{"id":147,"date":"2025-10-08T11:45:57","date_gmt":"2025-10-08T09:45:57","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=147"},"modified":"2025-10-05T20:53:58","modified_gmt":"2025-10-05T18:53:58","slug":"mathbio-aplicaciones-lineales","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=147","title":{"rendered":"MathBio: Aplicaciones lineales"},"content":{"rendered":"<p>En matem\u00e1ticas una aplicaci\u00f3n lineal, es una aplicaci\u00f3n entre dos espacios vectoriales, que preserva las operaciones de adici\u00f3n de vectores y multiplicaci\u00f3n por un escalar.<\/p>\n<p>Sean \\({\\displaystyle V}\\) y \\({\\displaystyle W}\\) espacios vectoriales sobre el mismo cuerpo \\({\\displaystyle K}\\), aunque en nuestro caso este cuerpo ser\u00e1 \\(\\mathbb{R}\\). Una aplicaci\u00f3n \\({\\displaystyle f}\\) de \\({\\displaystyle V}\\) en \\({\\displaystyle W}\\), es decir, \\({\\displaystyle f:V\\to W}\\), es una transformaci\u00f3n lineal si para todo par de vectores \\({\\displaystyle u,v\\in V}\\) y para todo escalar \\({\\displaystyle k\\in \\mathbb{R}}\\), se satisface que:<\/p>\n<ul>\n<li>\\({\\displaystyle f(u+v)=f(u)+f(v)\\,}\\)<\/li>\n<li>\\({\\displaystyle f(ku)=kf(u)\\,}\\).<\/li>\n<\/ul>\n<p>As\u00ed, por ejemplo, podemos definir la aplicaci\u00f3n \\(\\Phi:\\mathbb{R}^2\\to \\mathbb{R}^3\\), dada por \\(\\Phi(x,y)=(x,y,0)\\), que es lineal. Las propiedades que cumple \\(\\Phi\\) implican que \\(\\mathbf{Im}\\Phi\\) es un subespacio vectorial isomorfo a \\(\\mathbb{R}^2\\) y contenido en \\(\\mathbb{R}^3\\); es decir, \\(|\\mathbb{R}^2|<|\\mathbb{R}^3|\\). <\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea \\(\\mathbf{det}:\\mathcal{M}_n\\to\\mathbb{R}\\) que a cada matriz \\(A\\in\\mathcal{M}_n\\) le hace corresponder su determinante, \\(\\mathbf{det}(A)=|A|\\). Probar si es lineal. <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv162() {\n  var htmlShow162 = document.getElementById(\"html-show162\");\n  if (htmlShow162.style.display === \"none\") {\n    htmlShow162.style.display = \"block\";\n  } else {\n    htmlShow162.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv162()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show162\" style=\"display: none;\">\nConsideremos las dos matrices \\(\\begin{bmatrix}1&#038;0\\\\0&#038;0 \\end{bmatrix}\\) y \\(\\begin{bmatrix}0&#038;0\\\\0&#038;1 \\end{bmatrix}\\). Por la aplicaci\u00f3n \\(\\mathbf{det}()\\), es<br \/>\n\\[<br \/>\n\\mathbf{det}\\left(\\begin{bmatrix}1&#038;0\\\\0&#038;0 \\end{bmatrix}+\\begin{bmatrix}0&#038;0\\\\0&#038;1 \\end{bmatrix}\\right)=\\mathbf{det}\\left(\\begin{bmatrix}1&#038;0\\\\0&#038;1 \\end{bmatrix}\\right)=\\begin{vmatrix}1&#038;0\\\\0&#038;1 \\end{vmatrix}=1.<br \/>\n\\]<\/p>\n<p>Ahora,<br \/>\n\\[<br \/>\n\\mathbf{det}\\left(\\begin{bmatrix}1&#038;0\\\\0&#038;0 \\end{bmatrix}\\right)+\\mathbf{det}\\left(\\begin{bmatrix}0&#038;0\\\\0&#038;1 \\end{bmatrix}\\right)=\\begin{vmatrix}1&#038;0\\\\0&#038;0 \\end{vmatrix}+\\begin{vmatrix}0&#038;0\\\\0&#038;1 \\end{vmatrix}=0.