{"id":618,"date":"2025-12-01T08:30:37","date_gmt":"2025-12-01T07:30:37","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=618"},"modified":"2025-12-01T19:31:57","modified_gmt":"2025-12-01T18:31:57","slug":"alg-aplicaciones-y-matrices-ortogonales","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=618","title":{"rendered":"ALG: Aplicaciones y matrices ortogonales"},"content":{"rendered":"<h2>Aplicaciones y matrices ortogonales<\/h2>\n<p>Definimos las aplicaciones ortogonales a las aplicaciones de un espacio vectorial con producto escalar \\((\\mathcal{E},\\bullet)\\) que conservan el producto escalar; es decir, \\(f:\\mathcal{E}\\to \\mathcal{E}\\), es ortogonal si \\[f(\\vec{x})\\bullet f(\\vec{y})=\\vec{x}\\bullet \\vec{y},\\quad\\forall\\,\\vec{x}, \\vec{y}\\in\\mathcal{E}\\]<\/p>\n<p>Propiedades que cumple una aplicaci\u00f3n ortogonal:<\/p>\n<ul>\n<li><span style=\"line-height: 13px;\" data-mce-mark=\"1\">Es lineal<\/span><\/li>\n<li>Conserva la norma; es decir, \\(||f(\\vec{x})||=||\\vec{x}||\\)<\/li>\n<li>Dos vectores son ortogonales si, y solo si, sus im\u00e1genes son ortogonales<\/li>\n<li>La aplicaci\u00f3n es biyectiva<\/li>\n<li>Los vectores propios de valores propios distintos son ortogonales<\/li>\n<li>La imagen de una base ortonormal es ortonormal<\/li>\n<li>Su matriz asociada es ortogonal<\/li>\n<\/ul>\n<p>Hay varias formas de definir una matriz ortogonal. Nosotros emplearemos la que parte de la teor\u00eda de matrices. <\/p>\n<blockquote><p>\n<strong>Definici\u00f3n:<\/strong> Diremos que de una matriz cuadrada es ortogonal si su inversa coincide con su traspuesta; es decir,\\[A^{-1}=A^t.\\]\n<\/p><\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfLa matriz es \\(\\frac{1}{2}\\begin{bmatrix}\\sqrt{3}&#038;1\\\\ -1&#038;\\sqrt{3}\\end{bmatrix}\\) es ortogonal?\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfLa matriz es \\(\\frac{1}{5}\\begin{bmatrix}4&#038;0&#038;-3\\\\ 0&#038;5&#038;0\\\\ 3&#038;0&#038;4\\end{bmatrix}\\) es ortogonal?\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfLa matriz es \\(\\begin{bmatrix}1&#038;2&#038;0\\\\ 2&#038;1&#038;0\\\\ 0&#038;0&#038;-1\\end{bmatrix}\\) es ortogonal?\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejercicio:<\/strong> \u00bfLa matriz es \\(\\begin{bmatrix}0&#038;0&#038;1\\\\ 0&#038;-1&#038;0\\\\ -1&#038;0&#038;0\\end{bmatrix}\\) es ortogonal?\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> \u00bfLa matriz es \\(\\frac{1}{2}\\begin{bmatrix}2&#038;1&#038;-2\\\\ 1&#038;2&#038;2\\\\ -2&#038;2&#038;-1\\end{bmatrix}\\) es ortogonal? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9a1() {\n  var htmlShow9a1 = document.getElementById(\"html-show9a1\");\n  if (htmlShow9a1.style.display === \"none\") {\n    htmlShow9a1.style.display = \"block\";\n  } else {\n    htmlShow9a1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a1\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Matriz Ortogonal. Ej.2 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/FtfHtWZ18-c?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>Ejemplo:<\/strong> \u00bfPara qu\u00e9 valores de \\(\\alpha\\) y \\(\\beta\\), reales, la matriz \\(\\begin{bmatrix}0&#038;\\alpha&#038;\\beta\\\\ \\alpha&#038;\\beta&#038;0\\\\ \\beta&#038;0&#038;\\alpha\\end{bmatrix}\\) es ortogonal? <\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9a() {\n  var htmlShow9a = document.getElementById(\"html-show9a\");\n  if (htmlShow9a.style.display === \"none\") {\n    htmlShow9a.style.display = \"block\";\n  } else {\n    htmlShow9a.