{"id":266,"date":"2025-10-15T11:45:42","date_gmt":"2025-10-15T09:45:42","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=266"},"modified":"2025-10-15T17:30:34","modified_gmt":"2025-10-15T15:30:34","slug":"biomath-autovectores-y-autovalores","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=266","title":{"rendered":"MathBio: Autovectores y autovalores"},"content":{"rendered":"<p>Denominamos esta parte autovectores y autovalores, tambi\u00e9n conocidos como vectores y valores propios de una matriz. Su definici\u00f3n es simple:<\/p>\n<blockquote>\n<p>Dada una matriz, \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\), real porque es el principal cuerpo que trataremos, decimos que \\(\\vec{v} \\in\\mathbb{R}^n\\) es un autovector, o vector propio de la matriz, si \\[A\\vec{v}=\\lambda\\vec{v},\\] siendo \\(\\lambda\\in\\mathbb{R}\\). Al real \\(\\lambda\\) se le denomina autovalor o valor propio de la matriz.<\/p>\n<\/blockquote>\n<p>Nuestro principal prop\u00f3sito es saber determinar los autovectores y autovalores de una matriz real cuadrada.<\/p>\n<p>Para conseguir nuestro prop\u00f3sito necesitamos encontrar las soluciones de la ecuaci\u00f3n que plantea el determinante \\[det(A-\\lambda\\, I),\\] siendo \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\) la matriz cuadrada y \\(I\\) la indentidad en \\(\\mathcal{M}_n(\\mathbb{R})\\).<\/p>\n<p>El polinomio \\(p(\\lambda) = det(A -\\lambda\\,I)\\) es el polinomio caracter\u00edstico de A: los valores propios de una matriz son los ceros de su polinomio caracter\u00edstico (soluciones de la ecuaci\u00f3n caracter\u00edstica).<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovalores de la matriz \\[\\begin{bmatrix}5 &#038; 0 &#038; 4\\\\ 4 &#038; -1 &#038; 4\\\\ -8 &#038; 0 &#038; -7\\end{bmatrix}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2d() {\n  var htmlShow2d = document.getElementById(\"html-show2d\");\n  if (htmlShow2d.style.display === \"none\") {\n    htmlShow2d.style.display = \"block\";\n  } else {\n    htmlShow2d.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2d()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2d\" style=\"display: none;\">\nTenemos que determinar <\/p>\n<p>\\[\\begin{align*}\\mathbf{det}(A-\\lambda I_3)&#038;=\\begin{vmatrix}-\\lambda+5 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; -\\lambda-1 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -7-\\lambda\\end{vmatrix}\\\\ &#038;=-\\lambda^3-3\\lambda^2+\\lambda+3\\\\ &#038;=(\\lambda+3)(\\lambda+1)(\\lambda-1)\\end{align*}\\]<br \/>\nPor tanto, los autovalores son \\(\\lambda_1=3\\), \\(\\lambda_2=-1\\) y \\(\\lambda_3=1\\).\n<\/p><\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovalores de la matriz \\[\\begin{bmatrix}1 &#038; 1 &#038; 0\\\\<br \/>\n2 &#038; 0 &#038; 2\\\\ -3 &#038; -2 &#038; -1\\end{bmatrix}\\]<\/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 <!-- 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\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">matrix<\/span>([<span class=\"code_number\">1<\/span>,<span class=\"code_number\">1<\/span>,<span class=\"code_number\">0<\/span>],[<span class=\"code_number\">2<\/span>,<span class=\"code_number\">0<\/span>,<span class=\"code_number\">2<\/span>],[<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">3<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">2<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">1<\/span>])<span class=\"code_endofline\">$<\/span><br \/><span class=\"code_variable\">ecu_caract<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">expand<\/span>(<span class=\"code_function\">determinant<\/span>(<span class=\"code_variable\">A<\/span><span class=\"code_operator\">&#8211;<\/span><span