<br \/>\n\\]<br \/>\nPor tanto,<br \/>\n\\[<br \/>\n\\mathbf{det}\\left(\\begin{bmatrix}1&#038;0\\\\0&#038;0 \\end{bmatrix}+\\begin{bmatrix}0&#038;0\\\\0&#038;1 \\end{bmatrix}\\right)\\neq\\mathbf{det}\\left(\\begin{bmatrix}1&#038;0\\\\0&#038;0 \\end{bmatrix}\\right)+\\mathbf{det}\\left(\\begin{bmatrix}0&#038;0\\\\0&#038;1 \\end{bmatrix}\\right).<br \/>\n\\]<br \/>\nAs\u00ed pues, la aplicaci\u00f3n no puede ser lineal.\n<\/p><\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea \\(\\mathbf{tr}:\\mathcal{M}_n\\to\\mathbb{R}\\) que a cada matriz \\([a_{ij}]\\in\\mathcal{M}_n\\) le hace corresponder su traza, \\(\\mathbf{tr}([a_{ij}])=\\sum_{i=1}^na_{ii}\\). Probar si es lineal. <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1632() {\n  var htmlShow1632 = document.getElementById(\"html-show1632\");\n  if (htmlShow1632.style.display === \"none\") {\n    htmlShow1632.style.display = \"block\";\n  } else {\n    htmlShow1632.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1632()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1632\" style=\"display: none;\">\nConsideremos las dos matrices \\(\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22} \\end{bmatrix}\\) y \\(\\begin{bmatrix}b_{11}&#038;b_{12}\\\\ b_{21}&#038;b_{22} \\end{bmatrix}\\). Por la aplicaci\u00f3n \\(\\mathbf{tr}()\\), es<br \/>\n\\[\\begin{align*}<br \/>\n\\mathbf{tr}\\left(\\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}\\right)&#038;=\\mathbf{det}\\left(\\begin{bmatrix}a_{11}+b_{11}&#038;a_{12}+b_{12}\\\\ a_{21}+b_{21}&#038;a_{22}+b_{22}\\end{bmatrix}\\right)\\\\<br \/>\n&#038;=(a_{11}+b_{11})+(a_{22}+b_{22})\\\\<br \/>\n&#038;=(a_{11}+a_{22})+(b_{11}+b_{22})\\\\<br \/>\n&#038;=\\mathbf{tr}\\left(\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22} \\end{bmatrix}\\right)+\\mathbf{tr}\\left(\\begin{bmatrix}b_{11}&#038;b_{12}\\\\ b_{21}&#038;b_{22} \\end{bmatrix}\\right)<br \/>\n\\end{align*}<br \/>\n\\]<\/p>\n<p>Adem\u00e1s,<br \/>\n\\[\\begin{align*}<br \/>\n\\mathbf{tr}\\left(\\lambda\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22} \\end{bmatrix}\\right)&#038;=<br \/>\n\\mathbf{tr}\\left(\\begin{bmatrix}\\lambda a_{11}&#038;\\lambda a_{12}\\\\ \\lambda a_{21}&#038;\\lambda a_{22} \\end{bmatrix}\\right)\\\\<br \/>\n&#038;=\\lambda a_{11}+\\lambda a_{22} \\\\<br \/>\n&#038;=\\lambda(a_{11}+ a_{22}) \\\\<br \/>\n&#038;= \\lambda\\mathbf{tr}\\left(\\begin{bmatrix}a_{11}&#038;a_{12}\\\\ a_{21}&#038;a_{22} \\end{bmatrix}\\right)<br \/>\n\\end{align*}<br \/>\n\\]<br \/>\nPor tanto, se satisfacen las propiedades necesarias para ser lineal.\n<\/p><\/div>\n<hr \/>\n<p>Otra aplicaci\u00f3n lineal muy interesante es la que nos relaciona el conjunto de las matrices con las n-tuplas reales. As\u00ed la aplicaci\u00f3n de \\(\\Phi:\\mathcal{M}_{n\\times m}(\\mathbb{R})\\to \\mathbb{R}^{n}\\times \\overset{m}{\\cdots}\\times\\mathbb{R}^{n}\\), ofrece la posibilidad de trabajar las columnas de una matriz como vectores reales.<\/p>\n<p>Adem\u00e1s, toda aplicaci\u00f3n lineal tiene asociada una matriz, de modo que si \\(f:\\mathbb{R}^n\\to \\mathbb{R}^m\\), tal que \\[f(x_1,x_2,\\ldots,x_n)=(y_1,y_2,\\ldots,y_m),\\] ser\u00e1<br \/>\n\\[\\mathbf{M}_f \\begin{bmatrix}x_1\\\\ x_2\\\\ \\vdots\\\\ x_n\\end{bmatrix}=\\begin{bmatrix}y_1\\\\ y_2\\\\ \\vdots\\\\ y_m\\end{bmatrix}.