{"id":476,"date":"2025-11-19T08:15:54","date_gmt":"2025-11-19T07:15:54","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=476"},"modified":"2025-11-11T23:18:32","modified_gmt":"2025-11-11T22:18:32","slug":"alg-ortonormalizacion-de-gram-schmidt","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=476","title":{"rendered":"ALG: Ortonormalizaci\u00f3n de Gram-Schmidt"},"content":{"rendered":"<p>Recordemos que, dado un espacio eucl\u00eddeo, \\((\\mathcal{E},\\bullet)\\), dos vectores se dicen ortogonales si<br \/>\n\\[\\vec{x}\\perp  \\vec{y} \\Leftrightarrow \\vec{x}\\bullet \\vec{y}=0\\]<br \/>\nCon esta definici\u00f3n, decimos que \\(B=\\{\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_n\\}\\) es un conjunto ortogonal si dos a dos sus vectores son ortogonales; es decir, \\(\\vec{v}_i\\bullet\\vec{v}_j=0\\forall i\\neq j\\)<\/p>\n<p>Nuestro prop\u00f3sito ser\u00e1, a partir de un conjunto de vectores, de un subespacio de un espacio eucl\u00eddeo, \\((\\mathcal{E},\\bullet)\\), obtener otro conjunto ortonormal de vectores que genere el mismo subespacio vectorial. Este proceso se denomina <a href=\"https:\/\/es.wikipedia.org\/wiki\/Proceso_de_ortogonalizaci%C3%B3n_de_Gram-Schmidt\">ortonormalizaci\u00f3n de Gram-Schmidt<\/a>.<\/p>\n<p>El proceso de ortonormalizaci\u00f3n de Gram\u2013Schmidt nos dice que si \\(B=\\{\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_n\\}\\), los vectores<\/p>\n<ul>\n<li>\\(\\vec{u}_1=\\vec{v}_1\\)<\/li>\n<li>\\(\\vec{u}_2=\\vec{v}_2-\\displaystyle{\\frac{\\vec{u}_1\\bullet \\vec{v}_2}{||\\vec{u}_1||^2}\\vec{u}_1}\\)<\/li>\n<li>\\(\\vec{u}_3=\\vec{v}_3-\\displaystyle{\\frac{\\vec{u}_1\\bullet \\vec{v}_3}{||\\vec{u}_1||^2}\\vec{u}_1}-\\displaystyle{\\frac{\\vec{u}_2\\bullet \\vec{v}_3}{||\\vec{u}_2||^2}\\vec{u}_2}\\)<\/li>\n<li>\\(\\vec{u}_k=\\vec{v}_k-\\displaystyle\\sum_{i=1}^{k-1}\\frac{\\vec{u}_i\\bullet \\vec{v}_k}{||\\vec{u}_i||^2}\\vec{u}_i\\)<\/li>\n<\/ul>\n<p>forman una base ortogonal del subespacio. De este modo, el siguiente conjunto es una base ortonormal del subespacio \\[\\left\\{\\frac{\\vec{u}_1}{||\\vec{u}_1||},\\frac{\\vec{u}_2}{||\\vec{u}_2||},\\ldots,\\frac{\\vec{u}_n}{||\\vec{u}_n||}\\right\\}\\]<\/p>\n<blockquote><p><strong>Ejemplo:<\/strong> Dada la base de \\(\\mathbb{R}^2\\)<br \/>\n\\[ B = \\{ \\mathbf{v}_1, \\mathbf{v}_2 \\} = \\left\\{ (1, -2), (1, 1) \\right\\} \\]<br \/>\nconstruya una base ortonormal \\(\\mathcal{B} = \\{ \\mathbf{e}_1, \\mathbf{e}_2 \\}\\) para \\(\\mathbb{R}^2\\) mediante el proceso de Gram-Schmidt.