class=\"code_variable\">x<\/span><span class=\"code_operator\">*<\/span><span class=\"code_function\">ident<\/span>(<span class=\"code_number\">3<\/span>)))<span class=\"code_operator\">=<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">;<\/span><br \/><span class=\"code_function\">solve<\/span>(<span class=\"code_variable\">ecu_caract<\/span>,[<span class=\"code_variable\">x<\/span>])<span class=\"code_endofline\">;<\/span> <\/span>  <\/td>\n<\/tr>\n<\/table>\n<p> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable>  <mlabeledtr columnalign=\"left\"> <mtd><mtext>(ecu_caract)<\/mtext> <\/mtd> <mtd><mi>\u2212<\/mi><msup>  <mi>x<\/mi>  <mn>3<\/mn><\/msup><mi>\u2212<\/mi><mi>x<\/mi><mi>=<\/mi><mn>0<\/mn> <\/mtd>  <\/mlabeledtr><\/mtable> <\/math> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable>  <mlabeledtr columnalign=\"left\"> <mtd><mtext>(%o3) <\/mtext> <\/mtd> <mtd><mo>[<\/mo><mi>x<\/mi><mi>=<\/mi><mi>\u2212<\/mi><mi>%i<\/mi><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mi>%i<\/mi><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mn>0<\/mn><mo>]<\/mo> <\/mtd>  <\/mlabeledtr><\/mtable> <\/math>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovalores de la matriz \\[\\begin{bmatrix}4 &#038; 3 &#038; 0\\\\<br \/>\n-6 &#038; -5 &#038; 0\\\\ 3 &#038; 3 &#038; 1\\end{bmatrix}\\]\n<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2a() {\n  var htmlShow2a = document.getElementById(\"html-show2a\");\n  if (htmlShow2a.style.display === \"none\") {\n    htmlShow2a.style.display = \"block\";\n  } else {\n    htmlShow2a.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2a\" style=\"display: none;\">\n<!-- 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\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">matrix<\/span>([<span class=\"code_number\">4<\/span>,<span class=\"code_number\">3<\/span>,<span class=\"code_number\">0<\/span>],[<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">6<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">5<\/span>,<span class=\"code_number\">0<\/span>],[<span class=\"code_number\">3<\/span>,<span class=\"code_number\">3<\/span>,<span class=\"code_number\">1<\/span>])<span class=\"code_endofline\">$<\/span><br \/><span class=\"code_variable\">ecu_caract<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">expand<\/span>(<span class=\"code_function\">determinant<\/span>(<span class=\"code_variable\">A<\/span><span class=\"code_operator\">&#8211;<\/span><span class=\"code_variable\">x<\/span><span class=\"code_operator\">*<\/span><span class=\"code_function\">ident<\/span>(<span class=\"code_number\">3<\/span>)))<span class=\"code_operator\">=<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">;<\/span><br \/><span class=\"code_function\">solve<\/span>(<span class=\"code_variable\">ecu_caract<\/span>,[<span class=\"code_variable\">x<\/span>])<span class=\"code_endofline\">;<\/span> <\/span>  <\/td>\n<\/tr>\n<\/table>\n<p> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable>  <mlabeledtr columnalign=\"left\"> <mtd><mtext>(ecu_caract)<\/mtext> <\/mtd> <mtd><mi>\u2212<\/mi><msup>  <mi>x<\/mi>  <mn>3<\/mn><\/msup><mo>+<\/mo><mn>3<\/mn><mo>\u2062<\/mo><mi>x<\/mi><mi>\u2212<\/mi><mn>2<\/mn><mi>=<\/mi><mn>0<\/mn> <\/mtd>  <\/mlabeledtr><\/mtable> <\/math> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable>  <mlabeledtr columnalign=\"left\"> <mtd><mtext>(%o3) <\/mtext> <\/mtd> <mtd><mo>[<\/mo><mi>x<\/mi><mi>=<\/mi><mi>\u2212<\/mi><mn>2<\/mn><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mn>1<\/mn><mo>]<\/mo> <\/mtd>  <\/mlabeledtr><\/mtable> <\/math>\n<\/div>\n<hr \/>\n<p>Cada valor propio tiene asociado un conjunto \\(C_\\lambda=\\{\\vec{v}\\in\\mathbb{R}^n\\}\\), que se determina resolviendo el sistema homogeneo \\((A-\\lambda\\, I)\\vec{x}=\\vec{0}\\). Las soluciones de estos sistemas ser\u00e1n los vectores propios de la matriz, que veremos m\u00e1s adelante.