{"id":352,"date":"2025-10-24T10:15:54","date_gmt":"2025-10-24T08:15:54","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=352"},"modified":"2025-10-17T20:08:29","modified_gmt":"2025-10-17T18:08:29","slug":"alg-espacios-vectoriales-y-aplicaciones-lineales-con-maxima","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=352","title":{"rendered":"ALG: Espacios vectoriales y aplicaciones lineales con maxima"},"content":{"rendered":"<h2>Complecio\u0301n de un conjunto l.i. a una base<\/h2>\n<p>A veces, necesitamos completar un conjunto l.i. de vectores para que formen una base de todo el espacio vectorial. Para conseguirlo podemos obrar de la siguiente forma. <\/p>\n<p>Consideremos \\(\\{\\textbf{v}_1,\\textbf{v}_2,\\ldots,\\textbf{v}_m\\}\\) un conjunto linealmente independiente de vectores de una espacio vectorial, \\(\\mathcal{V}\\), y \\(\\{\\textbf{e}_1,\\textbf{e}_2,\\ldots,\\textbf{e}_n\\}\\) una base de \\(\\mathcal{V}\\). Sea la \\(A\\) la matriz cuyas filas son los vectores \\(\\textbf{v}_1,\\textbf{v}_2,\\ldots\\), \\(\\textbf{v}_m\\), y \\(B\\) la matriz de los vectores de la base, \\(\\textbf{e}_1,\\textbf{e}_2,\\ldots\\), \\(\\textbf{e}_n\\). Entonces realizamos operaciones elementales de forma que<br \/>\n\\[\\left[\\frac{A}{B}\\right] \\sim \\left[\\frac{E}{B^\\prime}\\right]\\] donde \\(E\\) es una matriz escalonada y \\({B^\\prime}\\) una matriz escalonada siguiente a \\(E\\), entonces los \\(n-m\\) escalones no nulos de \\({B^\\prime}\\) nos determinan los vectores que completan la base del espacio vectorial.<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea \\(S=\\left\\{\\begin{bmatrix}a&#038;b\\\\c&#038;d\\end{bmatrix}\\in\\mathcal{M}_2(\\mathbb{R});a=b+c,\\ d=2c\\right\\}\\), completar su base para obtener una del espacio vectorial.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv2b1() {\n  var htmlShow2b1 = document.getElementById(\"html-show2b1\");\n  if (htmlShow2b1.style.display === \"none\") {\n    htmlShow2b1.style.display = \"block\";\n  } else {\n    htmlShow2b1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2b1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2b1\" style=\"display: none;\">\nNecesitamos una base de \\(s\\) y para ello suficiente con observar que si \\(\\begin{bmatrix}a&#038;b\\\\c&#038;d\\end{bmatrix}\\in S\\) es:\\[ \\begin{bmatrix}a&#038;b \\\\c&#038;d\\end{bmatrix}= \\begin{bmatrix}b+c&#038;b\\\\ c&#038;2c\\end{bmatrix}= b\\begin{bmatrix}1&#038;1\\\\ 0&#038;0\\end{bmatrix}+c\\begin{bmatrix}1&#038;0\\\\ 1&#038;2\\end{bmatrix}\\]<br \/>\nLuego \\[\\left\\{\\begin{bmatrix}1&#038;1\\\\0&#038;0\\end{bmatrix},\\begin{bmatrix}1&#038;0\\\\1&#038;2\\end{bmatrix}\\right\\}\\] es una base del subespacio. Si consideramos la base can\u00f3nica de \\(\\mathcal{M}_2(\\mathbb{R})\\) podemos formar la matriz<br \/>\n\\[\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 1&#038;0&#038;1&#038;2\\\\ 1&#038;0&#038;0&#038;0\\\\ 0&#038;1&#038;0&#038;0\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\overset{f_2-f_1}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 1&#038;0&#038;0&#038;0\\\\ 0&#038;1&#038;0&#038;0\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\overset{f_3-f_1}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;-1&#038;0&#038;0\\\\ 0&#038;1&#038;0&#038;0\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\]<br \/>\n\\[\\overset{f_3-f_2}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\ 0&#038;1&#038;0&#038;0\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\overset{f_4+f_2}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\ 0&#038;0&#038;1&#038;2\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\overset{f_4+f_3}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\ 