<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a13b() {\n  var htmlShow1a13b = document.getElementById(\"html-show1a13b\");\n  if (htmlShow1a13b.style.display === \"none\") {\n    htmlShow1a13b.style.display = \"block\";\n  } else {\n    htmlShow1a13b.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1a13b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a13b\" style=\"display: none;\">\n## 1. Construcci\u00f3n de la Base Ortogonal \\(\\{ \\mathbf{u}_1, \\mathbf{u}_2 \\}\\)<\/p>\n<p>### Paso 1: Definir el primer vector ortogonal \\(\\mathbf{u}_1\\)<\/p>\n<p>El primer vector de la base ortogonal es el primer vector de la base original:<\/p>\n<p>\\[<br \/>\n\\mathbf{u}_1 = \\mathbf{v}_1 = (1, -2)<br \/>\n\\]<\/p>\n<p>### Paso 2: Calcular el segundo vector ortogonal \\(\\mathbf{u}_2\\)<\/p>\n<p>El vector \\(\\mathbf{u}_2\\) se calcula restando a \\(\\mathbf{v}_2\\) su proyecci\u00f3n sobre \\(\\mathbf{u}_1\\).<\/p>\n<p>F\u00f3rmula de Gram-Schmidt:<br \/>\n\\[<br \/>\n\\mathbf{u}_2 = \\mathbf{v}_2 &#8211; \\frac{\\mathbf{v}_2 \\cdot \\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|^2} \\mathbf{u}_1<br \/>\n\\]<\/p>\n<p><strong>C\u00e1lculo del producto escalar<\/strong> \\(\\mathbf{v}_2 \\cdot \\mathbf{u}_1\\):<\/p>\n<p>\\[<br \/>\n\\mathbf{v}_2 \\cdot \\mathbf{u}_1 = (1, 1) \\cdot (1, -2) = (1)(1) + (1)(-2) = 1 &#8211; 2 = -1<br \/>\n\\]<\/p>\n<p><strong>C\u00e1lculo de la norma cuadrada de<\/strong> \\(\\mathbf{u}_1\\):<\/p>\n<p>\\[<br \/>\n\\| \\mathbf{u}_1 \\|^2 = (1)^2 + (-2)^2 = 1 + 4 = 5<br \/>\n\\]<\/p>\n<p><strong>C\u00e1lculo de<\/strong> \\(\\mathbf{u}_2\\):<\/p>\n<p>Sustituimos los valores:<br \/>\n\\[<br \/>\n\\mathbf{u}_2 = (1, 1) &#8211; \\frac{-1}{5} (1, -2)<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\mathbf{u}_2 = (1, 1) + \\frac{1}{5} (1, -2)<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\mathbf{u}_2 = \\left( 1 + \\frac{1}{5}, 1 &#8211; \\frac{2}{5} \\right) = \\left( \\frac{5}{5} + \\frac{1}{5}, \\frac{5}{5} &#8211; \\frac{2}{5} \\right) = \\left( \\frac{6}{5}, \\frac{3}{5} \\right)<br \/>\n\\]<\/p>\n<p>La base <strong>ortogonal<\/strong> obtenida es:<br \/>\n\\[<br \/>\nB&#8217; = \\left\\{ \\mathbf{u}_1, \\mathbf{u}_2 \\right\\} = \\left\\{ (1, -2), \\left( \\frac{6}{5}, \\frac{3}{5} \\right) \\right\\}<br \/>\n\\]<\/p>\n<p>&#8212;<\/p>\n<p>## 2. Normalizaci\u00f3n de la Base Ortogonal<\/p>\n<p>Dividimos cada vector ortogonal por su norma para obtener una longitud de 1.