{"id":318,"date":"2026-10-21T08:15:49","date_gmt":"2026-10-21T06:15:49","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=318"},"modified":"2026-10-21T17:28:50","modified_gmt":"2026-10-21T15:28:50","slug":"alg-aplicaciones-lineales","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=318","title":{"rendered":"ALG: Aplicaciones lineales"},"content":{"rendered":"<p>Al hablar de grupos se introdujo la definici\u00f3n de <a href=\"http:\/\/es.wikipedia.org\/wiki\/Homomorfismo\" target=\"_blank\" rel=\"noopener noreferrer\">homomorfismo<\/a> y con ella la de <a href=\"http:\/\/en.wikipedia.org\/wiki\/Kernel_(algebra)\" target=\"_blank\" rel=\"noopener noreferrer\">n\u00facleo<\/a>. Ahora extendemos esta definici\u00f3n a espacios vectoriales para definir la <a href=\"http:\/\/es.wikipedia.org\/wiki\/Aplicaci%C3%B3n_lineal\" target=\"_blank\" rel=\"noopener noreferrer\">aplicaci\u00f3n lineal<\/a>: un homomorfismo entre espacios vectoriales. As\u00ed diremos que una aplicaci\u00f3n (en algunos libros le dicen Transformaci\u00f3n) entre dos espacios vectoriales, \\(f:V\\to W\\), sobre el mismo cuerpo\\(\\mathbb{K}\\), es lineal si se cumple que para todo par de vectores \\(\\vec{v},\\vec{u}\\in V\\) y todo par de escalares \\(\\lambda,\\mu\\in\\mathbb{K}\\) se verifica que: \\[f(\\lambda\\vec{v}+\\mu\\vec{u})=\\lambda f(\\vec{v})+\\mu f(\\vec{u}).\\]<\/p>\n<p>Para este tema pod\u00e9is consultar el cap\u00edtulo 6 del libro \u00c1LGEBRA LINEAL Definiciones, Teoremas y Resultados, Juan De Burgos Rom\u00e1n, ingebook<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n \\(tr:\\mathcal{M}_2(\\mathbb{R})\\to\\mathbb{R}\\) dada por \\[tr\\begin{bmatrix}a &#038; b \\\\<br \/>\nc &#038; d\\end{bmatrix}=a+d,\\] \u00bfes lineal?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a30() {\n  var htmlShow1a30 = document.getElementById(\"html-show1a30\");\n  if (htmlShow1a30.style.display === \"none\") {\n    htmlShow1a30.style.display = \"block\";\n  } else {\n    htmlShow1a30.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a30()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a30\" style=\"display: none;\">\n\\(\\forall \\begin{bmatrix}a_{11} &#038; a_{12} \\\\ a_{21} &#038; a_{22} \\end{bmatrix},\\begin{bmatrix}b_{11} &#038; b_{12} \\\\ b_{21} &#038; b_{22} \\end{bmatrix}\\in \\mathcal{M}_2(\\mathbb{R})\\) y \\(\\forall \\lambda,\\mu\\in\\mathbb{R}\\)<br \/>\n\\[tr\\left(\\lambda\\begin{bmatrix}a_{11} &#038; a_{12} \\\\ a_{21} &#038; a_{22} \\end{bmatrix}+\\mu\\begin{bmatrix}b_{11} &#038; b_{12} \\\\ b_{21} &#038; b_{22} \\end{bmatrix}\\right)=tr\\begin{bmatrix}\\lambda a_{11}+\\mu b_{11} &#038; \\lambda a_{12}+\\mu b_{12} \\\\ \\lambda a_{21}+\\mu b_{21} &#038; \\lambda a_{22}+\\mu b_{22} \\end{bmatrix}=\\]<\/p>\n<p>\\[\\begin{multline*}=\\lambda a_{11}+\\mu b_{11} + \\lambda a_{22}+\\mu b_{22}=\\lambda (a_{11}+a_{22}) + \\mu (b_{11}+b_{22})=\\\\ \\lambda\\ tr\\begin{bmatrix}a_{11} &#038; a_{12} \\\\ a_{21} &#038; a_{22} \\end{bmatrix}+\\mu \\ tr\\begin{bmatrix}b_{11} &#038; b_{12} \\\\ b_{21} &#038; b_{22} \\end{bmatrix}\\end{multline*}\\]\n<\/p><\/div>\n<hr>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n \\(f:\\mathbb{R}^3\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[f(x,y,z)=\\begin{bmatrix}x &#038; y \\\\ z &#038; x\\end{bmatrix}\\] \u00bfes lineal?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a30b() {\n  var htmlShow1a30b = document.getElementById(\"html-show1a30b\");\n  if (htmlShow1a30b.style.display === \"none\") {\n    htmlShow1a30b.style.display = \"block\";\n  } else {\n    htmlShow1a30b.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a30b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a30b\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Aplicaciones EJ.2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/I2hwyMxdCRY?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><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n \\(f:\\mathbb{R}_2[X]\\to\\mathbb{R}_2[X]\\) dada por \\[\\forall p\\in\\mathbb{R}_2[X];\\ f(p(X))=1-p(X)-\\frac{d}{dX}p(X)\\] \u00bfes lineal?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a30c() {\n  var htmlShow1a30c = document.getElementById(\"html-show1a30c\");\n  if (htmlShow1a30c.style.display === \"none\") {\n    htmlShow1a30c.style.display = \"block\";\n  } else {\n    htmlShow1a30c.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a30c()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a30c\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Aplicaci\u00f3n lineal Ej.3  - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/3Tee-xqNT6Q?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<h2>Matriz asociada a un aplicaci\u00f3n lineal<\/h2>\n<blockquote><p>\nDada una aplicaci\u00f3n lineal, \\(f:V\\to W\\), entre dos espacios vectoriales finitamente generados sobre el mismo cuerpo \\(\\mathbb{K}\\), con \\(\\mathbf{dim}(V)=n\\) y \\(\\mathbf{dim}(W)=m\\) entonces existe una matriz \\(M\\in\\mathcal{M}_{m\\times n}(\\mathbb{K})\\) tal que \\[\\forall \\mathbf{v}\\in V;\\ f(\\mathbf{v})\\sim M\\mathbf{v}\\]\n<\/p><\/blockquote>\n<p>Esta equivalencia se puede ver en el siguiente diagrama:<br \/>\n\\[\\begin{matrix}<br \/>\nf:V &#038;\\rightarrow   &#038; W \\\\<br \/>\n \\downarrow &#038;  &#038; \\downarrow \\\\<br \/>\nM:\\mathbb{K}^n &#038; \\rightarrow  &#038; \\mathbb{K}^m \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<p>Dada una aplicaci\u00f3n lineal, \\(f:V\\to W\\), entre dos espacios vectoriales definimos la matriz asociada de la aplicaci\u00f3n respecto de una base \\(B_V\\subseteq V\\) como la matriz cuyas columnas son las coordendas respecto de otra base \\(B_W\\subseteq W\\) de las im\u00e1genes de los vectores de \\(B_V\\); es decir, si \\(B_V=\\{\\vec{v}_1,\\ldots,\\vec{v}_n\\}\\), \\(B_W=\\{\\vec{w}_1,\\ldots,\\vec{w}_m\\}\\), y<br \/>\n\\[<br \/>\n\\begin{matrix}<br \/>\nf(\\vec{v}_1)=k_{11}\\vec{w}_1+k_{21}\\vec{w}_2+k_{31}\\vec{w}_3+\\ldots+k_{m1}\\vec{w}_m;\\\\<br \/>\nf(\\vec{v}_2)=k_{12}\\vec{w}_1+k_{22}\\vec{w}_2+k_{32}\\vec{w}_3+\\ldots+k_{m2}\\vec{w}_m;\\\\<br \/>\nf(\\vec{v}_3)=k_{13}\\vec{w}_1+k_{23}\\vec{w}_2+k_{33}\\vec{w}_3+\\ldots+k_{m3}\\vec{w}_m;\\\\<br \/>\n\\vdots \\quad \\vdots \\quad \\vdots\\\\<br \/>\nf(\\vec{v}_n)=k_{1n}\\vec{w}_1+k_{2n}\\vec{w}_2+k_{3n}\\vec{w}_n+\\ldots+k_{mn}\\vec{w}_m;<br \/>\n\\end{matrix}<br \/>\n\\]<br \/>\nllamamos matriz asociada de \\(f\\), a la matriz<br \/>\n\\[<br \/>\nM_f=\\begin{bmatrix}<br \/>\nk_{11} &amp; k_{12} &amp; k_{13} &amp;\\ldots &amp; k_{1n}\\\\<br \/>\nk_{21} &amp; k_{22} &amp; k_{23} &amp;\\ldots &amp; k_{2n}\\\\<br \/>\nk_{31} &amp; k_{32} &amp; k_{33} &amp;\\ldots &amp; k_{3n}\\\\<br \/>\n\\vdots &amp; \\vdots &amp; \\vdots &amp; \\ldots &amp; \\vdots \\\\<br \/>\nk_{m1} &amp; k_{n2} &amp; k_{n3} &amp;\\ldots &amp; k_{mn}\\\\<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}^4\\to\\mathbb{R}^4\\) dada por \\[f(x,y,z,t)=(x-y+t,y+3x-z,t-2x+y,t+z-y).