{"id":372,"date":"2025-10-27T08:15:41","date_gmt":"2025-10-27T07:15:41","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=372"},"modified":"2025-10-29T18:17:02","modified_gmt":"2025-10-29T17:17:02","slug":"alg-nucleo-e-imagen-de-una-aplicacion-lineal","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=372","title":{"rendered":"ALG: N\u00facleo e imagen de una aplicaci\u00f3n lineal"},"content":{"rendered":"<p>Recordemos es 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 forman 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\\) compuesto 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 consideremos 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 puede 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%\"><strong>Ejercicio:<\/strong> Sea la aplicaci\u00f3n lineal \\(f:\\mathbb{R}_3[X]\\to\\mathbb{R}_3[X]\\), dada por \\(\\forall p(X)\\in\\mathbb{R}_3[X]\\), es \\(f(p(X))=p(X)-2\\frac{d}{dX}p(X)\\). \u00bfCu\u00e1nto es la traza de la matriz asociada a la aplicaci\u00f3n \\(f\\)? <\/td>\n<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>-2<\/li>\n<li>2<\/li>\n<li>4<\/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><\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/uploads.jesussoto.es\/maxima\/ejrALGmatriz_apl01.html\" width=\"650\" height=\"300\" allow=\"fullscreen\"><\/iframe>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Recordemos es dada una aplicaci\u00f3n lineal, \\(T\\), se define el n\u00facleo (ker) y la imagen (Im) de \\(T:V\\to W\\) como: \\(\\mathbf{ker}(T)=\\{\\,v\\in V:T(v)=0_W\\,\\}\\) \\(\\mathbf{Im}(T)=\\{\\,w\\in W: \\exists v\\in V:T(v)=w\\,\\}\\) Es decir, que el n\u00facleo&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-372","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/372","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=372"}],"version-history":[{"count":7,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/372\/revisions"}],"predecessor-version":[{"id":408,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/372\/revisions\/408"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=372"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=372"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}