<\/p>\n<p>### Paso 3: Normalizar \\(\\mathbf{u}_1\\) (obteniendo \\(\\mathbf{e}_1\\))<\/p>\n<p><strong>C\u00e1lculo de la norma de<\/strong> \\(\\mathbf{u}_1\\):<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_1 \\| = \\sqrt{5}<br \/>\n\\]<\/p>\n<p><strong>C\u00e1lculo de <\/strong>\\(\\mathbf{e}_1\\):<br \/>\n\\[<br \/>\n\\mathbf{e}_1 = \\frac{\\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|} = \\frac{1}{\\sqrt{5}} (1, -2) = \\left( \\frac{1}{\\sqrt{5}}, &#8211; \\frac{2}{\\sqrt{5}} \\right)<br \/>\n\\]<\/p>\n<p>### Paso 4: Normalizar \\(\\mathbf{u}_2\\) (obteniendo \\(\\mathbf{e}_2\\))<\/p>\n<p><strong>C\u00e1lculo de la norma de<\/strong> \\(\\mathbf{u}_2\\):<\/p>\n<p>\\[<br \/>\n\\| \\mathbf{u}_2 \\| = \\sqrt{\\left( \\frac{6}{5} \\right)^2 + \\left( \\frac{3}{5} \\right)^2} = \\sqrt{\\frac{36}{25} + \\frac{9}{25}} = \\sqrt{\\frac{45}{25}} = \\frac{\\sqrt{9}\\sqrt{5}}{5} = \\frac{3\\sqrt{5}}{5} = \\frac{3}{\\sqrt{5}}<br \/>\n\\]<\/p>\n<p><strong>C\u00e1lculo de<\/strong> \\(\\mathbf{e}_2\\):<\/p>\n<p>\\[<br \/>\n\\mathbf{e}_2 = \\frac{\\mathbf{u}_2}{\\| \\mathbf{u}_2 \\|} = \\frac{1}{\\frac{3}{\\sqrt{5}}} \\left( \\frac{6}{5}, \\frac{3}{5} \\right)<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\mathbf{e}_2 = \\frac{\\sqrt{5}}{3} \\left( \\frac{6}{5}, \\frac{3}{5} \\right) = \\left( \\frac{\\sqrt{5}}{3} \\cdot \\frac{6}{5}, \\frac{\\sqrt{5}}{3} \\cdot \\frac{3}{5} \\right)<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\mathbf{e}_2 = \\left( \\frac{2\\sqrt{5}}{5}, \\frac{\\sqrt{5}}{5} \\right)<br \/>\n\\]<br \/>\n<em>Alternativamente, y sin racionalizar el denominador:<\/em><br \/>\n\\[<br \/>\n\\mathbf{e}_2 = \\left( \\frac{2}{\\sqrt{5}}, \\frac{1}{\\sqrt{5}} \\right)<br \/>\n\\]<\/p>\n<p>&#8212;<\/p>\n<p>## 3. Conclusi\u00f3n<\/p>\n<p>La base ortonormal de \\(\\mathbb{R}^2\\) obtenida mediante el proceso de Gram-Schmidt es:<\/p>\n<p>\\[<br \/>\n\\mathcal{B} = \\{ \\mathbf{e}_1, \\mathbf{e}_2 \\} = \\left\\{ \\left( \\frac{1}{\\sqrt{5}}, &#8211; \\frac{2}{\\sqrt{5}} \\right), \\left( \\frac{2}{\\sqrt{5}}, \\frac{1}{\\sqrt{5}} \\right) \\right\\}<br \/>\n\\]<\/p>\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Dada la base de \\(\\mathbb{R}^3\\)<br \/>\n\\[ B = \\{ \\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3 \\} = \\left\\{ (1, 0, 1), (1, 1, 0), (0, 2, -1) \\right\\} \\]<br \/>\nconstruya una base ortonormal \\(\\mathcal{B} = \\{ \\mathbf{e}_1, \\mathbf{e}_2, \\mathbf{e}_3 \\}\\) para \\(\\mathbb{R}^3\\) mediante el proceso de Gram-Schmidt.