\\] \u00bfCu\u00e1l es la traza de la matriz asociada?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a3() {\n  var htmlShow1a3 = document.getElementById(\"html-show1a3\");\n  if (htmlShow1a3.style.display === \"none\") {\n    htmlShow1a3.style.display = \"block\";\n  } else {\n    htmlShow1a3.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a3()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a3\" style=\"display: none;\">\nRecordemos de no indicar bases se considerar\u00e1 las bases can\u00f3nicas en ambos espacios vectoriales. Luego \\[\\begin{align*}<br \/>\n f(1,0,0,0)&#038;=(1,3,-2,0), &#038; f(0,1,0,0)&#038;=(-1,1,1,-1), \\\\<br \/>\n f(0,0,1,0)&#038;=(0,-1,0,1), &#038; f(0,0,0,1)&#038;=(1,0,1,1).<br \/>\n\\end{align*}\\]<br \/>\nPor tanto, \\[M_f=\\begin{bmatrix}1 &#038; -1 &#038; 0 &#038; 1\\\\<br \/>\n3 &#038; 1 &#038; -1 &#038; 0\\\\<br \/>\n-2 &#038; 1 &#038; 0 &#038; 1\\\\<br \/>\n0 &#038; -1 &#038; 1 &#038; 1\\end{bmatrix}\\]\n<\/div>\n<hr>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}^3\\to\\mathbb{R}^2\\) dada por \\[f(x,y,z)=(x-y,y+3x-z).\\] Sea la matriz asociada \\(M_f\\) respecto de la base \\(B=\\{(1,1,0),(0,-1,1),(3,1,1)\\}\\). \u00bfCu\u00e1l es la traza de \\(M_f.M_f^t\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a2() {\n  var htmlShow1a2 = document.getElementById(\"html-show1a2\");\n  if (htmlShow1a2.style.display === \"none\") {\n    htmlShow1a2.style.display = \"block\";\n  } else {\n    htmlShow1a2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a2\" style=\"display: none;\">\nRecordemos que planteamos el problema considerando la base \\(B=\\{(1,1,0),(0,-1,1),(3,1,1)\\}\\) como base del espacio vectorial de partida.<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\">(%i2) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">=<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">+<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">\u00b7<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">;<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">B<\/span><span class=\"code_operator\">:<\/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_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_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_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">3<\/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_operator\">]<\/span><span class=\"code_endofline\">$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\operatorname{ }\\operatorname{f}(v)\\operatorname{:=}\\left[ {v_1}-{v_2}\\operatorname{,}{v_2}+3 {v_1}-{v_3}\\right] \\]<\/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\">(%i3) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">M<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">matrix<\/span><span class=\"code_operator\">(<\/span><span class=\"code_endofline\"><br \/><\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">B<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_endofline\"><br \/><\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">B<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_endofline\"><br \/><\/span> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">B<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\"><br \/><\/span> \u00a0\u00a0 <span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\operatorname{ }\\begin{bmatrix}0 &amp; 1 &amp; 2\\\\4 &amp; -2 &amp; 9\\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\">(%i4) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">M<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">M<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\operatorname{ }106\\]<\/p>\n<\/div>\n<hr>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea 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).\\] \u00bfCu\u00e1l es la traza de la matriz asociada \\(M_f\\) respecto de la base \\(B=\\{1,1-X,1-X^2\\}\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a1() {\n  var htmlShow1a1 = document.getElementById(\"html-show1a1\");\n  if (htmlShow1a1.style.display === \"none\") {\n    htmlShow1a1.style.display = \"block\";\n  } else {\n    htmlShow1a1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a1\" style=\"display: none;\">\nRecordemos que planteamos el problema considerando la base \\(B=\\{1,1-X,1-X^2\\}\\) como base del espacio vectorial partida, y sus im\u00e1genes las consideramos respecto de la base can\u00f3nica.<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Matriz asociada de una aplicaci\u00f3n lineal: EJ.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/m3OvKN0xkHM?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>Acabamos de ver como una matriz puede representar una aplicaci\u00f3n lineal. De hecho podemos establecer una aplicaci\u00f3n entre el conjunto de aplicaciones lineales entre dos \\(\\mathbb{K}\\)-espacios vectoriales \\(V\\) y \\(W\\), de dimensiones \\(n\\) y \\(m\\) (respectivamente) y el espacio vectorial de las matrices \\(M_{m\\times n}(\\mathbb{K})\\) que sea un isomorfismo de espacios vectoriales; es decir, una aplicaci\u00f3n lineal biyectiva. Esto nos equipara las operaciones con aplicaciones a las operaciones con sus matrices asociadas.<\/p>\n<p>Sabemos que si \\(M_f\\) es la matriz asociada a la aplicaci\u00f3n lineal \\(f:V\\to W\\), entonces<br \/>\n\\[f(v_1,v_2,\\ldots,v_n)=(w_1,w_2,\\ldots,w_m)\\Leftrightarrow M_f \\begin{pmatrix}v_1\\\\v_2\\\\ \\vdots\\\\v_n\\end{pmatrix}=\\begin{pmatrix}w_1\\\\w_2\\\\ \\vdots\\\\w_n\\end{pmatrix}.\\]<\/p>\n<p>Esto nos permite deducir propiedades de la aplicaci\u00f3n con sus correspondientes en la matriz. Por ejemplo, una aplicaci\u00f3n lineal entre dos espacios vectoriales de la misma dimensi\u00f3n es un isomorfismo si, y solo si, su matriz asociada es regular.<\/p>\n<p>As\u00ed, podemos considerar la matriz asociada a una aplicaci\u00f3n lineal, \\(f:V\\to W\\), entre dos espacios vectoriales respecto de una base \\(B_V\\subseteq V\\) como la matriz cuyas columnas contiene las coordenadas respecto de otra base \\(B_W\\subseteq W\\) de las im\u00e1genes de los vectores de \\(B_V\\). <\/p>\n<h3>Ejemplo<\/h3>\n<p><iframe loading=\"lazy\" src=\"https:\/\/uploads.jesussoto.es\/maxima\/EjrALGmat_apl05.html\" width=\"650\" height=\"680\" allow=\"fullscreen\"><\/iframe><\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea \\(S=\\left\\{\\begin{bmatrix}a&#038;a-b\\\\b-a&#038;a+b\\end{bmatrix};a,b\\in\\mathbb{R}\\right\\}\\), determinar una base del subespacio vectorial.