\\]<\/p>\n<p>De este modo vemos que la imagen de un vector por una aplicaci\u00f3n es el producto de una matriz por el vector. As\u00ed se facilita el c\u00e1lculo de operaciones.<\/p>\n<blockquote>\n<p><strong>Proposici\u00f3n:<\/strong> Sea \\(f:\\mathbb{R}^n\\to \\mathbb{R}^m\\) una aplicaci\u00f3n lineal y \\(B=\\{\\mathbf{v}_1,\\mathbf{v}_2,\\ldots,\\mathbf{v}_n\\}\\) una base de \\(\\mathbb{R}^n\\). Entonces, la columna \\(i\\) de \\(\\mathbf{M}_f\\) ser\u00e1 \\(f(\\mathbf{v}_i)\\).\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfCu\u00e1l es la matriz asociada a la aplicaci\u00f3n lineal: \\(f(x,y,z)=(2x-y,x-z,y+2z)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv63() {\n  var htmlShow63 = document.getElementById(\"html-show63\");\n  if (htmlShow63.style.display === \"none\") {\n    htmlShow63.style.display = \"block\";\n  } else {\n    htmlShow63.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv63()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show63\" style=\"display: none;\">\nSi consideramos la base can\u00f3nica de \\(\\mathbb{R}^3\\) y sustituimos en \\(f\\), tendremos \\(f(1,0,0)=(2,1,0)\\), \\(f(0,1,0)=(-1,0,1)\\) y \\(f(0,0,1)=(0,-1,2)\\). Luego \\[M_f=\\begin{bmatrix}2 &#038; -1 &#038; 0\\\\1 &#038; 0 &#038; -1\\\\0 &#038; 1 &#038; 2\\end{bmatrix}\\]\n<\/div>\n<hr \/>\n<p>La matriz asociada a una aplicaci\u00f3n lineal depender\u00e1 de la base que elijamos del espacio vectorial. Salvo que digamos lo contrario, siempre nos referiremos con matriz asociada, o matriz can\u00f3nica, a la matriz asociada respecto de la base can\u00f3nica del espacio vectorial.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfCu\u00e1l es el determinante de la matriz can\u00f3nica de la aplicaci\u00f3n lineal: \\(f(x,y,z)=(x-2z,y-3z,2z)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv64() {\n  var htmlShow64 = document.getElementById(\"html-show64\");\n  if (htmlShow64.style.display === \"none\") {\n    htmlShow64.style.display = \"block\";\n  } else {\n    htmlShow64.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv64()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show64\" style=\"display: none;\">\nSi consideramos la base can\u00f3nica de \\(\\mathbb{R}^3\\) y sustituimos en \\(f\\), tendremos \\(f(1,0,0)=(1,0,0)\\), \\(f(0,1,0)=(0,1,0)\\) y \\(f(0,0,1)=(-2,-3,2)\\). Luego \\[\\begin{vmatrix}1 &#038; 0 &#038; -2\\\\0 &#038; 1 &#038; -3\\\\0 &#038; 0 &#038; 2\\end{vmatrix}=2\\]\n<\/div>\n<hr \/>\n<p>Dada una aplicaci\u00f3n lineal, \\(f\\), se define el <b>n\u00facleo<\/b> (ker) y la <b>imagen<\/b> (Im) de \\(f:V\\to W\\) como:<\/p>\n<blockquote>\n<dl>\n<dd>\\(\\mathbf{ker}(f)=\\{\\,v\\in V:f(v)=0_W\\,\\}\\)<\/dd>\n<dd>\\(\\mathbf{Im}(f)=\\{\\,w\\in W: \\exists v\\in V:f(v)=w\\,\\}\\)<\/dd>\n<\/dl>\n<\/blockquote>\n<p>Estas definiciones son muy importantes; por ejemplo, porque son la base de los sistemas de ecuaciones.