<\/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## 1. Construcci\u00f3n de la Base Ortogonal \\(\\{ \\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3 \\}\\)<\/p>\n<p>### Paso 1: Definir el primer vector ortogonal \\(\\mathbf{u}_1\\)<\/p>\n<p>Tomamos el primer vector de la base original:<br \/>\n\\[<br \/>\n\\mathbf{u}_1 = \\mathbf{v}_1 = (1, 0, 1)<br \/>\n\\]<br \/>\nCalculamos su norma cuadrada:<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_1 \\|^2 = 1^2 + 0^2 + 1^2 = 2<br \/>\n\\]<\/p>\n<p>### Paso 2: Calcular el segundo vector ortogonal \\(\\mathbf{u}_2\\)<\/p>\n<p>Restamos a \\(\\mathbf{v}_2\\) su proyecci\u00f3n sobre \\(\\mathbf{u}_1\\):<br \/>\n\\[<br \/>\n\\mathbf{u}_2 = \\mathbf{v}_2 &#8211; \\frac{\\mathbf{v}_2 \\cdot \\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|^2} \\mathbf{u}_1<br \/>\n\\]<\/p>\n<p>**C\u00e1lculos auxiliares:**<br \/>\n* Producto escalar: \\(\\mathbf{v}_2 \\cdot \\mathbf{u}_1 = (1, 1, 0) \\cdot (1, 0, 1) = 1\\)<\/p>\n<p>**Sustituci\u00f3n:**<br \/>\n\\[\\begin{split}<br \/>\n\\mathbf{u}_2 &#038;= (1, 1, 0) &#8211; \\frac{1}{2} (1, 0, 1)\\\\<br \/>\n &#038;= \\left( \\frac{1}{2}, 1, &#8211; \\frac{1}{2} \\right)<br \/>\n\\end{split}<br \/>\n\\]<br \/>\nCalculamos la norma cuadrada de \\(\\mathbf{u}_2\\):<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_2 \\|^2 = \\left( \\frac{1}{2} \\right)^2 + 1^2 + \\left( &#8211; \\frac{1}{2} \\right)^2 = \\frac{3}{2}<br \/>\n\\]<br \/>\n### Paso 3: Calcular el tercer vector ortogonal \\(\\mathbf{u}_3\\)<\/p>\n<p>Usaremos el nuevo vector $\\mathbf{v}_3 = (0, 2, -1)$ y la f\u00f3rmula:<br \/>\n\\[<br \/>\n\\mathbf{u}_3 = \\mathbf{v}_3 &#8211; \\frac{\\mathbf{v}_3 \\cdot \\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|^2} \\mathbf{u}_1 &#8211; \\frac{\\mathbf{v}_3 \\cdot \\mathbf{u}_2}{\\| \\mathbf{u}_2 \\|^2} \\mathbf{u}_2<br \/>\n\\]<\/p>\n<p>**C\u00e1lculos de Productos Escalares con el nuevo \\(\\mathbf{v}_3\\):**<\/p>\n<p>1.  Producto escalar \\(\\mathbf{v}_3 \\cdot \\mathbf{u}_1\\):<br \/>\n    \\[ \\mathbf{v}_3 \\cdot \\mathbf{u}_1 = (0, 2, -1) \\cdot (1, 0, 1) =  -1 \\]<\/p>\n<p>2.  Producto escalar \\(\\mathbf{v}_3 \\cdot \\mathbf{u}_2\\):<br \/>\n    \\[ \\mathbf{v}_3 \\cdot \\mathbf{u}_2 = (0, 2, -1) \\cdot \\left( \\frac{1}{2}, 1, &#8211; \\frac{1}{2} \\right) = \\frac{5}{2} \\]<\/p>\n<p>**C\u00e1lculo de los Coeficientes de Proyecci\u00f3n:**<\/p>\n<p>* Proyecci\u00f3n sobre \\(\\mathbf{u}_1\\): \\(\\frac{\\mathbf{v}_3 \\cdot \\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|^2} = \\frac{-1}{2}\\)<br \/>\n* Proyecci\u00f3n sobre \\(\\mathbf{u}_2\\): \\(\\frac{\\mathbf{v}_3 \\cdot \\mathbf{u}_2}{\\| \\mathbf{u}_2 \\|^2} = \\frac{5\/2}{3\/2} = \\frac{5}{3}\\)<\/p>\n<p>**Sustituci\u00f3n y C\u00e1lculo de \\(\\mathbf{u}_3\\):**<\/p>\n<p>\\[\\begin{split}<br \/>\n\\mathbf{u}_3 &#038;= (0, 2, -1) &#8211; \\left( &#8211; \\frac{1}{2} \\right) \\mathbf{u}_1 &#8211; \\left( \\frac{5}{3} \\right) \\mathbf{u}_2\\\\<br \/>\n &#038;= (0, 2, -1) + \\frac{1}{2} (1, 