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a() {\n  var htmlShow1a = document.getElementById(\"html-show1a\");\n  if (htmlShow1a.style.display === \"none\") {\n    htmlShow1a.style.display = \"block\";\n  } else {\n    htmlShow1a.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a\" style=\"display: none;\">\nConsideremos la aplicaci\u00f3n \\(f:\\mathbb{R}^2\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[f(a,b)=\\begin{bmatrix}a&#038;a-b\\\\b-a&#038;a+b\\end{bmatrix}\\]<\/p>\n<p>Esta aplicaci\u00f3n es lineal. Por tanto, se cumple que transforma bases de \\(f:\\mathbb{R}^2\\) en sistema generadores de las im\u00e1genes. Luego, dado la base \\(\\{(1,0),(0,1)\\}\\), vemos que <\/p>\n<p>\\[f(1,0)=\\begin{bmatrix}1&#038;1\\\\-1&#038;1\\end{bmatrix},\\mbox{ y, } f(0,1)=\\begin{bmatrix}0&#038;-1\\\\1&#038;1\\end{bmatrix}.\\]<br \/>\nPor tanto, \\(\\left\\{\\begin{bmatrix}1&#038;1\\\\-1&#038;1\\end{bmatrix},\\begin{bmatrix}0&#038;-1\\\\1&#038;1\\end{bmatrix}\\right\\}\\) es un sistema generador que contiene una base. Solo necesitamos probar cuales son l.i.\n<\/p><\/div>\n<hr>\n<p>&nbsp;<\/p>\n<h2>Matriz del cambio de base<\/h2>\n<p>\u00bfY si cambiamos las bases? Es decir, si tengo nuevas bases \\(B&#8217;_V\\) y \\(B&#8217;_W\\), y deseo encontrar la relaci\u00f3n entre la matriz asociada aplicaci\u00f3n \\(M_{f_{B_VB_W}}\\), y la matriz \\(M_{f_{B&#8217;_VB&#8217;_W}}\\). Veamos primero que ocurre cuando dentro de un mismo espacio vectorial cambiamos la base.<\/p>\n<p>Recordemos que todo espacio vectorial finitamente generado tiene una base. Sea \\(V\\) nuestro \\(\\mathbb{K}\\)-espacio vectorial y \\(\\mathcal{B}=\\{\\vec{v}_1,\\ldots, \\vec{v}_n\\}\\), una base del mismo. Para cualquier vector de \\(V\\), \\(\\vec{v}\\in V\\), existir\u00e1n unos \u00fanicos escalares \\(k_i\\in \\mathbb{K}\\), tales que \\[\\vec{v}=k_1\\vec{v}_1+k_2\\vec{v}_2+\\cdots+k_n\\vec{v}_n.\\] Pues bien, a esos escalares los denominamos coordenadas de \\(\\vec{v}\\) respecto de la base \\(\\mathcal{B}\\). As\u00ed, representado mediante sus coordenadas, expresamos que \\[\\vec{v}=\\begin{bmatrix}k_1\\\\ k_2\\\\ \\vdots \\\\ k_n\\end{bmatrix}_\\mathcal{B}\\]<br \/>\nQu\u00e9 ocurre si tenemos otra base \\(\\mathcal{B}&#8217;\\), entonces las coordenadas de \\(\\vec{v}\\) ser\u00e1n otras, pero habr\u00e1 una relaci\u00f3n entre ambas. Vamos a utilizar las matrices para encontrar la relaci\u00f3n entre ambas coordenadas.<\/p>\n<p>Cuando tenemos dos bases podemos calcular c\u00f3mo pasar de las coordenadas de una base a la otra. Para ello utilizamos la matriz del cambio de base.<br \/>\n<img decoding=\"async\" fetchpriority=\"high\" class=\"aligncenter\" src=\"http:\/\/uploads.jesussoto.es\/matriz_cambiobase.png\" alt=\"\" width=\"525\" height=\"383\" \/><\/p>\n<p>Veamos c\u00f3mo podemos calcular esta matriz del cambio de base. S\u00f3lo tenemos que darnos cuenta como representamos los vectores respecto de cada base. Pongamos dos bases \\(B=\\{\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_n\\}\\) y \\(B&#8217;=\\{\\vec{u}_1,\\vec{u}_2,\\ldots,\\vec{u}_n\\}\\). Que un vector cualquiera \\(\\vec{w}\\) tenga por coordenadas \\((c_1,c_2,\\ldots,c_n)_{B}\\) respecto de la base \\(B\\) significa que \\[\\vec{w}=c_1\\vec{v}_1+c_2\\vec{v}_2+\\ldots+c_n\\vec{v}_n\\]<\/p>\n<p>Si cada \\(\\vec{v}_i\\) tiene por coordenadas respecto de una base can\u00f3nica \\((v_{1i},v_{2i},\\ldots,v_{ni},)\\), podemos escribir lo anterior en forma matricial:<\/p>\n<p>\\[\\vec{w}=\\begin{bmatrix} v_{11} &amp;v_{12}&amp;v_{13}&amp;\\cdots &amp;v_{1n}\\\\ v_{21} &amp;v_{22}&amp;v_{23}&amp;\\cdots &amp;v_{2n}\\\\ \\vdots &amp; \\vdots &amp; \\vdots &amp;\\cdots &amp; \\vdots\\\\<br \/>\nv_{n1} &amp;v_{12}&amp;v_{13}&amp;\\cdots &amp;v_{1n}\\\\ \\end{bmatrix}\\begin{bmatrix}c_1\\\\c_2\\\\c_3\\\\ \\vdots \\\\c_n \\end{bmatrix}_{B}\\]<\/p>\n<p>Del mismo modo que \\(\\vec{w}\\) tenga por coordenadas \\((c&#8217;_1,c&#8217;_2,\\ldots,c&#8217;_n)_{B&#8217;}\\) respecto de la base \\(B&#8217;\\) significa que<\/p>\n<p>\\[\\vec{w}=c&#8217;_1\\vec{u}_1+c&#8217;_2\\vec{u}_2+\\ldots+c&#8217;_n\\vec{u}_n\\]<br \/>\nEscrito en forma matricial<br \/>\n\\[\\vec{w}=\\begin{bmatrix}<br \/>\nu_{11} &amp;u_{12}&amp;u_{13}&amp;\\cdots &amp;u_{1n}\\\\<br \/>\nu_{21} &amp;u_{22}&amp;u_{23}&amp;\\cdots &amp;u_{2n}\\\\<br \/>\n\\vdots &amp; \\vdots &amp; \\vdots &amp;\\cdots &amp; \\vdots\\\\<br \/>\nu_{n1} &amp;u_{12}&amp;u_{13}&amp;\\cdots &amp;u_{1n}\\\\<br \/>\n\\end{bmatrix}\\begin{bmatrix}c&#8217;_1\\\\c&#8217;_2\\\\c&#8217;_3\\\\ \\vdots \\\\c&#8217;_n \\end{bmatrix}_{B&#8217;}\\]<\/p>\n<p>La igualdad de ambos productos nos ofrece la posibilidad de conocer las coordenadas de un vector una base respecto de la otra:<\/p>\n<p>\\[\\begin{bmatrix}<br \/>\nv_{11} &amp;v_{12}&amp;v_{13}&amp;\\cdots &amp;v_{1n}\\\\<br \/>\nv_{21} &amp;v_{22}&amp;v_{23}&amp;\\cdots &amp;v_{2n}\\\\<br \/>\n\\vdots &amp; \\vdots &amp; \\vdots &amp;\\cdots &amp; \\vdots\\\\<br \/>\nv_{n1} &amp;v_{12}&amp;v_{13}&amp;\\cdots &amp;v_{1n}\\\\<br \/>\n\\end{bmatrix}\\begin{bmatrix}c_1\\\\c_2\\\\ \\vdots \\\\c_n \\end{bmatrix}_{B}=\\begin{bmatrix}<br \/>\nu_{11} &amp;u_{12}&amp;u_{13}&amp;\\cdots &amp;u_{1n}\\\\<br \/>\nu_{21} &amp;u_{22}&amp;u_{23}&amp;\\cdots &amp;u_{2n}\\\\<br \/>\n\\vdots &amp; \\vdots &amp; \\vdots &amp;\\cdots &amp; \\vdots\\\\<br \/>\nu_{n1} &amp;u_{12}&amp;u_{13}&amp;\\cdots &amp;u_{1n}\\\\<br \/>\n\\end{bmatrix}\\begin{bmatrix}c&#8217;_1\\\\c&#8217;_2\\\\ \\vdots \\\\c&#8217;_n \\end{bmatrix}_{B&#8217;}\\]<br \/>\nSi el sistema matricial lo escribimos as\u00ed: \\(P\\, C_{B}=Q\\,C&#8217;_{B&#8217;}\\), tendremos<br \/>\n\\[C_{B}=(P^{-1}Q)\\,C&#8217;_{B&#8217;},\\] o \\[(Q^{-1}P)\\,C_{B}=C&#8217;_{B&#8217;}.\\]<\/p>\n<p>A la matriz \\(Q^{-1}P\\), la llamamos matriz del cambio de bases de \\(B\\) a \\(B&#8217;\\), y la notamos como \\[C_{BB&#8217;}=Q^{-1}P.\\]<br \/>\nComo se observa \\(C_{B&#8217;B}\\) es la inversa de \\(C_{BB&#8217;}\\).<\/p>\n<h3>Ejemplo<\/h3>\n<p>D\u00e9mosle la vuelta a lo dicho antes y supongamos que ahora deseamos expresar el vector \\((-3,1,2)\\), dado en la base can\u00f3nica de \\(\\mathbb{R}^3\\), en sus coordenadas respecto de la base \\(B=\\{(1,1,1),(1,1,0),(1,0,-1)\\}\\). La matriz que expresa cualquier coordenada dada respecto de \\(B\\) en la base can\u00f3nica es:<br \/>\n\\[M_B=\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\ 1 &#038; 1 &#038; 0\\\\ 1 &#038; 0 &#038; -1\\end{bmatrix}.\\] Luego buscamos \\((a,b,c)\\) tal que<br \/>\n\\[\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\ 1 &#038; 1 &#038; 0\\\\ 1 &#038; 0 &#038; -1\\end{bmatrix}\\begin{bmatrix}a\\\\ b\\\\ c\\end{bmatrix}=\\begin{bmatrix}-3\\\\ 1\\\\ 2\\end{bmatrix}.