{"id":157,"date":"2025-10-01T08:15:36","date_gmt":"2025-10-01T06:15:36","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=157"},"modified":"2025-10-09T20:08:41","modified_gmt":"2025-10-09T18:08:41","slug":"alg-inversa-de-una-matriz-y-pseudoinversa","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=157","title":{"rendered":"ALG: Inversa de una matriz y pseudoinversa"},"content":{"rendered":"<p>Definimos la inversa de una matriz cuadrada \\(A=[a_{ij}]\\in \\mathcal{M}_{n}(\\mathbb{R} o \\mathbb{C})\\) como la matriz \\(B=[b_{ij}]\\in \\mathcal{M}_{n}(\\mathbb{R} o \\mathbb{C})\\) tal que \\[AB=BA=I_n.\\]<\/p>\n<p>El procedimiento que damos para calcular la inversa, es el de realizar operaciones elementales entre filas o columnas, que conoc\u00e9is como m\u00e9todo de Gauss. Ser\u00eda el siguiente: Sea \\(A\\) la matriz, y consideremos la matriz formada por \\([A\\, |\\, I_n]\\). Si conseguimos mediante semejanza por transformaciones elementales una matriz tal que<\/p>\n<p>\\[[A\\, |\\, I_n] \\sim [I_n\\, |\\, B],\\]<\/p>\n<p>entonces \\(B\\) es la inversa de \\(A\\).<\/p>\n<blockquote><p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots &#038; 0 &#038; 0\\\\<br \/>\n\\vdots &#038; \\vdots  &#038; \\vdots  &#038;\\vdots   &#038; \\cdots  &#038; \\vdots  &#038; \\vdots \\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; 0\\\\<br \/>\n1 &#038; 2 &#038; 3 &#038; 4 &#038; \\cdots  &#038; n-1 &#038; n<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_n(R),\\] y \\(B=[b_{ij}]\\) su matriz inversa. \u00bfCu\u00e1nto es la traza de \\(B\\)? <\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv6() {\n  var htmlShow6 = document.getElementById(\"html-show6\");\n  if (htmlShow6.style.display === \"none\") {\n    htmlShow6.style.display = \"block\";\n  } else {\n    htmlShow6.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv6()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show6\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra lineal - Matriz inversa EJ. 3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/22e9j5XLi0Q?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n2 &#038; 1 &#038; 0 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n3 &#038; 2 &#038; 1 &#038; 0 &#038; \\cdots  &#038; 0 &#038; 0\\\\<br \/>\n4 &#038; 3 &#038; 2 &#038; 1 &#038; \\cdots &#038; 0 &#038; 0\\\\<br \/>\n\\vdots &#038; \\vdots  &#038; \\vdots  &#038;\\vdots   &#038; \\cdots  &#038; \\vdots  &#038; \\vdots \\\\<br \/>\nn-1 &#038; n-2 &#038; n-3 &#038; n-4 &#038; \\cdots  &#038; 1 &#038; 0\\\\<br \/>\nn &#038; n-1 &#038; n-2 &#038; n-3 &#038; \\cdots  &#038; 2 &#038; 1<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_n(R),\\] y \\(B=[b_{ij}]\\) su matriz inversa. \u00bfCu\u00e1ntos \\(b_{ij}=0\\) hay? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv1() {\n  var htmlShow1 = document.getElementById(\"html-show1\");\n  if (htmlShow1.style.display === \"none\") {\n    htmlShow1.style.display = \"block\";\n  } else {\n    htmlShow1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Matriz inversa. Ej. 2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/Wry0pXShQGg?