\\<br \/>\n0 &#038; -1 &#038; -2 &#038; 0\\\\<br \/>\n-4 &#038; 7 &#038; 0 &#038; -2\\end{bmatrix}=4=\\mathbf{rank}\\ \\begin{bmatrix}1 &#038; -1 &#038; -1 &#038; 0 &#038; 0\\\\<br \/>\n-3 &#038; 2 &#038; 0 &#038; -1 &#038; -4\\\\<br \/>\n0 &#038; -1 &#038; -2 &#038; 0 &#038; -1\\\\<br \/>\n-4 &#038; 7 &#038; 0 &#038; -2 &#038; -11\\end{bmatrix}\\]<\/p>\n<p>Luego estamos ante dos variedades que se cortan o son incidentes. Como \\(\\mathbf{rank}\\ A=4>\\mathbf{max}\\{\\mathbf{dim}\\ r_1,\\mathbf{dim}\\ r_2\\}=2\\), resulta que las variedades se cortan. Observar que en este caso la intersecci\u00f3n es un punto.\n<\/p><\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar la posici\u00f3n relativa de las variedades \\[\\begin{align*} r_1&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ x+y=0,\\ y+2z=0,\\ z+3=1\\}\\\\ r_2&#038;:\\{(x,y,z,t)\\in\\mathbb{R}^4;\\ \\frac{x-2}{2}= y=\\frac{z}{-1}=t+1\\}\\end{align*}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1s() {\n  var htmlShow1s = document.getElementById(\"html-show1s\");\n  if (htmlShow1s.style.display === \"none\") {\n    htmlShow1s.style.display = \"block\";\n  } else {\n    htmlShow1s.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1s()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1s\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Posici\u00f3n Relativa de dos Variedades. Ej.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/4Lu25H0LASE?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar la posici\u00f3n relativa de las variedades \\[\\begin{align*} S&#038;=\\{(x,y,z,t,u)\\in\\mathbb{R}^5;\\ -z-y+3x+t+1=0,\\ 2y+x+u-t=0\\}\\\\ T&#038;=\\left\\{(x,y,z,t,u)\\in\\mathbb{R}^5;\\ \\begin{array}{l}-2z+y-x-u=0\\\\ 3z+2t+1=0\\\\ -z+y+4x+u+4=0\\end{array}\\right\\}\\end{align*}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2() {\n  var htmlShow2 = document.getElementById(\"html-show2\");\n  if (htmlShow2.style.display === \"none\") {\n    htmlShow2.style.display = \"block\";\n  } else {\n    htmlShow2.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Posici\u00f3n relativa de dos variedades. Ejercicio 4 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/0nQfMpINLAM?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<p>&nbsp;<\/p>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\"><strong>Ejercicio:<\/strong> La ecuaci\u00f3n \\(x+z+t=0\\) es una de las ecuaciones impl\u00edcitas de la imagen de la aplicaci\u00f3n \\(f:\\mathbb{R}^3\\to\\mathcal{M}_2(\\mathbb{R})\\), dada por \\[f(a,b,c)=\\begin{bmatrix}-2c+b+a&amp; -c-b\\\\ b+c &amp; c-2b-a\\end{bmatrix}\\]<\/td>\n<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>Verdadero<\/li>\n<li>Falso<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv() {\n  var htmlShow = document.getElementById(\"html-show\");\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>A.)<\/strong><\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/uploads.jesussoto.es\/maxima\/EjrALGimagen01.html\" width=\"650\" height=\"300\" allow=\"fullscreen\"><\/iframe>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Vamos a tratar la posici\u00f3n relativa de dos variedades afines: \\(L_1=P+C_1\\) y \\(L_2=Q+C_2\\). Diremos que se cortan si el conjunto \\(L_1\\cap L_2\\) no es vac\u00edo. Si \\(L_1\\cap L_2=\\phi\\); es decir, si no&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-457","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/457","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=457"}],"version-history":[{"count":4,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/457\/revisions"}],"predecessor-version":[{"id":459,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/457\/revisions\/459"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=457"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=457"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=457"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}