\\]<br \/>\nNos basta con<br \/>\n\\[\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\ 1 &#038; 1 &#038; 0\\\\ 1 &#038; 0 &#038; -1\\end{bmatrix}^{-1}\\begin{bmatrix}-3\\\\ 1\\\\ 2\\end{bmatrix}=\\begin{bmatrix}-2\\\\ 3\\\\ -4\\end{bmatrix};\\]<br \/>\nEs decir, \\[-2(1,1,1)+3(1,1,0)-4(1,0,-1)=(-3,1,2).\\]<\/p>\n<p>Supongamos que \\((2,-1,4)\\), est\u00e1 dado respecto de la base \\(B^\\prime=\\{(1,-1,2),(0,2,1),(-1,0,1)\\}\\), entonces las coordenadas de \\((2,-1,4)\\) respecto de la base \\(B\\) ser\u00e1n<br \/>\n\\[\\begin{bmatrix}1 &#038; 0 &#038; -1\\\\ -1 &#038; 2 &#038; 0\\\\ 2 &#038; 1 &#038; 1\\end{bmatrix}\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\ 1 &#038; 1 &#038; 0\\\\ 1 &#038; 0 &#038; -1\\end{bmatrix}^{-1}\\begin{bmatrix}2\\\\ -1\\\\ 4\\end{bmatrix}=\\begin{bmatrix}9\\\\ -13\\\\ 2\\end{bmatrix};\\]<\/p>\n<hr \/>\n<p>Para calcular la matriz del cambio de bases podemos utilizar un resultado que nos dice: si a la matriz ampliada \\([Q|P]\\) le hacemos transformaciones elementales por fila, de modo que obtengamos<br \/>\n\\[[Q|P]\\to [I_n|C],\\]<br \/>\nentonces la matriz \\(C=C_{BB&#8217;}\\).<\/p>\n<h3>Matriz del cambio de base para aplicaciones<\/h3>\n<\/p>\n<p>Ahora veamos c\u00f3mo afecta el cambio de base  entre la matriz asociada aplicaci\u00f3n \\(M_{f_{B_VB_W}}\\), y la matriz \\(M_{f_{B&#8217;_VB&#8217;_W}}\\). Esta relaci\u00f3n nos la ofrece el siguiente gr\u00e1fico:<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"size-full wp-image-164 aligncenter\" src=\"http:\/\/uploads.jesussoto.es\/2014\/12\/cambio_base_apli.png\" alt=\"cambio_base_apli\" width=\"247\" height=\"166\" \/><\/p>\n<p>En este diagrama \\(A=M_{f_{B_VB_W}}\\) y \\(C=M_{f_{B&#8217;_VB&#8217;_W}}\\) es la matriz que desconocemos y buscamos. \\(P=M_{B&#8217;_VB_V}\\) es la matriz del cambio de base de \\(B&#8217;_V\\) a \\(B_V\\) y \\(Q=M_{B&#8217;_WB_W}\\). As\u00ed la matriz que buscamos es \\[C=Q^{-1}\\,A\\,P.\\]<\/p>\n<h3>Ejemplo<\/h3>\n<p>En el pasado ejercicio consideramos 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).\\] Ahora buscamos su matriz asociada \\(M_f\\) respecto de la base \\(B=\\{1,1-X,1+X-X^2\\}\\). Si consideramos la misma aplicaci\u00f3n sobre las bases can\u00f3nicas, en este caso la misma al ser el mismo espacio vectorial, tendremos:<br \/>\n\\[M_{fc}=\\begin{bmatrix} f(1)\\\\ f(X)\\\\ f(X^2)\\end{bmatrix}^t=\\begin{bmatrix}1 &#038; -1 &#038; 0\\\\0 &#038; 1 &#038; -2\\\\ 0 &#038; 0 &#038; 1\\end{bmatrix}\\]<br \/>\nDel mismo modo,<br \/>\n\\[M_{B}=\\begin{bmatrix} 1\\\\ 1-X\\\\ 1+X-X^2\\end{bmatrix}^t=\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\<br \/>\n0 &#038; -1 &#038; 1\\\\ 0 &#038; 0 &#038; -1\\end{bmatrix}\\]<\/p>\n<p>Luego la matriz buscada es: \\[M_{f_B}=M_{fc}M_B=\\begin{bmatrix}1 &#038; 2 &#038; 0\\\\ 0 &#038; -1 &#038; 3\\\\ 0 &#038; 0 &#038; -1\\end{bmatrix}.\\]<\/p>\n<p>Esto significa que si \\((3,-2,5)\\) son las coordenadas de un polinomio respecto de la base \\(B=\\{1,1-X,1+X-X^2\\}\\), cuando calculamos su imagen \\(f(3,-2,5)=-5X^2+17X-1\\) respecto de la base can\u00f3nica, ya que \\[\\begin{bmatrix}1 &#038; 2 &#038; 0\\\\ 0 &#038; -1 &#038; 3\\\\ 0 &#038; 0 &#038; -1\\end{bmatrix}\\begin{bmatrix}3\\\\ -2\\\\ 5\\end{bmatrix}=\\begin{bmatrix}-1\\\\ 17\\\\ -5\\end{bmatrix}.\\]<\/p>\n<p>Consideremos que queremos expresar la imagen respecto de una base \\(B^\\prime=\\{1+2X,X-X^2,1+X^2\\}\\). Calculamos \\[M_{B^\\prime}=\\begin{bmatrix} 1+2X\\\\ X-X^2\\\\ 1+X^2\\end{bmatrix}^t=\\begin{bmatrix}1 &#038; 0 &#038; 1\\\\ 2 &#038; 1 &#038; 0\\\\ 0 &#038; -1 &#038; 1\\end{bmatrix}\\]<\/p>\n<p>Ahora la imagen respecto de esta base \\(B^\\prime\\) ser\u00e1 \\[\\begin{bmatrix}1 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; 1 &#038; 0\\\\ 0 &#038; -1 &#038; 1\\end{bmatrix}^{-1}\\begin{bmatrix}1 &#038; 2 &#038; 0\\\\ 0 &#038; -1 &#038; 3\\\\ 0 &#038; 0 &#038; -1\\end{bmatrix}\\begin{bmatrix}3\\\\ -2\\\\ 5\\end{bmatrix}=\\begin{bmatrix}13\\\\ -9\\\\ -14\\end{bmatrix}\\]<\/p>\n<p>Supongamos que queremos saber qu\u00e9 polinomio, con coordenadas \\((a,b,c)\\) en la base \\(B\\), tiene por imagen \\(f(a,b,c)=1\\cdot(1+2X)+(-2)\\cdot (X-X^2)+(-1)\\cdot(1+X^2)\\); es decir, tiene por coordenadas \\((1,-2,-1)\\) en la base \\(B^\\prime\\). Entonces \\[\\begin{bmatrix}1 &#038; 1 &#038; 1\\\\<br \/>\n0 &#038; -1 &#038; 1\\\\ 0 &#038; 0 &#038; -1\\end{bmatrix}^{-1} \\begin{bmatrix}1 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; 1 &#038; 0\\\\ 0 &#038; -1 &#038; 1\\end{bmatrix} \\begin{bmatrix}1\\\\ -2\\\\ -1\\end{bmatrix}=\\begin{bmatrix}5 &#038; -4 &#038; 7\\\\ -2 &#038; 2 &#038; -3\\\\ 0 &#038; 1 &#038; -1\\end{bmatrix}\\begin{bmatrix}1\\\\ -2\\\\ -1\\end{bmatrix}=\\begin{bmatrix}6\\\\ -3\\\\ -1\\end{bmatrix}\\]<\/p>\n<hr>\n<p>&nbsp;<\/p>\n<p>Como habitualmente tratamos los espacios vectoriales \\(\\mathbb{R}^n\\) (recordad que todo espacio vectorial finitamente generado, de dimensi\u00f3n \\(n\\), es isomorfo a \\(\\mathbb{R}^n\\)), este gr\u00e1fico se representar\u00eda como<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" class=\"size-medium wp-image-166 aligncenter\" src=\"http:\/\/uploads.jesussoto.es\/2014\/12\/matriz_aplic_base-300x199.png\" alt=\"matriz_aplic_base\" width=\"300\" height=\"199\" \/><\/p>\n<p>donde \\(E_n\\) y \\(E_m\\) son las bases can\u00f3nicas respectivas.<\/p>\n<h2>N\u00facleo e imagen de una aplicaci\u00f3n lineal<\/h2>\n<p>Recordemos que, dada una aplicaci\u00f3n lineal, \\(T\\), se define el <b>n\u00facleo<\/b> (ker) y la <b>imagen<\/b> (Im) de \\(T:V\\to W\\) como:<\/p>\n<blockquote>\n<dl>\n<dd>\\(\\mathbf{ker}(T)=\\{\\,v\\in V:T(v)=0_W\\,\\}\\)<\/dd>\n<dd>\\(\\mathbf{Im}(T)=\\{\\,w\\in W: \\exists v\\in V:T(v)=w\\,\\}\\)<\/dd>\n<\/dl>\n<\/blockquote>\n<p>Es decir, que el n\u00facleo de una aplicaci\u00f3n lineal est\u00e1 formado por el conjunto de todos los vectores del dominio que tienen por imagen al vector nulo del codominio.<\/p>\n<p>El n\u00facleo de toda aplicaci\u00f3n lineal es un subespacio vectorial del dominio.<\/p>\n<p>As\u00ed, si \\(M_f\\) es la matriz asociada a la aplicaci\u00f3n lineal \\(f:V\\to W\\), entonces el sistema dado por \\[M_f\\,\\textbf{X}=\\textbf{0},\\] \\(\\textbf{X}\\) la matriz columna de inc\u00f3gnitas y \\(\\textbf{0}\\) la matriz columna de coordenadas de \\(0_W\\), nos proporciona los elementos del n\u00facleo de \\(f\\) que buscamos.