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea \\(f(x,y,z)=(2x-y,x-z,y+2z)\\), determinar \\(\\mathbf{ker}(f)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv63a() {\n  var htmlShow63a = document.getElementById(\"html-show63a\");\n  if (htmlShow63a.style.display === \"none\") {\n    htmlShow63a.style.display = \"block\";\n  } else {\n    htmlShow63a.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv63a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show63a\" style=\"display: none;\">\nPor el ejercicio anterior sabemos que  \\[M_f=\\begin{bmatrix}2 &#038; -1 &#038; 0\\\\1 &#038; 0 &#038; -1\\\\0 &#038; 1 &#038; 2\\end{bmatrix}.\\]<br \/>\nLuego el n\u00facleo vendr\u00e1 determinado por \\[\\mathbf{ker}(f)=\\left\\{[x,y,z];\\begin{bmatrix}2 &#038; -1 &#038; 0\\\\1 &#038; 0 &#038; -1\\\\0 &#038; 1 &#038; 2\\end{bmatrix}\\begin{bmatrix}x\\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0\\\\ 0\\\\ 0\\end{bmatrix}\\right\\}.\\]\n<\/div>\n<hr \/>\n<p>Dos resultados muy \u00fatiles:<\/p>\n<blockquote>\n<p><strong>Proposici\u00f3n:<\/strong> \\(dim(\\mathbf{Im}(f))=\\mathbf{rank}(\\mathbf{M}_f)\\)\n<\/p>\n<\/blockquote>\n<p>Es decir, para obtener \\(\\mathbf{Im}(f)\\), basta 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>\n<p><strong>Ejercicio:<\/strong> Sea \\(f(x,y)=(2x-y,x-y,y)\\), determinar una base de  \\(\\mathbf{Im}(f)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv63b1() {\n  var htmlShow63b1 = document.getElementById(\"html-show63b1\");\n  if (htmlShow63b1.style.display === \"none\") {\n    htmlShow63b1.style.display = \"block\";\n  } else {\n    htmlShow63b1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv63b1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show63b1\" style=\"display: none;\">\nSabemos que  \\[M_f^t=\\begin{bmatrix}2 &#038; 1 &#038; 0\\\\ -1 &#038; -1 &#038; 1\\end{bmatrix}.\\]<br \/>\nComo \\[\\mathbf{rank}\\begin{bmatrix}2 &#038; 1 &#038; 0\\\\ -1 &#038; -1 &#038; 1\\end{bmatrix}=2,\\]<br \/>\n\\(\\mathbf{Im}(f)=\\mathbf{Gen}\\{[2,1,0],[-1,-1,1]\\}\\), y, por tanto, base.\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<p>Esto nos permite deducir si un vector pertenece a la imagen.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea \\(f(x,y,z,t)=(2x-y+t,x-z-t,x+y+2z-2t)\\), determinar si \\([1,0,3]\\in\\mathbf{Im}(f)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv63a1() {\n  var htmlShow63a1 = document.getElementById(\"html-show63a1\");\n  if (htmlShow63a1.style.display === \"none\") {\n    htmlShow63a1.style.display = \"block\";\n  } else {\n    htmlShow63a1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv63a1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show63a1\" style=\"display: none;\">\nSabemos que  \\[M_f=\\begin{bmatrix}2 &#038; -1 &#038; 0 &#038; 1\\\\ 1 &#038; 0 &#038; -1 &#038; -1\\\\1 &#038; 1 &#038; 2 &#038; -2\\end{bmatrix}.