0, 1) &#8211; \\frac{5}{3} \\left( \\frac{1}{2}, 1, &#8211; \\frac{1}{2} \\right)\\\\<br \/>\n &#038;= \\left( &#8211; \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3} \\right)<br \/>\n\\end{split}<br \/>\n\\]<\/p>\n<p>## 2. Normalizaci\u00f3n de la Base Ortogonal \\(\\{ \\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3 \\}\\)<\/p>\n<p>### Paso 4: Normalizar \\(\\mathbf{u}_1\\) (obteniendo \\(\\mathbf{e}_1\\))<\/p>\n<p>**C\u00e1lculo de la norma de \\(\\mathbf{u}_1\\):**<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_1 \\| = \\sqrt{2}<br \/>\n\\]<br \/>\n**C\u00e1lculo de \\(\\mathbf{e}_1\\):**<br \/>\n\\[<br \/>\n\\mathbf{e}_1 = \\frac{\\mathbf{u}_1}{\\| \\mathbf{u}_1 \\|} = \\frac{1}{\\sqrt{2}} (1, 0, 1)<br \/>\n\\]<\/p>\n<p>### Paso 5: Normalizar \\(\\mathbf{u}_2\\) (obteniendo \\(\\mathbf{e}_2\\))<\/p>\n<p>**C\u00e1lculo de la norma de \\(\\mathbf{u}_2\\):**<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_2 \\| = \\sqrt{\\frac{3}{2}}<br \/>\n\\]<br \/>\n**C\u00e1lculo de \\(\\mathbf{e}_2\\):**<br \/>\n\\[\\begin{split}<br \/>\n\\mathbf{e}_2 = \\frac{\\mathbf{u}_2}{\\| \\mathbf{u}_2 \\|} &#038;= \\frac{1}{\\sqrt{3\/2}} \\left( \\frac{1}{2}, 1, &#8211; \\frac{1}{2} \\right)\\\\<br \/>\n &#038;= \\sqrt{\\frac{2}{3}} \\cdot \\frac{1}{2} (1, 2, -1) \\\\ &#038;= \\frac{1}{\\sqrt{6}} (1, 2, -1)<br \/>\n\\end{split}<br \/>\n\\]<\/p>\n<p>### Paso 6: Normalizar \\(\\mathbf{u}_3\\) (obteniendo \\(\\mathbf{e}_3\\))<\/p>\n<p>Primero, calculamos la norma cuadrada de \\(\\mathbf{u}_3\\):<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_3 \\|^2 = \\left( &#8211; \\frac{1}{3} \\right)^2 + \\left( \\frac{1}{3} \\right)^2 + \\left( \\frac{1}{3} \\right)^2 = \\frac{1}{3}<br \/>\n\\]<\/p>\n<p>**C\u00e1lculo de la norma de \\(\\mathbf{u}_3\\):**<br \/>\n\\[<br \/>\n\\| \\mathbf{u}_3 \\| = \\sqrt{\\frac{1}{3}} = \\frac{1}{\\sqrt{3}}<br \/>\n\\]<\/p>\n<p>**C\u00e1lculo de \\(\\mathbf{e}_3\\):**<br \/>\n\\[\\begin{split}<br \/>\n\\mathbf{e}_3 = \\frac{\\mathbf{u}_3}{\\| \\mathbf{u}_3 \\|} &#038;= \\frac{1}{1\/\\sqrt{3}} \\left( &#8211; \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3} \\right)\\\\<br \/>\n &#038;= \\sqrt{3} \\left( &#8211; \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3} \\right)\\\\ &#038;= \\left( &#8211; \\frac{\\sqrt{3}}{3}, \\frac{\\sqrt{3}}{3}, \\frac{\\sqrt{3}}{3} \\right)<br \/>\n\\end{split}<br \/>\n\\]<br \/>\n*(Alternativamente, sin racionalizar, podemos ver que $\\mathbf{e}_3 = \\frac{1}{\\sqrt{3}} (-1, 1, 1)$)*<\/p>\n<p>&#8212;<\/p>\n<p>## 3. Conclusi\u00f3n: Base Ortonormal para \\(\\mathbb{R}^3\\)<\/p>\n<p>La base **ortonormal** de $\\mathbb{R}^3$, \\(\\mathcal{B} = \\{ \\mathbf{e}_1, \\mathbf{e}_2, \\mathbf{e}_3 \\}\\), es la siguiente.