0&#038;0&#038;0&#038;0\\\\ 0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix}\\]<br \/>\n\\[\\overset{f_4\\leftrightarrow f_5}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\  0&#038;0&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;0\\\\0&#038;0&#038;0&#038;1 \\end{bmatrix}\\overset{f_5\\leftrightarrow f_6}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\  0&#038;0&#038;1&#038;0\\\\0&#038;0&#038;0&#038;1\\\\ 0&#038;0&#038;0&#038;0 \\end{bmatrix}\\overset{f_4+ f_3}{\\sim}\\begin{bmatrix}1&#038;1&#038;0&#038;0\\\\ 0&#038;-1&#038;1&#038;2\\\\ 0&#038;0&#038;-1&#038;-2\\\\  0&#038;0&#038;0&#038;-2\\\\0&#038;0&#038;0&#038;1\\\\ 0&#038;0&#038;0&#038;0 \\end{bmatrix}\\]<br \/>\nPor comodidad podr\u00edamos permutar las filas 4 y 5 y tendr\u00edamos las cuatro primeras filas linealmente independientes. Luego \\[\\left\\{\\begin{bmatrix}1&#038;1\\\\0&#038;0\\end{bmatrix},\\begin{bmatrix}1&#038;0\\\\1&#038;2\\end{bmatrix},\\begin{bmatrix}0&#038;0\\\\-1&#038;-2\\end{bmatrix},\\begin{bmatrix}0&#038;0\\\\0&#038;1\\end{bmatrix}\\right\\}\\]\n<\/div>\n<hr\/>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea \\(S=\\left\\{a+bX+cX^2+dX^3+eX^4\\in\\mathbb{R}_4[X];b=a-d,\\ c=e-2d\\right\\}\\), completar su base para obtener una del espacio vectorial \\(\\mathbb{R}_4[X]\\).\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv2b13e() {\n  var htmlShow2b13e = document.getElementById(\"html-show2b13e\");\n  if (htmlShow2b13e.style.display === \"none\") {\n    htmlShow2b13e.style.display = \"block\";\n  } else {\n    htmlShow2b13e.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2b13e()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2b13e\" style=\"display: none;\">\nPrimero debemos sacar una base:<br \/>\n\\[S=\\left\\{a+bX+cX^2+dX^3+eX^4\\in\\mathbb{R}_4[X];b=a-d,\\ c=e-2d\\right\\}=\\] \\[=\\left\\{a+(a-d)X+(e-2d)X^2+dX^3+eX^4;a,d,e\\in\\mathbb{R}\\right\\}\\]<br \/>\nLuego \\[a+(a-d)X+(e-2d)X^2+dX^3+eX^4=a(1-X)+d(-X-2X^2+X^3)+e(X^2+X^4).\\] Es decir,<br \/>\n\\[S=\\mathbf{Gen}\\left\\{1-X,-X-2X^2+X^3,X^2+X^4\\right\\}\\]<br \/>\nPas\u00e9moslo a sus vectores de coordenadas y realicemos el proceso de completaci\u00f3n.<br \/>\n<!-- Code cell --><\/p>\n<table>\n<tr style=\"border: 0px;\">\n<td style=\"width: 70px;vertical-align: top;padding: 1mm;\"><span class=\"prompt\">(%i4)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">v<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">u<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">w<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">addrow<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">matrix<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">v<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">u<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">w<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_function\">ident<\/span><span class=\"code_operator\">(<\/span><span class=\"code_number\">5<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[{ }\\begin{bmatrix}1 &amp; -1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; -1 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\\\1 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 1\\end{bmatrix}\\]<\/p>\n<p><!-- Code cell --><\/p>\n<table>\n<tr style=\"border: 0px;\">\n<td style=\"width: 70px;vertical-align: top;padding: 1mm;\"><span class=\"prompt\">(%i5)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowop<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">4<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[{ }\\begin{bmatrix}1 &amp; -1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; -1 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\\\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; 1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 1\\end{bmatrix}\\]<\/p>\n<p><!