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\[A=\\begin{bmatrix}<br \/>\ni &#038; 1 &#038; -1 &#038; i\\\\<br \/>\n0 &#038; i &#038; 1 &#038; 1 \\\\<br \/>\n0 &#038; 0 &#038; i &#038; -1 \\\\<br \/>\n0 &#038; 0 &#038; 0 &#038; i<br \/>\n\\end{bmatrix}\\in\\mathcal{M}_4(\\mathbb{C}),\\]  \u00bfcu\u00e1nto es la traza de la inversa? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv4() {\n  var htmlShow4 = document.getElementById(\"html-show4\");\n  if (htmlShow4.style.display === \"none\") {\n    htmlShow4.style.display = \"block\";\n  } else {\n    htmlShow4.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv4()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show4\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Matriz Inversa. Ej. 2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/Njt9dLxNbpk?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>Proposici\u00f3n:<\/strong> Dadas las matrices cuadradas regulares del mismo orden \\(A\\) y \\(B\\)<\/p>\n<ol>\n<li>\\((A^t)^{-1}=(A^{-1})^t\\)<\/li>\n<li>\\((A\\,B)^{-1}=B^{-1}\\cdot A^{-1}\\)<\/li>\n<\/ol>\n<\/blockquote>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Para todo \\(p\\in\\mathbb{Z}^+\\) y  \\(A\\in\\mathcal{M}_n(\\mathbb{K})\\) un matriz regular; entonces, la inversa de \\(A^p\\) es \\(\\left(A^{-1}\\right)^p\\). \u00bfVerdadero o falso? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv5() {\n  var htmlShow5 = document.getElementById(\"html-show5\");\n  if (htmlShow5.style.display === \"none\") {\n    htmlShow5.style.display = \"block\";\n  } else {\n    htmlShow5.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv5()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show5\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal -  Propiedades de las matrices regulares -Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/v8zybNqJ7dA?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>pseudoinversa<\/h2>\n<p>No siempre podemos conseguir la inversa, bien por que la matriz no sea cuadrada o por que no tenga. Entonces tenemos que plantearnos la posibilidad de encontrar una matriz, para cualquier matriz \\(A\\in\\mathcal{M}_{m\\times n}(\\mathbb{R})\\), talque<br \/>\n\\[AR=I_m\\] o \\[LA=I_n.\\]<br \/>\nEn caso de existir, denominamos a \\(R\\in\\mathcal{M}_{n\\times m}(\\mathbb{R})\\), matriz pseudoinversa por la derecha de la matriz \\(A\\); y a \\[L\\in\\mathcal{M}_{n\\times m}(\\mathbb{R}),\\] matriz pseudoinversa por la izquierda de la matriz \\(A\\).<\/p>\n<p>Un resultado que utilizaremos:<\/p>\n<blockquote>\n<p><strong>Proposici\u00f3n:<\/strong> Una matriz \\(A\\in\\mathcal{M}_{m\\times n}(\\mathbb{R})\\) tiene pseudoinversa por la derecha(izquierda) si, y s\u00f3lo si, \\(rang(A)=m\\) (\\(rang(A)=n\\))<\/p>\n<\/blockquote>\n<p>En caso de existir la pseudoinversa, entonces esta la calcularemos mediante \\[R=A^t(AA^t)^{-1},\\]<br \/>\no<br \/>\n\\[L=(A^tA)^{-1}A^t.\\]<\/p>\n<p>&nbsp;<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\(A=\\begin{bmatrix} 1&#038;0\\\\ 0&#038;1 \\\\ -1&#038;1<br \/>\n\\end{bmatrix},\\) si \\(B=[b_{ij}]\\) es su pseudoinversa por la izquierda \u00bfcu\u00e1nto es \\(\\sum b_{ii}\\)?