<\/p>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovectores del autovalor \\(\\lambda=-3\\) de la matriz \\[\\begin{bmatrix}5 &#038; 0 &#038; 4\\\\ 4 &#038; -1 &#038; 4\\\\ -8 &#038; 0 &#038; -7\\end{bmatrix}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2d45() {\n  var htmlShow2d45 = document.getElementById(\"html-show2d45\");\n  if (htmlShow2d45.style.display === \"none\") {\n    htmlShow2d45.style.display = \"block\";\n  } else {\n    htmlShow2d45.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2d45()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2d45\" style=\"display: none;\">\nPara calcular los autovectores del autovalor \\(\\lambda=-3\\), planteamos el sistema<\/p>\n<p>\\[(A-\\lambda I_3)X=0\\to \\begin{bmatrix}-3+5 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; -3-1 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -7-3\\end{bmatrix}\\begin{bmatrix}x\\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0\\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Utilicemos las matrices como hecho en casos anteriores para resolver este sistema:<br \/>\n\\[\\begin{bmatrix}8 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; 2 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -4\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}8 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; 2 &#038; 4\\\\<br \/>\n0 &#038; 0 &#038; 0\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}2 &#038; 0 &#038; 1\\\\<br \/>\n2 &#038; 1 &#038; 2\\\\<br \/>\n0 &#038; 0 &#038; 0\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}2 &#038; 0 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 1\\\\<br \/>\n0 &#038; 0 &#038; 0\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Luego, \\(x=-\\frac{1}{2}\\mu\\), \\(y=-\\mu\\) y \\(z=\\mu\\), por tanto el subespacio asociado al autovalor \\(\\lambda=-3\\) es \\(\\mathcal{C}_{-3}=\\mathbf{Gen}\\{[-1,-2,2]\\}\\) <\/p>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovectores del autovalor \\(\\lambda=-1\\) de la matriz \\[\\begin{bmatrix}5 &#038; 0 &#038; 4\\\\ 4 &#038; -1 &#038; 4\\\\ -8 &#038; 0 &#038; -7\\end{bmatrix}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2d4() {\n  var htmlShow2d4 = document.getElementById(\"html-show2d4\");\n  if (htmlShow2d4.style.display === \"none\") {\n    htmlShow2d4.style.display = \"block\";\n  } else {\n    htmlShow2d4.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2d4()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2d4\" style=\"display: none;\">\nPara calcular los autovectores del autovalor \\(\\lambda=-1\\), planteamos el sistema<\/p>\n<p>\\[(A-\\lambda I_3)X=0\\to \\begin{bmatrix}1+5 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; -1+1 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -7+1\\end{bmatrix}\\begin{bmatrix}x\\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0\\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Utilicemos las matrices como hecho en casos anteriores para resolver este sistema:<br \/>\n\\[\\begin{bmatrix}6 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; 0 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -6\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}3 &#038; 0 &#038; 2\\\\<br \/>\n1 &#038; 0 &#038; 1\\\\<br \/>\n-4 &#038; 0 &#038; -3\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}0 &#038; 0 &#038; -1\\\\<br \/>\n1 &#038; 0 &#038; 1\\\\<br \/>\n-4 &#038; 0 &#038; -3\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Luego, \\(x=0\\) y \\(z=0\\), por