<\/p>\n<p>Para obtener \\(\\mathbf{Im}(f)\\), es suficiente con ver que, dada una base \\(B=\\{v_1\\,\\ldots,v_r\\}\\subset V \\), la imagen de los vectores de la base forma un sistema generador de \\(\\mathbf{Im}(f)\\); es decir, \\(\\mathbf{Im}(f)=\\mathbf{Gen}\\{f(v_1)\\,\\ldots,f(v_r)\\}\\). As\u00ed pues, solo necesitamos los vectores l.i. de \\(\\{f(v_1)\\,\\ldots,f(v_r)\\}\\). <\/p>\n<p>Otra forma bastar\u00eda con determinar el rango de la matriz asociada y elegir un n\u00famero igual al rango, de vectores columna de la matriz, linealmente independientes. Dicho subconjunto formar\u00e1 una base de \\(\\mathbf{Im}(f)\\)<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea \\(f : \\mathbb{R}^2\\to\\mathbb{R}^3\\) la aplicaci\u00f3n lineal definida por \\(f(x, y) = (\u2212x + 2y, x, 2x \u2212 y)\\). Encontrar su n\u00facleo y su imagen.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a21w3() {\n  var htmlShow1a21w3 = document.getElementById(\"html-show1a21w3\");\n  if (htmlShow1a21w3.style.display === \"none\") {\n    htmlShow1a21w3.style.display = \"block\";\n  } else {\n    htmlShow1a21w3.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a21w3()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a21w3\" style=\"display: none;\">\nPara encontrar su n\u00facleo vemos que \\(\\mathbf{ker}(f)=\\{\\,(x,y)\\in \\mathbb{R}^2;f(x,y)=(0,0,0)\\,\\}\\); es decir<br \/>\n\\[\\begin{matrix}<br \/>\n\u2212x + 2y=0\\\\<br \/>\nx=0\\\\<br \/>\n2x \u2212 y=0<br \/>\n\\end{matrix}\\]<br \/>\nComo vemos, la soluci\u00f3n del sistema es (0,0), luego \\(\\mathbf{ker}(f)=(0,0)\\).<\/p>\n<p>Para calcular la imagen, calculamos la imagen de una base de \\(\\mathbb{R}^2\\), si no nos dicen lo contrario: la can\u00f3nica. De esta forma \\(\\mathbf{Im}(f)=\\mathbf{Gen}\\{f(1,0),f(0,1)\\}\\). Como<br \/>\n\\[\\mathbf{rank}\\begin{bmatrix} -1 &#038; 1&#038; 2\\\\  2 &#038;0&#038; -1 \\end{bmatrix}=2,\\]<br \/>\nentonces ambos vectores son l.i. y forman base de \\(\\mathbf{Im}(f)\\).\n<\/p><\/div>\n<hr>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea 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).\\] Determinar el n\u00facleo de la aplicaci\u00f3n.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a1() {\n  var htmlShow1a1 = document.getElementById(\"html-show1a1\");\n  if (htmlShow1a1.style.display === \"none\") {\n    htmlShow1a1.style.display = \"block\";\n  } else {\n    htmlShow1a1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a1\" style=\"display: none;\">\nComo no nos dicen bases, consideramos que trabajamos respecto de la base can\u00f3nica. Luego, la matriz asociada ser\u00e1 la obtenida por las im\u00e1genes de los vectores de la base can\u00f3nica: \\[ f(1)=1,\\  f(X)=X-1,\\ f(X^2)=X^2-2X.\\] Por tanto,<br \/>\n\\[\\begin{bmatrix}<br \/>\n1 &#038; -1 &#038; 0\\\\<br \/>\n0 &#038; 1 &#038; -2\\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}\\]<br \/>\nAs\u00ed pues, \\[\\mathbf{Ker}(f)=\\left\\{p_0+p_1X+p_2X^2\\in\\mathbb{R}_2[X];\\ \\begin{bmatrix}<br \/>\n1 &#038; -1 &#038; 0\\\\<br \/>\n0 &#038; 1 &#038; -2\\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}\\begin{bmatrix}<br \/>\np_1\\\\<br \/>\np_2\\\\<br \/>\np_3<br \/>\n\\end{bmatrix}=\\begin{bmatrix}<br \/>\n0\\\\<br \/>\n0\\\\<br \/>\n0<br \/>\n\\end{bmatrix}\\right\\}\\]\n<\/div>\n<hr>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}_2[X]\\to\\mathbb{R}_2[X]\\) dada por \\[\\forall p(X)=p_0+p_1X+p_2X^2\\in\\mathbb{R}_2[X];\\] \\[f(p(X))=(p_0+p_1+p_2)+(p_0+p_1)X+p_2X^2.\\] Determinar el n\u00facleo de la aplicaci\u00f3n.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a1d() {\n  var htmlShow1a1d = document.getElementById(\"html-show1a1d\");\n  if (htmlShow1a1d.style.display === \"none\") {\n    htmlShow1a1d.style.display = \"block\";\n  } else {\n    htmlShow1a1d.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a1d()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a1d\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - N\u00facleo de una Aplicaci\u00f3n Lineal. Ej.9 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/dK-NEWL_5iY?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><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}_3[X]\\to\\mathbb{R}_2[X]\\) dada por \\[\\forall p(X)=p_0+p_1X+p_2X^2+p_3X^3\\in\\mathbb{R}_3[X];\\] \\[f(p(X))=(p_0+p_1+p_2)+(p_0-p_3)X^2.\\] \u00bfCu\u00e1l es la dimensi\u00f3n de su n\u00facleo m\u00e1s tres veces la dimensi\u00f3n de su imagen? Es decir, \u00bf\\(\\mathbf{dim}\\,\\mathbf{Ker}(f)+3\\,\\mathbf{dim}\\,\\mathbf{Im}(f)\\)?.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv13a1w() {\n  var htmlShow13a1w = document.getElementById(\"html-show13a1w\");\n  if (htmlShow13a1w.style.display === \"none\") {\n    htmlShow13a1w.style.display = \"block\";\n  } else {\n    htmlShow13a1w.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv13a1w()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show13a1w\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - N\u00facleo de una Aplicaci\u00f3n Lineal. Ej.11 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/JYmzm6rnjEI?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><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathcal{M}_2(\\mathbb{R})\\to\\mathbb{R}^2\\) dada por \\[f\\begin{bmatrix}a&#038;b\\\\ c&#038;d\\end{bmatrix}=(a-b,c-d).\\] Determinar el n\u00facleo de la aplicaci\u00f3n.\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a2() {\n  var htmlShow1a2 = document.getElementById(\"html-show1a2\");\n  if (htmlShow1a2.style.display === \"none\") {\n    htmlShow1a2.style.display = \"block\";\n  } else {\n    htmlShow1a2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a2\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - N\u00facleo de una aplicaci\u00f3n lineal: EJ.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/gmYYw2ajAAc?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><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}^3\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[f(x,y,z)=<br \/>\n\\begin{bmatrix}x-y&#038;y-z\\\\ z-x&#038;x-y\\end{bmatrix}.\\] \u00bfCu\u00e1l \\(\\mathbf{dim}\\,\\mathbf{Im}(f)\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a2x() {\n  var htmlShow1a2x = document.getElementById(\"html-show1a2x\");\n  if (htmlShow1a2x.style.display === \"none\") {\n    htmlShow1a2x.style.display = \"block\";\n  } else {\n    htmlShow1a2x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a2x()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a2x\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Imagen de una aplicaci\u00f3n lineal. Ejemplo 3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/x89dKF86_Dw?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>Corolario:<\/strong> Sea \\(\\mathbf{Im}(f)=\\mathbf{Gen}\\{\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_r\\}\\), vectores linealmente independientes. Entonces \\(\\vec{u}\\in\\mathbf{Im}(f)\\Leftrightarrow \\mathbf{rank}(\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_r,\\vec{u})=r \\)\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Definici\u00f3n:<\/strong> Llamamos rango de una aplicaci\u00f3n lineal \\(f\\), \\(\\mathbf{rank}\\ (f)\\), al rango de su matriz asociada.