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a\" style=\"display: none;\">\n<iframe loading=\"lazy\" title=\"\u00c1lgebra Lineal - Matriz ortogonal. Ej 1 - Jes\u00fas Soto\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/3ClgOe5WQHk?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>Propiedad:<\/strong> Una matriz \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\) ortogonal, entonces \\(\\det(A)=\\pm 1\\).\n<\/p>\n<\/blockquote>\n<p>En efecto, basta con observar que \\[{\\displaystyle \\det(A\\cdot A^{t})=\\det A\\ \\det A^{t}=\\det A\\ \\det A=(\\det A)^{2}=\\det I=1.}\\]<\/p>\n<blockquote>\n<p><strong>Propiedad:<\/strong> Una matriz \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\) ortogonal, para todo \\(\\mathbf{u},\\mathbf{v}\\in\\mathbb{R}^n\\) es \\(A\\mathbf{u}\\bullet A\\mathbf{v}=\\mathbf{u}\\bullet \\mathbf{v}\\) y  \\(\\|A\\mathbf{u}\\|=\\|\\mathbf{u}\\|\\).\n<\/p>\n<\/blockquote>\n<p>Observemos que dados \\(\\mathbf{u},\\mathbf{v}\\in\\mathbb{R}^n\\), \\[A\\mathbf{u}\\bullet  A\\mathbf{v}=\\left( A\\mathbf{u}\\right) ^{t}\\left( A\\mathbf{v}\\right) =\\left( \\mathbf{u}^{t}A^{t}\\right) A\\mathbf{v}=\\mathbf{u}^{t}\\left( A^{t}A\\right) \\mathbf{v}=\\mathbf{u}^{t}\\;I_{n}\\;\\mathbf{v}=\\mathbf{u}^{t}\\;\\mathbf{v}=\\mathbf{u}\\bullet  \\mathbf{v}.\\]<br \/>\nDe aqu\u00ed, vemos que \\[\\|A\\mathbf{u}\\|=\\sqrt{A\\mathbf{u}\\bullet  A\\mathbf{u}}=\\sqrt{\\mathbf{u}\\bullet  \\mathbf{u}}=\\|\\mathbf{u}\\|\\]<\/p>\n<p>Para nosotros ser\u00e1 muy \u00fatil el siguiente resultado:<\/p>\n<blockquote>\n<p><strong>Teorema:<\/strong> Una matriz \\(A\\in\\mathcal{M}_n(\\mathbb{R})\\) es ortogonal si y s\u00f3lo si sus vectores filas o vectores columna son cada uno un conjunto ortonormal de vectores.\n<\/p>\n<\/blockquote>\n<blockquote>\n<p><strong>Propiedad:<\/strong> Si \\(A,B\\in\\mathcal{M}_n(\\mathbb{R})\\) son ortogonales, entonces el producto \\(AB\\) es ortogonal.\n<\/p>\n<\/blockquote>\n<p>La relaci\u00f3n entre los concepto de aplicaci\u00f3n ortogonal y matriz ortogonal es muy sencilla:<\/p>\n<blockquote>\n<p><strong>Teorema:<\/strong>  Si tenemos un endomorfismo ortogonal sobre una base ortonormal, entonces su matriz asociada es una matriz ortogonal.<\/p>\n<\/blockquote>\n<p>Dicho de otro modo, las aplicaciones ortogonales, aquellas que conservan el producto escalar, tienen por matrices asociadas a matrices ortogonales (matrices cuadradas que cumplen que su inversa coincide con la traspuesta). Adem\u00e1s se cumple, que en una matriz ortogonal las filas o columnas, consideradas como vectores, son ortonormales.<\/p>\n<p>&nbsp;<\/p>\n<h2>Clasificaci\u00f3n de las aplicaciones ortogonales en \\(\\mathbb{R}^2\\)<\/h2>\n<blockquote>\n<p><strong>Teorema:<\/strong>  Si \\(f:\\mathbb{R}^2\\to\\mathbb{R}^2\\) es un endomorfismo ortogonal respecto una base ortonormal, entonces su matriz asociada es \\[A=\\begin{bmatrix}a&#038; -b|A|\\\\ b&#038; a|A|\\end{bmatrix},\\] con \\(|A|=\\pm 1\\).<\/p>\n<\/blockquote>\n<h3>Giros<\/h3>\n<p>Si resulta \\(|A|=1\\), ser\u00e1 \\[A=\\begin{bmatrix}a&#038; -b\\\\ b&#038; a\\end{bmatrix}.\\]<br \/>\nDe este modo, ex\u00edste un \u00fanico \\(\\alpha\\in [0,2\\pi)\\), tal que \\(a=\\cos(\\alpha)\\) y \\(b=\\sin(\\alpha)\\), tal que<br \/>\n\\[A=\\begin{bmatrix}\\cos(\\alpha)&#038; -\\sin(\\alpha)\\\\ \\sin(\\alpha)&#038; \\cos(\\alpha)\\end{bmatrix}.