\n<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2() {\n  var htmlShow2 = document.getElementById(\"html-show2\");\n  if (htmlShow2.style.display === \"none\") {\n    htmlShow2.style.display = \"block\";\n  } else {\n    htmlShow2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Pseudoinversa. Ej.2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/xSjyb5d8zi0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dada la matriz \\(A=\\begin{bmatrix} 1&#038;-1&#038;0\\\\ 0&#038;1 &#038;1<br \/>\n\\end{bmatrix},\\) si \\(B=[b_{ij}]\\) es su pseudoinversa por la derecha \u00bfcu\u00e1nto es \\(\\sum b_{ii}\\)?\n<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv3() {\n  var htmlShow3 = document.getElementById(\"html-show3\");\n  if (htmlShow3.style.display === \"none\") {\n    htmlShow3.style.display = \"block\";\n  } else {\n    htmlShow3.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv3()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show3\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Pseudoinversa de una matriz. Ej.3 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/zKzXVfzhmaY?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<h2>Interpretaci\u00f3n de la Pseudoinversa<\/h2>\n<p>Como $A$ tiene una inversa por la derecha, su rango debe ser igual al n\u00famero de filas (es de rango completo por filas). Por el teorema del rango-nulidad, esto implica que el <strong>espacio nulo<\/strong> (o n\u00facleo) de $A$, denotado como $\\text{Nul}(A)$, tiene una dimensi\u00f3n $\\text{dim}(\\text{Nul}(A)) > 0$ si $A$ no es cuadrada (m\u00e1s columnas que filas).<\/p>\n<p>La soluci\u00f3n general de la ecuaci\u00f3n lineal $A \\cdot X = b$ se expresa como:<\/p>\n<p>$$\\mathbf{X} = \\mathbf{X_p} + \\mathbf{X_h}$$<\/p>\n<p>Donde:<\/p>\n<ul>\n<li>$\\mathbf{X_p}$ es una soluci\u00f3n particular (como $R \\cdot b$).<\/li>\n<li>$\\mathbf{X_h}$ es la soluci\u00f3n general de la ecuaci\u00f3n homog\u00e9nea $A \\cdot X = 0$, es decir, $\\mathbf{X_h} \\in \\text{Nul}(A)$.<\/li>\n<\/ul>\n<p>Sustituyendo $\\mathbf{X_p} = R \\cdot b$, la <strong>soluci\u00f3n general<\/strong> es:<\/p>\n<p>\\[\\mathbf{X} = R \\cdot b + \\text{Nul}(A)\\]<\/p>\n<p>Esto significa que:<\/p>\n<ul>\n<li> La matriz $R$ te da un punto de partida (una soluci\u00f3n particular) para el sistema.<\/li>\n<li><strong>Existe un n\u00famero infinito de soluciones<\/strong> (siempre que $\\text{dim}(\\text{Nul}(A)) > 0$), ya que puedes sumar cualquier vector del espacio nulo de $A$ a la soluci\u00f3n particular $R \\cdot b$. <\/li>\n<\/ul>\n<p>La existencia de una pseudoinversa por la izquierda $L$ implica que la matriz $A$ es de <strong>rango completo por columnas<\/strong>. Esto, a su vez, significa que la aplicaci\u00f3n lineal representada por $A$ es <strong>inyectiva<\/strong> (o uno a uno).<br \/>\n\\[A \\cdot X = b\\to L\\cdot (A \\cdot X) = L\\cdot b\\to (L\\cdot A) \\cdot X = L\\cdot b\\to\\mathbf{X} = L \\cdot b\\]<br \/>\nEl resultado $\\mathbf{X} = L \\cdot b$ tiene dos implicaciones cr\u00edticas:<\/p>\n<ul>\n<li>Si la ecuaci\u00f3n $A \\cdot X = b$ tiene una soluci\u00f3n, <strong>esta soluci\u00f3n es \u00fanica<\/strong>. sto se debe a que la inyectividad de $A$ significa que el <strong>espacio nulo<\/strong> (n\u00facleo) de $A$ solo contiene el vector cero, es decir, $\\text{Nul}(A) = \\{\\mathbf{0}\\}$. No hay vectores no triviales para sumar a la soluci\u00f3n particular.