tanto el subespacio asociado al autovalor \\(\\lambda=-1\\) es \\(\\mathcal{C}_{-1}=\\mathbf{Gen}\\{[0,1,0]\\}\\) <\/p>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovectores del autovalor \\(\\lambda=1\\) de la matriz \\[\\begin{bmatrix}5 &#038; 0 &#038; 4\\\\ 4 &#038; -1 &#038; 4\\\\ -8 &#038; 0 &#038; -7\\end{bmatrix}\\]<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2d4w() {\n  var htmlShow2d4w = document.getElementById(\"html-show2d4w\");\n  if (htmlShow2d4w.style.display === \"none\") {\n    htmlShow2d4w.style.display = \"block\";\n  } else {\n    htmlShow2d4w.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv2d4w()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2d4w\" style=\"display: none;\">\nPara calcular los autovectores del autovalor \\(\\lambda=1\\), planteamos el sistema<\/p>\n<p>\\[(A-\\lambda I_3)X=0\\to \\begin{bmatrix}5-1 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; -1-1 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -7-1\\end{bmatrix}\\begin{bmatrix}x\\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0\\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Utilicemos las matrices como hecho en casos anteriores para resolver este sistema:<br \/>\n\\[\\begin{bmatrix}4 &#038; 0 &#038; 4\\\\<br \/>\n4 &#038; -2 &#038; 4\\\\<br \/>\n-8 &#038; 0 &#038; -8\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}4 &#038; 0 &#038; 4\\\\<br \/>\n0 &#038; -2 &#038; 0\\\\<br \/>\n0 &#038; 0 &#038; 0\\end{bmatrix}\\sim<br \/>\n\\begin{bmatrix}1 &#038; 0 &#038; 1\\\\<br \/>\n0 &#038; 1 &#038; 0\\\\<br \/>\n0 &#038; 0 &#038; 0\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Luego, \\(y=0\\) y \\(x=-z\\), por tanto el subespacio asociado al autovalor \\(\\lambda=1\\) es \\(\\mathcal{C}_{1}=\\mathbf{Gen}\\{[-1,0,1]\\}\\) <\/p>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Determinar los autovectores de la matriz \\[\\begin{bmatrix}4 &#038; 0 &#038; -3\\\\<br \/>\n3 &#038; -1 &#038; 3\\\\<br \/>\n6 &#038; 0 &#038; 5\\end{bmatrix}\\] correspondientes a su autovalor real.<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2b() {\n  var htmlShow2b = document.getElementById(\"html-show2b\");\n  if (htmlShow2b.style.display === \"none\") {\n    htmlShow2b.style.display = \"block\";\n  } else {\n    htmlShow2b.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2b\" style=\"display: none;\">\n<!-- 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\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">matrix<\/span>([<span class=\"code_number\">4<\/span>,<span class=\"code_number\">0<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">3<\/span>],[<span class=\"code_number\">3<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">1<\/span>,<span class=\"code_number\">3<\/span>],[<span class=\"code_number\">6<\/span>,<span class=\"code_number\">0<\/span>,<span class=\"code_number\">5<\/span>])<span class=\"code_endofline\">$<\/span><br \/><span class=\"code_variable\">ecu_caract<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">expand<\/span>(<span class=\"code_function\">determinant<\/span>(<span class=\"code_variable\">A<\/span><span class=\"code_operator\">&#8211;<\/span><span class=\"code_variable\">x<\/span><span class=\"code_operator\">*<\/span><span class=\"code_function\">ident<\/span>(<span class=\"code_number\">3<\/span>)))<span class=\"code_operator\">=<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">;<\/span><br \/><span class=\"code_function\">solve<\/span>(<span class=\"code_variable\">ecu_caract<\/span>,[<span class=\"code_variable\">x<\/span>])<span class=\"code_endofline\">;<\/span> <\/span><\/td>\n<\/tr>\n<\/table>\n<p> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable><mlabeledtr