<br \/>\n\\[<br \/>\n\\mathcal{B} = \\left\\{ \\frac{1}{\\sqrt{2}} (1, 0, 1), \\frac{1}{\\sqrt{6}} (1, 2, -1), \\frac{1}{\\sqrt{3}} (-1, 1, 1) \\right\\}<br \/>\n\\]\n<\/p><\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Dada la base de un subesapcio vectorial de \\(S \\subseteq\\mathcal{M}_2(\\mathbb{R})\\)<br \/>\n\\[ B = \\{ \\mathbf{V}_1, \\mathbf{V}_2, \\mathbf{V}_3 \\} = \\left\\{ \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; 1 \\\\ 1 &#038; 0 \\end{bmatrix} \\right\\} \\]<br \/>\nconstruya una base ortonormal \\(\\mathcal{B} = \\{ \\mathbf{e}_1, \\mathbf{e}_2, \\mathbf{e}_3 \\}\\) para \\(S\\) mediante el proceso de Gram-Schmidt.<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1a2b() {\n  var htmlShow1a2b = document.getElementById(\"html-show1a2b\");\n  if (htmlShow1a2b.style.display === \"none\") {\n    htmlShow1a2b.style.display = \"block\";\n  } else {\n    htmlShow1a2b.style.display = \"none\";\n  }\n}\n<\/script> <\/p>\n<p><button onclick=\"showHtmlDiv1a2b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1a2b\" style=\"display: none;\">\nUsaremos el producto interno de la traza: $\\langle A, B \\rangle = \\text{tr}(A^t B)= \\text{tr}(B^t A)$.<br \/>\n## 1. Construcci\u00f3n de la Base Ortogonal \\(\\{ \\mathbf{U}_1, \\mathbf{U}_2, \\mathbf{U}_3 \\}\\)<\/p>\n<p>### Paso 1: Definir la primera matriz ortogonal \\(\\mathbf{U}_1\\)<\/p>\n<p>\\[<br \/>\n\\mathbf{U}_1 = \\mathbf{V}_1 = \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix}<br \/>\n\\]<br \/>\nCalculamos su **norma cuadrada**:<br \/>\n\\[<br \/>\n\\| \\mathbf{U}_1 \\|^2 = 1^2 + (-1)^2 + 0^2 + 2^2 = 6<br \/>\n\\]<\/p>\n<p>### Paso 2: Calcular la segunda matriz ortogonal \\(\\mathbf{U}_2\\)<\/p>\n<p>F\u00f3rmula: $\\mathbf{U}_2 = \\mathbf{V}_2 &#8211; \\frac{\\langle \\mathbf{V}_2, \\mathbf{U}_1 \\rangle}{\\| \\mathbf{U}_1 \\|^2} \\mathbf{U}_1$<\/p>\n<p>**C\u00e1lculos auxiliares:**<br \/>\n* Producto escalar $\\langle \\mathbf{V}_2, \\mathbf{U}_1 \\rangle$:<br \/>\n    \\[<br \/>\n    \\left\\langle \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix} \\right\\rangle = (1)(1) + (1)(-1) + (0)(0) + (0)(2) = 0<br \/>\n    \\]<\/p>\n<p>**Resultado de $\\mathbf{U}_2$:**<br \/>\nPuesto que $\\langle \\mathbf{V}_2, \\mathbf{U}_1 \\rangle = 0$, la proyecci\u00f3n es nula. Esto significa que $\\mathbf{V}_1$ y $\\mathbf{V}_2$ ya eran **ortogonales** por casualidad.