\n<\/p>\n<p><strong>Proposici\u00f3n:<\/strong> \\(\\mathbf{rank}\\ (f)=\\mathbf{dim}\\,\\mathbf{Im}(f)\\)<\/p>\n<\/blockquote>\n<p>Un resultado importante nos dice que si \\(f:V\\to W\\), es lineal entre dos espacios vectoriales finitos sobre el mismo cuerpo, entonces<\/p>\n<p>\\[dim\\,\\mathbf{Ker}(f) + dim\\,\\mathbf{Im}(f)=dim\\, V\\]<\/p>\n<p>&nbsp;<\/p>\n<h1>Propiedades de la matriz asociada a una aplicaci\u00f3n<\/h1>\n<p>Hemos visto que si \\(M_f\\) es la matriz asociada a la aplicaci\u00f3n lineal \\(f:V\\to W\\), entonces<br \/>\n\\[f(v_1,v_2,\\ldots,v_n)=(w_1,w_2,\\ldots,w_m)\\Leftrightarrow M_f \\begin{pmatrix}v_1\\\\v_2\\\\ \\vdots\\\\v_n\\end{pmatrix}=\\begin{pmatrix}w_1\\\\w_2\\\\ \\vdots\\\\w_n\\end{pmatrix}.\\]<\/p>\n<p>Esto nos permite deducir interesantes propiedades de la aplicaci\u00f3n analizando sus correspondientes matrices asociadas.<\/p>\n<p>Llamamos rango de una aplicaci\u00f3n lineal \\(f\\) al rango de su matriz asociada. Propiedades para aplicar. Si \\(f:V\\to W\\) es lineal<\/p>\n<ol>\n<li>\\(f\\) es inyectiva si, y s\u00f3lo si, \\(\\mathbf{rank}\\, f=\\mathbf{dim}(V)\\)<\/li>\n<li>\\(f\\) es sobreyectiva si, y s\u00f3lo si, \\(\\mathbf{rank}\\, f=\\mathbf{dim}(W)\\)<\/li>\n<li>\\(\\mathbf{dim}(\\mathbf{Im}\\,f)=\\mathbf{rank}\\, f\\)<\/li>\n<\/ol>\n<p>Otra aplicaci\u00f3n es en la composici\u00f3n: <\/p>\n<blockquote>\n<p>Dadas dos aplicaciones lineales \\(f:V\\to V&#8217;\\) y \\(g:V&#8217;\\to W\\) se define la aplicaci\u00f3n lineal \\(f\\) compuesta con \\(g\\), \\((g\\circ f):V\\to W\\), como \\[(g\\circ f)(\\vec{v})=g(f(\\vec{v})),\\quad \\forall\\vec{v}\\in V.\\]<\/p>\n<\/blockquote>\n<p>De este modo la composici\u00f3n de aplicaciones se puede realizar mediante multiplicaci\u00f3n de matrices<\/p>\n<p>\\[(g\\circ f)(\\vec{v})=g(f(\\vec{v}))\\Leftrightarrow M_g(M_f\\vec{v})\\Leftrightarrow (M_g\\cdot M_f)\\vec{v}\\]<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sean las aplicaciones lineales \\(f:\\mathbb{R}^3\\to\\mathbb{R}^4\\) dada por \\[f(x,y,z)=(x+2y,y-z,x-z,2z-x-y),\\] y \\(g:\\mathbb{R}^4\\to\\mathbb{R}^2\\) dada por \\[g(x,y,z,t)=(x-3y,2t-5z).\\] \u00bfCu\u00e1l es la \\(\\sum a_{ii}\\) de la matriz asociada a \\(g\\circ f\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a1s() {\n  var htmlShow1a1s = document.getElementById(\"html-show1a1s\");\n  if (htmlShow1a1s.style.display === \"none\") {\n    htmlShow1a1s.style.display = \"block\";\n  } else {\n    htmlShow1a1s.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1a1s()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a1s\" style=\"display: none;\">\n\\[\\begin{split}(g\\circ f)(x,y,z)&#038;=\\begin{bmatrix}1&#038;0&#038;2&#038;-5\\\\ 0&#038;-3&#038;0&#038;0\\end{bmatrix}\\begin{bmatrix}1&#038;2&#038;0\\\\ 0&#038;1&#038;-1\\\\ 1&#038;0&#038;-1\\\\ -1&#038;-1&#038;2\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}\\\\ &#038;=\\begin{bmatrix}8&#038;7&#038;-12\\\\ 0&#038;-3&#038;3\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end{bmatrix}\\end{split}\\]<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Composici\u00f3n de Aplicaciones Lineales. Ej.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/LGm_NxJzc7M?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>Si consideramos lo que hemos visto, al hecho de que podemos establecer un isomorfismo entre un \\(\\mathbb{R}\\)-espacios vectoriales \\(V\\), de dimensi\u00f3n \\(n\\), y \\(\\mathbb{R}^n\\), resultar\u00e1 que podremos tratar los elementos del \\(\\mathbb{R}\\)-espacio vectorial como si fuesen vectores de \\(\\mathbb{R}^n\\). Esto nos ayudar\u00e1 a resolver problemas diversos; por ejemplo, determinar la independencia lineal de un conjunto de polinomios mediante su matriz como vectores en \\(\\mathbb{R}^n\\).<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sean las aplicaciones lineales \\(f:\\mathbb{R}_{2}[X]\\to\\mathbb{R}^2\\) dada por \\[f(p_0+p_1X+p_2X^2)=(p_0-p_2,p_1-p_2),\\] y \\(g:\\mathbb{R}^2\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[g(x,y)=\\begin{bmatrix}x&#038;x-y\\\\ y-x&#038;y\\end{bmatrix}.\\] \u00bfCu\u00e1l es el valor del determinante de \\((g\\circ f)(1-2X)\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1bs() {\n  var htmlShow1bs = document.getElementById(\"html-show1bs\");\n  if (htmlShow1bs.style.display === \"none\") {\n    htmlShow1bs.style.display = \"block\";\n  } else {\n    htmlShow1bs.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1bs()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1bs\" style=\"display: none;\">\n\\[\\begin{split}(g\\circ f)(p_0+p_1X+p_2X^2)&#038;=\\begin{bmatrix}1&#038;0\\\\ 1&#038;-1\\\\ -1&#038;1\\\\ 0&#038; 1\\end{bmatrix}\\begin{bmatrix}1&#038;0&#038;-1\\\\ 0&#038;1&#038;-1 \\end{bmatrix} \\begin{bmatrix}p_0\\\\p_1\\\\p_2\\end{bmatrix}\\\\ &#038;=\\begin{bmatrix}1&#038;0&#038;-1\\\\ 1&#038;-1&#038;0\\\\ -1&#038;1&#038;0\\\\ 0&#038;1&#038;-1 \\end{bmatrix}\\begin{bmatrix}p_0\\\\p_1\\\\p_2\\end{bmatrix}\\end{split}\\]<br \/>\nRecordad que el resultado est\u00e1 dado en \\(\\mathbf{R}^4\\), la expresi\u00f3n matricial ser\u00eda:<br \/>\n\\[\\begin{bmatrix}<br \/>\n\\begin{bmatrix}1&#038;0\\\\0&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;1\\\\0&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;0\\\\1&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;0\\\\0&#038;1\\end{bmatrix}<br \/>\n\\end{bmatrix}(g\\circ f)(p_0+p_1X+p_2X^2)\\]<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Composici\u00f3n de Aplicaciones Lineales. Ej.2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/wZoMwGVb-Z0?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><p><strong>Ejercicio:<\/strong> Sean las aplicaciones lineales \\(f:\\mathbb{R}^3\\to\\mathbb{R}_{3}[X]\\) dada por \\[f(a,b,c)=a+(b-a)X+(c-a)X^2+(2a-b)X^3,\\] y \\(g:\\mathbb{R}_3[X]\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[g(p_0+p_1X+p_2X^2+p_3X^3)= \\begin{bmatrix}p_0&#038;p_2-p_1\\\\ p_1-p_2&#038;p_3\\end{bmatrix}.\\] \u00bfCu\u00e1l es el valor del determinante de \\((g\\circ f)(-1,3,1)\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1cs() {\n  var htmlShow1cs = document.getElementById(\"html-show1cs\");\n  if (htmlShow1cs.style.display === \"none\") {\n    htmlShow1cs.style.display = \"block\";\n  } else {\n    htmlShow1cs.