columnalign=\"left\"> <mtd><mtext>(ecu_caract)<\/mtext> <\/mtd> <mtd><mi>\u2212<\/mi><msup><mi>x<\/mi><mn>3<\/mn><\/msup><mo>+<\/mo><mn>8<\/mn><mo>\u2062<\/mo><msup><mi>x<\/mi><mn>2<\/mn><\/msup><mi>\u2212<\/mi><mn>29<\/mn><mo>\u2062<\/mo><mi>x<\/mi><mi>\u2212<\/mi><mn>38<\/mn><mi>=<\/mi><mn>0<\/mn> <\/mtd><\/mlabeledtr><\/mtable> <\/math> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable><mlabeledtr columnalign=\"left\"> <mtd><mtext>(%o3) <\/mtext> <\/mtd> <mtd><mo>[<\/mo><mi>x<\/mi><mi>=<\/mi><mi>\u2212<\/mi><mfrac><mrow> <msqrt><mn>71<\/mn> <\/msqrt> <mo>\u2062<\/mo> <mi>%i<\/mi> <mi>\u2212<\/mi> <mn>9<\/mn><\/mrow><mn>2<\/mn><\/mfrac><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mfrac><mrow> <msqrt><mn>71<\/mn> <\/msqrt> <mo>\u2062<\/mo> <mi>%i<\/mi> <mo>+<\/mo> <mn>9<\/mn><\/mrow><mn>2<\/mn><\/mfrac><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mi>\u2212<\/mi><mn>1<\/mn><mo>]<\/mo> <\/mtd><\/mlabeledtr><\/mtable> <\/math><\/p>\n<p>El \u00fanico autovalor real es \\(\\lambda=-1\\), luego el autovector ser\u00e1 la base del subespacio dado por \\[\\begin{bmatrix}5 &#038; 0 &#038; -3\\\\ 3 &#038; 0 &#038; 3\\\\ 6 &#038; 0 &#038; 6\\end{bmatrix}\\begin{bmatrix}x \\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0 \\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Mediante operaciones elementales, el sistema anterior es equivalente a \\[\\begin{bmatrix}1 &#038; 0 &#038; 0\\\\<br \/>\n1 &#038; 0 &#038; 1\\\\ 0 &#038; 0 &#038; 0\\end{bmatrix}\\begin{bmatrix}x \\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0 \\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Luego, \\(x=0\\) y \\(z=0\\), por tanto el subespacio asociado al autovalor \\(\\lambda=-1\\) es \\(\\mathcal{C}_{-1}=\\mathbf{Gen}\\{[0,1,0]\\}\\) <\/p>\n<\/div>\n<hr \/>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> Dados los autovectores de la matriz \\[\\begin{bmatrix}-4 &#038; 0 &#038; -3\\\\<br \/>\n3 &#038; -1 &#038; 3\\\\<br \/>\n6 &#038; 0 &#038; 5\\end{bmatrix}\\] del mayor autovalor real, \u00bfcu\u00e1l es el producto escalar de [1,1,1] por uno de dichos vectores unitarios?.<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv2c() {\n  var htmlShow2c = document.getElementById(\"html-show2c\");\n  if (htmlShow2c.style.display === \"none\") {\n    htmlShow2c.style.display = \"block\";\n  } else {\n    htmlShow2c.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv2c()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show2c\" style=\"display: none;\">\n<!-- 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\">A<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">matrix<\/span>([<span class=\"code_number\">-4<\/span>,<span class=\"code_number\">0<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">3<\/span>],[<span class=\"code_number\">3<\/span>,<span class=\"code_number\">&#8211;<\/span><span class=\"code_number\">1<\/span>,<span class=\"code_number\">3<\/span>],[<span class=\"code_number\">6<\/span>,<span class=\"code_number\">0<\/span>,<span class=\"code_number\">5<\/span>])<span class=\"code_endofline\">$<\/span><br \/><span class=\"code_variable\">ecu_caract<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">expand<\/span>(<span class=\"code_function\">determinant<\/span>(<span class=\"code_variable\">A<\/span><span class=\"code_operator\">&#8211;<\/span><span class=\"code_variable\">x<\/span><span class=\"code_operator\">*<\/span><span class=\"code_function\">ident<\/span>(<span class=\"code_number\">3<\/span>)))<span class=\"code_operator\">=<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">;<\/span><br \/><span class=\"code_function\">solve<\/span>(<span class=\"code_variable\">ecu_caract<\/span>,[<span