-- Code cell --><\/p>\n<table>\n<tr style=\"border: 0px;\">\n<td style=\"width: 70px;vertical-align: top;padding: 1mm;\"><span class=\"prompt\">(%i7)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowop<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">4<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowop<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">4<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[{ }\\begin{bmatrix}1 &amp; -1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; -1 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\\\0 &amp; 0 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 1\\end{bmatrix}\\]<\/p>\n<p><!-- Code cell --><\/p>\n<table>\n<tr style=\"border: 0px;\">\n<td style=\"width: 70px;vertical-align: top;padding: 1mm;\"><span class=\"prompt\">(%i10) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowswap<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">6<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowswap<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">6<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">7<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowswap<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">7<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">8<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[{ }\\begin{bmatrix}1 &amp; -1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; -1 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\\\0 &amp; 0 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 1\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\end{bmatrix}\\]<\/p>\n<p><!-- Code cell --><\/p>\n<table>\n<tr style=\"border: 0px;\">\n<td style=\"width: 70px;vertical-align: top;padding: 1mm;\"><span class=\"prompt\">(%i12) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowop<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">4<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">rowop<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">A<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[{ }\\begin{bmatrix}1 &amp; -1 &amp; 0 &amp; 0 &amp; 0\\\\0 &amp; -1 &amp; -2 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 1 &amp; 0 &amp; 1\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 2\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; -1\\\\0 &amp; 0 &amp; 0 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 1\\\\0 &amp; 0 &amp; 0 &amp; 0 &amp; 0\\end{bmatrix}\\]<\/p>\n<p>Aqui podemos parar, pues los cinco primero vectores son l.i.; es decir, una base de \\(\\mathbb{R}^4\\). Por tanto, para completar la base solo necesitamos los dos vectores que continuan a los que tenemos: \\[\\left\\{1-X,-X-2X^2+X^3,X^2+X^4, X^3+2X^4, -X^4\\right\\}\\]\n<\/p><\/div>\n<hr\/>\n<p>La complementaci\u00f3n que acabamos de realizar se fundamenta en el siguiente teorema:<\/p>\n<blockquote><p><strong>Teorema:<\/strong> Sea \\(S\\) un subespacio vectorial de un espacio vectorial \\(V\\), finitamente generado(e.v.f.g). Si \\(B_S\\) es una base de \\(S\\), existen \\(\\{u_1,\\ldots,u_r\\}\\subset V-B_S\\), de modo que \\(B_S\\cap \\{u_1,\\ldots,u_r\\}\\) es una base de \\(V\\).\n<\/p><\/blockquote>\n<p>As\u00ed podemos dar la siguiente definici\u00f3n:<\/p>\n<blockquote><p><strong>Definici\u00f3n:<\/strong> Sea \\(S\\) un subespacio vectorial de un espacio vectorial \\(V\\), finitamente generado(e.v.f.g). Se llama subespacio vectorial <strong>suplementario<\/strong> de \\(S\\), con base \\(B_S\\), al subespacio generado por los vectores \\(\\{u_1,\\ldots,u_r\\}\\subset V-B_S\\), tales que \\(B_S\\cap \\{u_1,\\ldots,u_r\\}\\) es una base de \\(V\\).