\\]<br \/>\nComo \\[\\begin{vmatrix}-1 &#038; 0 &#038; 1\\\\ 0 &#038; -1 &#038; -1\\\\ 1 &#038; 2 &#038; -2\\end{vmatrix}\\neq 0,\\] podemos considerar los vectores columna como base de \\(\\mathbf{Im}(f)\\). Luego \\([1,0,3]\\in\\mathbf{Im}(f)\\) si \\[\\mathbf{rank}\\begin{bmatrix}-1 &#038; 0 &#038; 1 &#038; 1\\\\ 0 &#038; -1 &#038; -1 &#038; 0\\\\  1 &#038; 2 &#038; -2 &#038; 3\\end{bmatrix}=3\\]\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Sea \\(f(x,y,z)=(2x-y,x-z,x+y+2z,2y-z)\\), determinar si \\([1,0,3,-1]\\in\\mathbf{Im}(f)\\) <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv73a1() {\n  var htmlShow73a1 = document.getElementById(\"html-show73a1\");\n  if (htmlShow73a1.style.display === \"none\") {\n    htmlShow73a1.style.display = \"block\";\n  } else {\n    htmlShow73a1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv73a1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show73a1\" style=\"display: none;\">\nSabemos que  \\[M_f=\\begin{bmatrix}2 &#038; -1 &#038; 0\\\\1 &#038; 0 &#038; -1\\\\1 &#038; 1 &#038; 2\\\\0 &#038; 2 &#038; -1\\end{bmatrix}.\\] El rango de la matriz es 3, luego los vectores columna forman una base de \\(\\mathbf{Im}(f)\\). Para saber si \\([1,0,3,-1]\\in\\mathbf{Im}(f)\\) es suficiente verificar \\[\\mathbf{rank}\\begin{bmatrix}2 &#038; 1 &#038; 1 &#038; 0\\\\<br \/>\n-1 &#038; 0 &#038; 1 &#038; 2\\\\0 &#038; -1 &#038; 2 &#038; -1\\\\ 1&#038;0&#038;3&#038;-1\\end{bmatrix}=3\\]\n<\/div>\n<hr \/>\n<p>As\u00ed, defenimos el rango de una aplicaci\u00f3n lineal como el rango de su matriz asociada: \\(\\mathbf{rank}(f)=\\mathbf{rank}(\\mathbf{M}_f)\\) <\/p>\n<hr \/>\n<h3>Bibliograf\u00eda<\/h3>\n<ul>\n<li>Cap\u00edtulo 2 de \u00c1lgebra lineal y sus aplicaciones. 5\u00ba edici\u00f3n, David C. Lay. Pearson. 2016.<\/li>\n<\/ul>\n<hr \/>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\"><strong>Ejercicio:<\/strong><br \/>\nLa matriz \\(\\begin{bmatrix}1&#038;0&#038;0\\\\0&#038;1&#038;-2\\\\ 0&#038;1&#038;-1\\end{bmatrix}\\), es la matriz asociada a la aplicaci\u00f3n:<\/td>\n<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>f(x,y,z)=(x+y,2z,x-z,y-x)<\/li>\n<li>g(x,y,z)=(x,y-2z,y-z)<\/li>\n<li>h(x,y,z)=(y,x+2z,z-y)<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv() {\n  var htmlShow = document.getElementById(\"html-show\");\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>B.)<\/strong><\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/uploads.jesussoto.es\/maxima\/ejrALGapl01.html\" width=\"650\" height=\"300\" allow=\"fullscreen\"><\/iframe>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>En matem\u00e1ticas una aplicaci\u00f3n lineal, es una aplicaci\u00f3n entre dos espacios vectoriales, que preserva las operaciones de adici\u00f3n de vectores y multiplicaci\u00f3n por un escalar. Sean \\({\\displaystyle V}\\) y \\({\\displaystyle W}\\) 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-147","post","type-post","status-publish","format-standard","hentry","category-mathbio"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/147","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=147"}],"version-history":[{"count":6,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/147\/revisions"}],"predecessor-version":[{"id":234,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/147\/revisions\/234"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=147"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=147"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=147"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}