class=\"code_variable\">x<\/span>])<span class=\"code_endofline\">;<\/span> <\/span><\/td>\n<\/tr>\n<\/table>\n<p><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable><mlabeledtr columnalign=\"left\"> <mtd><mtext>(ecu_caract)<\/mtext> <\/mtd> <mtd><mi>\u2212<\/mi><msup><mi>x<\/mi><mn>3<\/mn><\/msup> <mo>+<\/mo><mn>3<\/mn><mo>\u2062<\/mo><mi>x<\/mi><mo>+<\/mo><mn>2<\/mn><mi>=<\/mi><mn>0<\/mn> <\/mtd><\/mlabeledtr><\/mtable> <\/math> <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><mtable><mlabeledtr columnalign=\"left\"> <mtd><mtext>(%o3) <\/mtext> <\/mtd> <mtd><mo>[<\/mo><mi>x<\/mi><mi>=<\/mi><mn>2<\/mn><mo><mo>,<\/mo><mi>x<\/mi><mi>=<\/mi><mi>\u2212<\/mi><mn>1<\/mn><mo>]<\/mo> <\/mtd><\/mlabeledtr><\/mtable> <\/math><\/p>\n<p>El autovalor real mayor es \\(\\lambda=2\\), luego los autovectores ser\u00e1n la base del subespacio dado por \\[\\begin{bmatrix}-6 &#038; 0 &#038; -3\\\\ 3 &#038; -3 &#038; 3\\\\ 6 &#038; 0 &#038; 3\\end{bmatrix}\\begin{bmatrix}x \\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0 \\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Mediante operaciones elementales, el sistema anterior es equivalente a \\[\\begin{bmatrix}1 &#038; 0 &#038; \\frac{1}{2}\\\\<br \/>\n0 &#038; -1 &#038; \\frac{1}{2}\\\\ 0 &#038; 0 &#038; 0\\end{bmatrix}\\begin{bmatrix}x \\\\ y\\\\ z\\end{bmatrix}=\\begin{bmatrix}0 \\\\ 0\\\\ 0\\end{bmatrix}\\]<\/p>\n<p>Luego, \\(x=\\mu\\frac{-1}{2}\\), \\(y=\\mu\\frac{1}{2}\\) y \\(z=\\mu\\), por tanto el subespacio asociado al autovalor \\(\\lambda=2\\) es \\(\\mathcal{C}_{2}=\\mathbf{Gen}\\{[-1,1,2]\\}\\)<\/p>\n<p>Ahora solo tenemos que calcular \\[\\frac{1}{\\|[-1,1,2]\\|}([1,1,1]\\bullet[-1,1,2])=\\frac{\\sqrt{6}}{3}\\]<\/p>\n<\/div>\n<hr \/>\n<p>Pod\u00e9is ver m\u00e1s ejemplos en <a href=\"http:\/\/en.wikibooks.org\/wiki\/Linear_Algebra\/Eigenvalues_and_Eigenvectors\" target=\"_blank\" rel=\"noopener noreferrer\">Linear Algebra\/Eigenvalues and Eigenvectors<\/a>.<\/p>\n<p>Dado \\({A} \\in\\mathcal{C}_n(\\mathbb {R} )\\), una matriz cuadrada con valores sobre un cuerpo \\(\\mathbb {R}\\), decimos que \\({A}\\) es diagonalizable si, y s\u00f3lo si, \\(\\mathbf{A}\\) se puede descomponer de la forma: \\[{A}=\\mathbf{P}{D}\\mathbf{P}^{-1},\\]<br \/>\ndonde \\({D}\\) es una matriz diagonal.<\/p>\n<p>El proceso de diagonalizaci\u00f3n de una matriz necesita conocer los autovalores y autovectores de la matriz.<\/p>\n<p>Cada valor propio tiene asociado un conjunto \\(\\mathcal{C}_\\lambda=\\{\\vec{v}\\in\\mathbb{R}^n|A\\vec{v}=\\lambda\\vec{v}\\}\\), que se determina resolviendo el sistema homog\u00e9neo \\((A-\\lambda\\, I)\\vec{x}=\\vec{0}\\). Las soluciones de estos sistemas ser\u00e1n los vectores propios de la matriz.<\/p>\n<p>As\u00ed al n\u00famero de veces que un autovalor \\(\\lambda\\) se repite como ra\u00edz del polinomio caracter\u00edstico se le llama multiplicidad algebraica y se representa por \\(m_a(\\lambda)\\). Y al n\u00famero m\u00e1ximo de autovectores linealmente independientes que tiene asociado un autovalor \\(\\lambda\\), es decir la dimensi\u00f3n del subespacio propio \\(\\mathcal{C}_\\lambda\\), se le llama multiplicidad geom\u00e9trica de \\(\\lambda\\) y se representa por \\(m_g(\\lambda)\\). Estos dos n\u00fameros est\u00e1n relacionados por una desigualdad: \\[m_g(\\lambda)\\leqslant m_a(\\lambda)\\]<\/p>\n<p>El proceso de diagonalizaci\u00f3n de una matriz necesita conocer los autovalores y autovectores de la matriz. Sea, por tanto, \\(\\mathbf{A}\\) una matriz cuadrada  de orden \\(n\\), y sean \\(\\lambda_i\\) los autovalores de dicha matriz. Entonces<\/p>\n<blockquote>\n<p>La matriz \\(\\mathbf{A}\\) es diagonalizable si, y s\u00f3lo si, se cumple: \\(a)\\) el n\u00famero de soluciones de la