<\/li>\n<li>a soluci\u00f3n $X = L \\cdot b$ es v\u00e1lida <strong>si y solo si<\/strong> el vector $b$ se encuentra en el <strong>espacio columna<\/strong> (o imagen) de $A$, denotado como $\\text{Col}(A)$.<\/li>\n<\/ul>\n<h2>Resumen<\/h2>\n<table border=\"1\" style=\"width: 100%; border-collapse: collapse; text-align: center;\">\n<thead>\n<tr style=\"background-color: #f2f2f2;\">\n<th>Caso<\/th>\n<th>Pseudoinversa<\/th>\n<th>Ecuaci\u00f3n $A \\cdot X = b$<\/th>\n<th>Implicaci\u00f3n y Existencia de Soluciones<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>Izquierda<\/strong><\/td>\n<td>$L \\cdot A = I$<\/td>\n<td>$\\mathbf{X = L \\cdot b}$ (soluci\u00f3n <strong>\u00fanica<\/strong>)<\/td>\n<td>$A$ es <strong>inyectiva<\/strong> (columna completa). $A \\cdot X = b$ tiene <strong>como m\u00e1ximo<\/strong> una soluci\u00f3n (si $b$ est\u00e1 en la Columna de $A$).<\/td>\n<\/tr>\n<tr>\n<td><strong>Derecha<\/strong><\/td>\n<td>$A \\cdot R = I$<\/td>\n<td>$\\mathbf{X = R \\cdot b} + \\text{Nul}(A)$ (soluciones <strong>infinitas<\/strong>)<\/td>\n<td>$A$ es <strong>sobreyectiva<\/strong> (fila completa). $A \\cdot X = b$ tiene <strong>siempre<\/strong> al menos una soluci\u00f3n.<\/td>\n<\/tr>\n<tr>\n<td><strong>Ambos<\/strong><\/td>\n<td>$L = R = A^{-1}$<\/td>\n<td>$\\mathbf{X = A^{-1} \\cdot b}$ (soluci\u00f3n <strong>\u00fanica<\/strong>)<\/td>\n<td>$A$ es <strong>cuadrada y regular<\/strong>. Soluci\u00f3n \u00fanica y siempre existente.<\/td>\n<\/tr>\n<tr>\n<td><strong>Ninguno(*)<\/strong><\/td>\n<td>$A^+$ (Moore-Penrose)<\/td>\n<td>$\\mathbf{X = A^+ \\cdot b}$ (soluci\u00f3n de <strong>m\u00ednimos cuadrados<\/strong>)<\/td>\n<td>$A$ es de rango distinto a columnas o filas. Se busca la soluci\u00f3n que minimiza el error $\\|A \\cdot X &#8211; b\\|$.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>(*)Este caso no lo trataremos, aunque m\u00e1s adelante veremos la soluci\u00f3n por <strong>m\u00ednimos cuadrados<\/strong>.<\/p>\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 \/>\nSea \\(A=\\left[\\begin{smallmatrix}1&#038;3&#038;-2&#038;0&#038;2&#038;0\\\\ 2&#038;6&#038;-5&#038;-2&#038;4&#038;-3\\\\ 0&#038;0&#038;5&#038;10&#038;0&#038;15\\\\2&#038;6&#038;0&#038;8&#038;4&#038;8\\end{smallmatrix}\\right]\\), el rango de la matriz es<\/tr>\n<tr>\n<td>\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() {\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>B.)<\/strong>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Definimos la inversa de una matriz cuadrada \\(A=[a_{ij}]\\in \\mathcal{M}_{n}(\\mathbb{R} o \\mathbb{C})\\) como la matriz \\(B=[b_{ij}]\\in \\mathcal{M}_{n}(\\mathbb{R} o \\mathbb{C})\\) tal que \\[AB=BA=I_n.\\] El procedimiento que damos para calcular la inversa, es el de&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"class_list":["post-157","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/157","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=157"}],"version-history":[{"count":9,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/157\/revisions"}],"predecessor-version":[{"id":253,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/157\/revisions\/253"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}