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1cs()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1cs\" style=\"display: none;\">\n\\[\\begin{split}(g\\circ f)(a,b,c)&#038;=\\begin{bmatrix}1&#038;0&#038;0&#038;0\\\\ 0&#038;1&#038;-1&#038;0\\\\0&#038;-1&#038;1&#038;0\\\\ 0&#038;0&#038;0&#038;1 \\end{bmatrix} \\begin{bmatrix}1&#038;0&#038;0\\\\ -1&#038;1&#038;0\\\\ -1&#038;0&#038;1\\\\ 2&#038; -1 &#038; 0\\end{bmatrix}\\begin{bmatrix}a\\\\b\\\\b\\end{bmatrix}\\\\ &#038;=\\begin{bmatrix}1&#038;0&#038;0\\\\ 0&#038;1&#038;-1\\\\ 0&#038;-1&#038;1\\\\ 2&#038;-1&#038;0 \\end{bmatrix}\\begin{bmatrix}a\\\\b\\\\c\\end{bmatrix}\\end{split}\\]<br \/>\nRecordad que el resultado est\u00e1 dado en \\(\\mathbf{R}^4\\), la expresi\u00f3n matricial ser\u00eda:<br \/>\n\\[\\begin{bmatrix}<br \/>\n\\begin{bmatrix}1&#038;0\\\\0&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;1\\\\0&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;0\\\\1&#038;0\\end{bmatrix} &#038;<br \/>\n\\begin{bmatrix}0&#038;0\\\\0&#038;1\\end{bmatrix}<br \/>\n\\end{bmatrix}(g\\circ f)(a,b,c)\\]<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Composici\u00f3n de Aplicaciones Lineales. Ej.4 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/hbu5m-Z9rGo?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><p><strong>Ejercicio:<\/strong> Sean las aplicaciones lineales \\(f:\\mathbb{R}_2[X]\\to\\mathbb{R}_{2}[X]\\) dada por \\[f(p(X))=\\frac{d}{dX}p(X),\\] y \\(g:\\mathbb{R}_2[X]\\to\\mathcal{M}_2(\\mathbb{R})\\) dada por \\[g(p_0+p_1X+p_2X^2)= \\begin{bmatrix}p_0&#038;p_1\\\\ p_2&#038;p_0\\end{bmatrix}.\\] \u00bfCu\u00e1l es el valor del determinante de \\((g\\circ f)(1-X+X^2)\\)?\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1c2ws() {\n  var htmlShow1c2ws = document.getElementById(\"html-show1c2ws\");\n  if (htmlShow1c2ws.style.display === \"none\") {\n    htmlShow1c2ws.style.display = \"block\";\n  } else {\n    htmlShow1c2ws.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1c2ws()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1c2ws\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Composici\u00f3n de aplicaciones lineales: EJ.1 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/asF4EZXmumY?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<h2>Imagen rec\u00edproca<\/h2>\n<p>Para terminar, tratamos la imagen rec\u00edproca de un vector.<\/p>\n<p>Si tenemos una aplicaci\u00f3n lineal \\(f:V\\to W\\), entre dos espacios vectoriales sobre el mismo cuerpo, y consideramos un vector fijo \\(\\vec{w}\\in W\\), llamamos conjunto imagen rec\u00edproca al conjunto \\[f^{-1}(\\vec{w})=\\{\\vec{v}\\in V;\\,f(\\vec{v})=\\vec{w}\\}\\subset V.\\]<br \/>\nPara este conjunto pueden ocurrirle dos propiedades interesantes.<\/p>\n<ol>\n<li>Si \\(\\vec{w}\\notin \\operatorname{Im}(f)\\), entonces \\(f^{-1}(\\vec{w})=\\varnothing\\)<\/li>\n<li>Si \\(\\vec{w}\\in \\operatorname{Im}(f)\\); es decir, existe alg\u00fan \\(\\vec{v}_0\\in V\\) tal que \\(f(\\vec{v}_0)=\\vec{w}\\), entonces \\(f^{-1}(\\vec{w})\\) es la variedad lineal dada por \\[f^{-1}(\\vec{w})=\\vec{v}_0+\\operatorname{ker}(f)\\]<\/li>\n<\/ol>\n<p>Veamos c\u00f3mo aplicamos esto. Consideremos la aplicaci\u00f3n \\(f(x,y,z)=(2x-y,-x+z)\\). La imagen rec\u00edproca del vector \\((1,3)\\in\\mathbb{R}^2\\) est\u00e1 formada por los vectores de \\((x,y,z)\\in\\mathbb{R}^3\\) tales que<br \/>\n\\[\\left.\\begin{array}{r}<br \/>\n2x-y=1 \\\\ -x+z=3<br \/>\n\\end{array}\\right\\}<br \/>\n\\]<br \/>\nSi resolvemos el sistema tendremos<br \/>\n\\[\\left\\{\\begin{array}{l}<br \/>\nx=k \\\\ y=-1+2k \\\\z=3+k<br \/>\n\\end{array}\\right.\\]<br \/>\nPor tanto, la imagen rec\u00edproca la podremos poner como<\/p>\n<p> \\[f^{-1}(1,3)=\\{(k,-1+2k,3+k);k\\in\\mathbb{R}\\}=(0,1,3)+\\{(k,2k,k);k\\in\\mathbb{R}\\},\\]<br \/>\ndonde \\[\\operatorname{ker}(f)=\\{(k,2k,k);k\\in\\mathbb{R}\\}.\\]<\/p>\n<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}^3\\to\\mathbb{R}^3\\) dada por \\[f(x,y,z)=(2x-y+z,3x-2y+z,2x+2y-2z),\\] determinar la imagen rec\u00edproca de (-5,-9,-8).\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1d2() {\n  var htmlShow1d2 = document.getElementById(\"html-show1d2\");\n  if (htmlShow1d2.style.display === \"none\") {\n    htmlShow1d2.style.display = \"block\";\n  } else {\n    htmlShow1d2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1d2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1d2\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Imagen Rec\u00edproca. Ej. 2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/y3mT4R1a4DA?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><p><strong>Ejercicio:<\/strong> Dada la aplicaci\u00f3n lineal \\(f\\begin{bmatrix} a &#038; b \\\\ c &#038; d\\end{bmatrix}=a+(b-c)X+dX^2\\), \u00bfcu\u00e1l de las matrices dadas pertenece a la imagen rec\u00edproca del vector \\(5+X-X^2\\in\\mathbb{R}_2[X]\\)?<br \/>\n\\[A:\\begin{bmatrix} 5 &#038; 2 \\\\ 1 &#038; 1\\end{bmatrix},\\,B:\\begin{bmatrix} 5 &#038; 3 \\\\ 2 &#038; -1\\end{bmatrix},\\, C:\\begin{bmatrix} 2 &#038; 1 \\\\ 0 &#038; -1\\end{bmatrix} \\]\n<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1d2s() {\n  var htmlShow1d2s = document.getElementById(\"html-show1d2s\");\n  if (htmlShow1d2s.style.display === \"none\") {\n    htmlShow1d2s.style.display = \"block\";\n  } else {\n    htmlShow1d2s.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1d2s()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1d2s\" style=\"display: none;\">\nPodemos resolverlo por el m\u00e9todo directo:<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\">(%i1)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">=<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">v<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">4<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">X<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">X<\/span><span class=\"code_operator\">^<\/span><span class=\"code_number\">2<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\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\">(%i3)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">lt<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/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\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">[<\/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\">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 class=\"code_operator\">[<\/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_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_function\">makelist<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">f<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">lt<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">i<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">i<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\left[ {{X}^{2}}+X+5\\operatorname{,}-{{X}^{2}}+X+5\\operatorname{,}-{{X}^{2}}+X+2\\right] \\]<\/p>\n<p><!-- Text cell --><\/p>\n<div class=\"comment\">Como vemos el segundo polinomio es el vector del enunciado, con lo cual la matriz B es la soluci\u00f3n.