<br \/>\n\\[<br \/>\n\\mathbf{U}_2 = \\mathbf{V}_2 &#8211; \\frac{0}{6} \\mathbf{U}_1 = \\mathbf{V}_2 = \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Calculamos la **norma cuadrada** de $\\mathbf{U}_2$:<br \/>\n\\[<br \/>\n\\| \\mathbf{U}_2 \\|^2 = 1^2 + 1^2 + 0^2 + 0^2 = 2<br \/>\n\\]<\/p>\n<p>### Paso 3: Calcular la tercera matriz ortogonal \\(\\mathbf{U}_3\\)<\/p>\n<p>F\u00f3rmula: $\\mathbf{U}_3 = \\mathbf{V}_3 &#8211; \\frac{\\langle \\mathbf{V}_3, \\mathbf{U}_1 \\rangle}{\\| \\mathbf{U}_1 \\|^2} \\mathbf{U}_1 &#8211; \\frac{\\langle \\mathbf{V}_3, \\mathbf{U}_2 \\rangle}{\\| \\mathbf{U}_2 \\|^2} \\mathbf{U}_2$<\/p>\n<p>**C\u00e1lculos de Productos Escalares:**<br \/>\n1.  Producto escalar $\\langle \\mathbf{V}_3, \\mathbf{U}_1 \\rangle$:<br \/>\n    \\[<br \/>\n    \\left\\langle \\begin{bmatrix} 1 &#038; 1 \\\\ 1 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix} \\right\\rangle = (1)(1) + (1)(-1) + (1)(0) + (0)(2) = 1 &#8211; 1 + 0 + 0 = 0<br \/>\n    \\]<br \/>\n2.  Producto escalar $\\langle \\mathbf{V}_3, \\mathbf{U}_2 \\rangle$:<br \/>\n    \\[<br \/>\n    \\left\\langle \\begin{bmatrix} 1 &#038; 1 \\\\ 1 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix} \\right\\rangle = (1)(1) + (1)(1) + (1)(0) + (0)(0) = 1 + 1 + 0 + 0 = 2<br \/>\n    \\]<\/p>\n<p>**C\u00e1lculo de los Coeficientes de Proyecci\u00f3n:**<br \/>\n* Proyecci\u00f3n sobre $\\mathbf{U}_1$: $\\frac{\\langle \\mathbf{V}_3, \\mathbf{U}_1 \\rangle}{\\| \\mathbf{U}_1 \\|^2} = \\frac{0}{6} = 0$<br \/>\n* Proyecci\u00f3n sobre $\\mathbf{U}_2$: $\\frac{\\langle \\mathbf{V}_3, \\mathbf{U}_2 \\rangle}{\\| \\mathbf{U}_2 \\|^2} = \\frac{2}{2} = 1$<\/p>\n<p>**Sustituci\u00f3n y C\u00e1lculo de \\(\\mathbf{U}_3\\):**<br \/>\n\\[<br \/>\n\\mathbf{U}_3 = \\mathbf{V}_3 &#8211; 0 \\cdot \\mathbf{U}_1 &#8211; 1 \\cdot \\mathbf{U}_2<br \/>\n\\]<br \/>\n\\[<br \/>\n\\mathbf{U}_3 = \\begin{bmatrix} 1 &#038; 1 \\\\ 1 &#038; 0 \\end{bmatrix} &#8211; \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}<br \/>\n\\]<br \/>\n\\[<br \/>\n\\mathbf{U}_3 = \\begin{bmatrix} 1 &#8211; 1 &#038; 1 &#8211; 1 \\\\ 1 &#8211; 0 &#038; 0 &#8211; 0 \\end{bmatrix} = \\begin{bmatrix} 0 &#038; 0 \\\\ 1 &#038; 0 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>La base **ortogonal** obtenida es:<br \/>\n\\[<br \/>\nB&#8217; = \\left\\{ \\mathbf{U}_1, \\mathbf{U}_2, \\mathbf{U}_3 \\right\\} = \\left\\{ \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix}, \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 0 &#038; 0 \\\\ 1 &#038; 0 \\end{bmatrix} \\right\\}<br \/>\n\\]<\/p>\n<p>&#8212;<\/p>\n<p>## 2. Normalizaci\u00f3n de la Base Ortogonal \\(\\{ \\mathbf{U}_1, \\mathbf{U}_2, \\mathbf{U}_3 \\}\\)<\/p>\n<p>Dividimos cada matriz ortogonal por su norma: $\\mathbf{E}_i = \\frac{\\mathbf{U}_i}{\\| \\mathbf{U}_i \\|}$<\/p>\n<p>### Paso 4: Normalizar \\(\\mathbf{U}_1\\) (obteniendo \\(\\mathbf{E}_1\\))<br \/>\n\\[<br \/>\n\\| \\mathbf{U}_1 \\| = \\sqrt{6}<br \/>\n\\]<br \/>\n\\[<br \/>\n\\mathbf{E}_1 = \\frac{1}{\\sqrt{6}} \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>### Paso 5: Normalizar \\(\\mathbf{U}_2\\) (obteniendo \\(\\mathbf{E}_2\\))<br \/>\n\\[<br \/>\n\\| \\mathbf{U}_2 \\| = \\sqrt{2}<br \/>\n\\]<br \/>\n\\[<br \/>\n\\mathbf{E}_2 = \\frac{1}{\\sqrt{2}} \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>### Paso 6: Normalizar \\(\\mathbf{U}_3\\) (obteniendo \\(\\mathbf{E}_3\\))<br \/>\n\\[<br \/>\n\\| \\mathbf{U}_3 \\| = \\sqrt{0^2 + 0^2 + 1^2 + 0^2} = 1<br \/>\n\\]<br \/>\n\\[<br \/>\n\\mathbf{E}_3 = \\frac{1}{1} \\begin{bmatrix} 0 &#038; 0 \\\\ 1 &#038; 0 \\end{bmatrix} = \\begin{bmatrix} 0 &#038; 0 \\\\ 1 &#038; 0 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>&#8212;<\/p>\n<p>## 3. Conclusi\u00f3n<\/p>\n<p>La base **ortonormal** de las tres matrices de $\\mathbb{R}^{2 \\times 2}$ es:<\/p>\n<p>\\[<br \/>\n\\mathcal{B} = \\{ \\mathbf{E}_1, \\mathbf{E}_2, \\mathbf{E}_3 \\} = \\left\\{ \\frac{1}{\\sqrt{6}} \\begin{bmatrix} 1 &#038; -1 \\\\ 0 &#038; 2 \\end{bmatrix} , \\frac{1}{\\sqrt{2}} \\begin{bmatrix} 1 &#038; 1 \\\\ 0 &#038; 0 \\end{bmatrix}, \\begin{bmatrix} 0 &#038; 0 \\\\ 1 &#038; 0 \\end{bmatrix} \\right\\}<br \/>\n\\]\n<\/p><\/div>\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\u00e1l es el valor de \\(\\begin{bmatrix}1&#038;2\\\\ 0&#038; -1\\end{bmatrix}\\bullet\\begin{bmatrix}0&#038;-1\\\\ 1&#038; 3\\end{bmatrix}\\)<\/td>\n<\/tr>\n<tr>\n<td>\n<div id=\"menu-a\">\n<ul>\n<li>-5<\/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>\n<button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>A.)<\/strong><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Recordemos que, dado un espacio eucl\u00eddeo, \\((\\mathcal{E},\\bullet)\\), dos vectores se dicen ortogonales si \\[\\vec{x}\\perp \\vec{y} \\Leftrightarrow \\vec{x}\\bullet \\vec{y}=0\\] Con esta definici\u00f3n, decimos que \\(B=\\{\\vec{v}_1,\\vec{v}_2,\\ldots,\\vec{v}_n\\}\\) es un conjunto ortogonal si dos a dos&#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-476","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/476","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=476"}],"version-history":[{"count":19,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/476\/revisions"}],"predecessor-version":[{"id":479,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/476\/revisions\/479"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}