<\/p>\n<p>Observar que \\[S+\\mathbf{Gen}\\{u_1,\\ldots,u_r\\}=V\\]<\/p>\n<p>Se dice que la suma anterio es directa \\[S\\oplus\\mathbf{Gen}\\{u_1,\\ldots,u_r\\}=V,\\] cuando para cada \\(v\\in V\\) existen dos \u00fanicos \\(s\\in S\\) y \\(u\\in \\mathbf{Gen}\\{u_1,\\ldots,u_r\\}\\), tales que \\[s+u=v.\\]\n<\/p><\/blockquote>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea \\(S=\\left\\{\\begin{bmatrix}a&#038;2b\\\\a-b&#038;a+b\\end{bmatrix};a,b\\in\\mathbb{R}\\right\\}\\subset\\mathcal{M}_2(\\mathbb{R})\\), obtener una base para su suplementario.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b2w() {\n  var htmlShow1b2w = document.getElementById(\"html-show1b2w\");\n  if (htmlShow1b2w.style.display === \"none\") {\n    htmlShow1b2w.style.display = \"block\";\n  } else {\n    htmlShow1b2w.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b2w()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b2w\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Compleci\u00f3n de una base de un subespacio vectorial Ej.1 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/OK9mV29OrDA?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<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\"><strong>Ejercicio:<\/strong> Dada la aplicaci\u00f3n lineal \\(f:\\mathbb{R}_2[X]\\to\\mathbb{R}_2[X]\\) dada por \\[\\forall p(X)\\in\\mathbb{R}_2[X];\\ f(p(X))=p(X)-\\frac{d}{dX}p(X),\\] y \\(M_f\\) su matriz asociada respecto de la base \\(B=\\{2+X-X^2,2X-1,1+3X^2\\}\\). \u00bfCu\u00e1l es el valor de \\(\\mathbf{tr}(M_fM_f^t)\\) ?<\/p>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>58<\/li>\n<li>64<\/li>\n<li>70<\/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<strong>C)<\/strong>.<br \/>\nComo \\[\\forall p(X)\\in\\mathbb{R}_2[X];\\ f(p(X))=p(X)-\\frac{d}{dX}p(X),\\]<br \/>\naplicamos cada uno de los polinomios de \\(B=\\{p=2+X-X^2,q=2X-1,r=1+3X^2\\}\\) tendremos<br \/>\n\\[f(p)=-{{X}^{2}}+3 X+1\\operatorname{,}f(q)=2 X-3\\operatorname{,}f(r)=3 {{X}^{2}}-6 X+1 \\]<br \/>\nAhora solo tenemos que colocar las coordenadas en columnas:<br \/>\n\\[M_f\\begin{bmatrix}1 &#038; -3 &#038; 1\\\\<br \/>\n3 &#038; 2 &#038; -6\\\\<br \/>\n-1 &#038; 0 &#038; 3\\end{bmatrix}\\]<br \/>\nResolvemos el ejercicio<br \/>\n\\[\\begin{bmatrix}1 &#038; -3 &#038; 1\\\\<br \/>\n3 &#038; 2 &#038; -6\\\\<br \/>\n-1 &#038; 0 &#038; 3\\end{bmatrix}\\begin{bmatrix}1 &#038; 3 &#038; -1\\\\<br \/>\n-3 &#038; 2 &#038; 0\\\\<br \/>\n1 &#038; -6 &#038; 3\\end{bmatrix}=\\begin{bmatrix}11 &#038; -9 &#038; 2\\\\<br \/>\n-9 &#038; 49 &#038; -21\\\\<br \/>\n2 &#038; -21 &#038; 10\\end{bmatrix}\\]<br \/>\nAs\u00ed la traza es 70.\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Complecio\u0301n de un conjunto l.i. a una base A veces, necesitamos completar un conjunto l.i. de vectores para que formen una base de todo el espacio vectorial. Para conseguirlo podemos obrar de&#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":[7],"class_list":["post-352","post","type-post","status-publish","format-standard","hentry","category-algebra","tag-practicas-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/352","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=352"}],"version-history":[{"count":5,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/352\/revisions"}],"predecessor-version":[{"id":354,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/352\/revisions\/354"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=352"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=352"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=352"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}