<\/p>\n<p>Resolv\u00e1moslo viendo si la matriz pertenece a la imagen. Como la imagen rec\u00edproca es \\[f^{-1}(5+X-X^2)=\\left\\{M+\\lambda v\\right\\}=\\left\\{\\begin{bmatrix}5&amp;2\\\\ 1&amp;-1\\end{bmatrix}+\\lambda\\begin{bmatrix}0&amp;1\\\\ 1&amp;0\\end{bmatrix}\\right\\}(*)\\] <br \/>la matriz, D, que pertenezca verificar\u00e1 que el rango de las coordenadas de \\[rank[M-D,v]=1\\]<\/div>\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\">(%i6)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">M<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/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_operator\">\u2212<\/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\">v<\/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\">1<\/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\">$<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">lt<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">[<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">5<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">$<\/span><\/span><\/td>\n<\/tr>\n<\/table>\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_function\">makelist<\/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\">M<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">lt<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">i<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">i<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\left[ \\begin{bmatrix}0 &amp; 1 &amp; 1 &amp; 0\\\\0 &amp; 0 &amp; 0 &amp; -2\\end{bmatrix}\\operatorname{,}\\begin{bmatrix}0 &amp; 1 &amp; 1 &amp; 0\\\\0 &amp; -1 &amp; -1 &amp; 0\\end{bmatrix}\\operatorname{,}\\begin{bmatrix}0 &amp; 1 &amp; 1 &amp; 0\\\\3 &amp; 1 &amp; 1 &amp; 0\\end{bmatrix}\\right] \\]<\/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\">(%i8)<\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_function\">makelist<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">rank<\/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\">M<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">lt<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">i<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">i<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\left[ 2\\operatorname{,}1\\operatorname{,}2\\right] \\]<\/p>\n<p><!-- Text cell --><br \/>\nEsto nos confirma que B es la matriz buscada.<\/p>\n<p>(*) Aqu\u00ed se explica c\u00f3mo obtenemos la imagen rec\u00edproca.<br \/>\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Imagen rec\u00edproca: EJ.1 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/_bEpQlzvRzI?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%\">\n<p><strong>Ejercicio:<\/strong>  Sean los subespacios vectoriales \\(S=\\textbf{Gen}\\{[[1,2],[2,1]],[[0,-1],[1,1]]\\}\\in\\mathcal{M}_2(\\mathbb{R})\\) y \\(T=\\textbf{Gen}\\{[[-1,0],[3,-1]],\\)\\([[1,9],[9,-2]],\\)\\([[3,10],[2,-1]]\\}\\in\\mathcal{M}_2(\\mathbb{R})\\). \u00bf\\(\\textbf{dim}(S+T)\\)?<\/p>\n<div id=\"menu-a\">\n<ul>\n<li>2<\/li>\n<li>3<\/li>\n<li>4<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv(id) {\n  var htmlShow = document.getElementById(id);\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script><br \/>\n<p><button onclick=\"showHtmlDiv('html-show-1')\">Soluci\u00f3n:<\/button><\/p><div id=\"html-show-1\" style=\"display: none;\"><\/p>\n<p><strong>B.)<\/strong> y <strong>D.)<\/strong><\/p>\n<p>En efecto, nos basta con determinar el rango de:<\/p>\n<p><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\">\n  <mtable>\n      <mtd>\n        <mrow>\n          <mo>[<\/mo>\n          <mrow>\n            <mtable>\n              <mtr>\n                <mtd>\n                  <mn>1<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>2<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>2<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>1<\/mn>\n                <\/mtd>\n              <\/mtr>\n              <mtr>\n                <mtd>\n                  <mn>0<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mrow>\n                    <mi>&#x2212;<\/mi>\n                    <mn>1<\/mn>\n                  <\/mrow>\n                <\/mtd>\n                <mtd>\n                  <mn>1<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>1<\/mn>\n                <\/mtd>\n              <\/mtr>\n              <mtr>\n                <mtd>\n                  <mrow>\n                    <mi>&#x2212;<\/mi>\n                    <mn>1<\/mn>\n                  <\/mrow>\n                <\/mtd>\n                <mtd>\n                  <mn>0<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>3<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mrow>\n                    <mi>&#x2212;<\/mi>\n                    <mn>1<\/mn>\n                  <\/mrow>\n                <\/mtd>\n              <\/mtr>\n              <mtr>\n                <mtd>\n                  <mn>3<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>10<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>2<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mrow>\n                    <mi>&#x2212;<\/mi>\n                    <mn>1<\/mn>\n                  <\/mrow>\n                <\/mtd>\n              <\/mtr>\n              <mtr>\n                <mtd>\n                  <mn>1<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>9<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mn>9<\/mn>\n                <\/mtd>\n                <mtd>\n                  <mrow>\n                    <mi>&#x2212;<\/mi>\n                    <mn>2<\/mn>\n                  <\/mrow>\n                <\/mtd>\n              <\/mtr>\n            <\/mtable>\n          <\/mrow>\n          <mo>]<\/mo>\n        <\/mrow>\n      <\/mtd>\n  <\/mtable>\n<\/math><br \/>\n<\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Al hablar de grupos se introdujo la definici\u00f3n de homomorfismo y con ella la de n\u00facleo. Ahora extendemos esta definici\u00f3n a espacios vectoriales para definir la aplicaci\u00f3n lineal: un homomorfismo entre espacios&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-318","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/318","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=318"}],"version-history":[{"count":4,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/318\/revisions"}],"predecessor-version":[{"id":428,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/318\/revisions\/428"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=318"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=318"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=318"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}