ecuaci\u00f3n caracter\u00edstica es igual a \\(n\\); \\(b)\\) para todo autovalor \\(\\lambda_i\\), la dimensi\u00f3n del subespacio \\(\\mathcal{C}_{\\lambda_i}\\) coincide con la multiplicidad del autovalor \\(\\lambda_i\\) como soluci\u00f3n de la ecuaci\u00f3n caracter\u00edstica de \\(A\\); es decir, \\(m_g(\\lambda_i)= m_a(\\lambda_i)\\)<\/p>\n<\/blockquote>\n<p>As\u00ed pues si  \\(\\mathbf{A}\\) es diagonalizable, ser\u00e1  \\(\\mathbf{D}\\) una matriz cuya diagonal principal est\u00e1 formada por los autovalores de \\(\\mathbf{A}\\) pareciendo cada uno tantas veces como indique su multiplicidad algebraica, y \\(\\mathbf{P}\\) es la matriz cuyas columnas son los autovectores; es decir, los vectores que constituyen una base del subespacio propio asociado a cada autovalor siguiendo el orden establecido en \\(\\mathbf{D}\\).<\/p>\n<blockquote>\n<p><strong>Propiedades<\/strong><\/p>\n<p>Toda matriz sim\u00e9trica de coeficientes reales es diagonalizable y sus autovalores son reales.<\/p>\n<p>Dadas dos matrices diagonalizables \\(\\mathbf{A}\\) y \\(\\mathbf{B}\\), son conmutables (\\(\\mathbf{AB}=\\mathbf{BA}\\)) si y solo si son simult\u00e1neamente diagonalizables (comparten la misma base ortonormal).<\/p>\n<p>Toda matriz \\(\\mathbf{A}\\) de dimensi\u00f3n \\(n\\) y coeficientes reales es diagonalizable si, y s\u00f3lo si, existe una base de \\(\\mathbb{R}^{n}\\) formada por autovectores de \\(\\mathbf{A}\\)<\/p>\n<\/blockquote>\n<p>El resultado anterior nos permite formular la definici\u00f3n de diagonalizaci\u00f3n ortogonal o matriz ortogonalmente diagonalizable: Una matriz cuadrada se dice que es ortogonalmente diagonalizable si y s\u00f3lo si es diagonalizable mediante una matriz de \\(\\mathbf{P}\\) ortogonal. Por tanto, si una matriz es ortogonalmente diagonalizable si y s\u00f3lo si se puede encontrar una base de \\(\\mathbb{R}^{n}\\) formada por autovectores ortonormales de \\(\\mathbf{A}\\) (que compondr\u00e1n las columnas de la matriz \\(\\mathbf{P}\\)).<\/p>\n<blockquote>\n<p><strong>Teorema<\/strong>: Una matriz real es ortogonalmente diagonalizable si, y solo si, es sim\u00e9trica.<\/p>\n<\/blockquote>\n<p>Observemos que, aunque podemos hacer ortogonales los autovectores conseguidos, no nos garantiza que la matriz formada por ellos sea ortogonal. Eso solo se producir\u00e1 en el caso de ser una matriz sim\u00e9trica.<\/p>\n<p>&nbsp;<\/p>\n<hr>\n<h3>Bibliograf\u00eda<\/h3>\n<ul>\n<li>Cap\u00edtulo 5 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>  \u00bfCu\u00e1ntos autovalores distintos tiene la matriz \\(A=\\begin{bmatrix}2&#038;1&#038;0\\\\ 1&#038;2&#038;0 \\\\ 0&#038;0&#038;1\\end{bmatrix}\\)?<\/p>\n<div id=\"menu-a\">\n<ul>\n<li>1<\/li>\n<li>2<\/li>\n<li>3<\/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\/Ejer_autovalores.html\" width=\"650\" height=\"300\" allow=\"fullscreen\"><\/iframe>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Denominamos esta parte autovectores y autovalores, tambi\u00e9n conocidos como vectores y valores propios de una matriz. Su definici\u00f3n es simple: Dada una matriz, \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\), real porque es el principal cuerpo que trataremos,&#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-266","post","type-post","status-publish","format-standard","hentry","category-mathbio"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/266","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=266"}],"version-history":[{"count":16,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/266\/revisions"}],"predecessor-version":[{"id":301,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/266\/revisions\/301"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}