\\]<br \/>\nPor tanto, la aplicaci\u00f3n ortogonal es un giro, centrado en el origen, de \u00e1ngulo \\(\\alpha\\) con<br \/>\n\\[Tr(A)=2a=2\\cos(\\alpha)\\Rightarrow \\alpha=\\arccos\\left(\\tfrac{1}{2}Tr(A)\\right)\\]<\/p>\n<blockquote><p>\n<strong>Proposici\u00f3n:<\/strong> Sean \\(v:(a, b)\\) y \\(v&#8217;:(a&#8217;, b&#8217;)\\) dos puntos en el plano af\u00edn \\(\\mathbb{R}^2\\). Si se cumple que \\(a^2 + b^2 = (a&#8217;)^2 + (b&#8217;)^2\\), entonces<br \/>\n\\[<br \/>\nR = \\begin{bmatrix} \\frac{a a&#8217; + b b&#8217;}{a^2 + b^2} &#038; -\\frac{a b&#8217; &#8211; b a&#8217;}{a^2 + b^2} \\\\ \\frac{a b&#8217; &#8211; b a&#8217;}{a^2 + b^2} &#038; \\frac{a a&#8217; + b b&#8217;}{a^2 + b^2} \\end{bmatrix}<br \/>\n\\]<br \/>\nrealiza un giro de \\((a, b)\\) a \\((a&#8217;, b&#8217;)\\)\n<\/p><\/blockquote>\n<h3>Giro m\u00e1s desplazamiento<\/h3>\n<p>La construcci\u00f3n de un \u00abgiro m\u00e1s desplazamiento\u00bb se conoce formalmente como <strong>movimiento r\u00edgido<\/strong> o <strong>isometr\u00eda directa<\/strong> en \\(\\mathbb{R}^2\\). En \u00c1lgebra Lineal, esto se modela de manera eficiente utilizando <strong>coordenadas homog\u00e9neas<\/strong>.<\/p>\n<p>Un movimiento r\u00edgido es una transformaci\u00f3n af\u00edn y se define como la combinaci\u00f3n de una rotaci\u00f3n y una traslaci\u00f3n (desplazamiento):<\/p>\n<p>\\[<br \/>\nT(v) = R v + t<br \/>\n\\]<\/p>\n<p>Donde:<\/p>\n<ol>\n<li> \\(v\\) es el vector de coordenadas originales \\(\\begin{bmatrix} x \\\\ y \\end{bmatrix}\\).<\/li>\n<li>\\(R\\) es la <strong>matriz de giro<\/strong> \\(2 \\times 2\\) (ortogonal, \\(\\det(R)=1\\)).<\/li>\n<li>\\(t\\) es el <strong>vector de traslaci\u00f3n<\/strong> (desplazamiento) \\(\\begin{bmatrix} t_x \\\\ t_y \\end{bmatrix}\\).<\/li>\n<\/ol>\n<p>De este modo, el punto girado \\(v_{giro}\\) se traslada por el vector \\(t\\):<br \/>\n\\[<br \/>\nv&#8217; = v_{giro} + t = \\begin{bmatrix} \\cos \\theta &#038; -\\sin \\theta \\\\ \\sin \\theta &#038; \\cos \\theta \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} + \\begin{bmatrix} t_x \\\\ t_y \\end{bmatrix},<br \/>\n\\]<\/p>\n<p>Recordad que las transformaciones afines no son transformaciones lineales, por lo que no pueden representarse con una simple multiplicaci\u00f3n de matriz \\(2 \\times 2\\).<\/p>\n<p>Para manejar tanto la rotaci\u00f3n (operaci\u00f3n lineal) como la traslaci\u00f3n (operaci\u00f3n no lineal) con una <strong>\u00fanica multiplicaci\u00f3n matricial<\/strong>, se utiliza la t\u00e9cnica de <strong>coordenadas homog\u00e9neas<\/strong> en \\(\\mathbb{R}^3\\):<\/p>\n<p>\\[<br \/>\n\\begin{bmatrix} a&#8217; \\\\ b&#8217; \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} \\cos \\theta &#038; -\\sin \\theta &#038; t_x \\\\ \\sin \\theta &#038; \\cos \\theta &#038; t_y \\\\ 0 &#038; 0 &#038; 1 \\end{bmatrix} \\begin{bmatrix} a \\\\ b \\\\ 1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>La matriz, de \\(3 \\times 3\\), que hemos visto se denomina matriz de transformaci\u00f3n, que combina el giro \\(R\\) y la traslaci\u00f3n \\(t\\). En Gr\u00e1ficos por Computadora y Rob\u00f3tica, la representaci\u00f3n con coordenadas homog\u00e9neas es el est\u00e1ndar porque permite concatenar (componer) m\u00faltiples giros, traslaciones y escalados con solo multiplicar las matrices de transformaci\u00f3n homog\u00e9neas resultantes.<\/p>\n<blockquote>\n<p><strong>Ejemplo:<\/strong> Determinar el vector resultado de un giro de $90^\\circ$ (\\(\\pi\/2\\) radianes) en sentido antihorario, del vector [3,2] y una traslaci\u00f3n (desplazamiento) de [4, -1].<\/p>\n<\/blockquote>\n<p><script>\nfunction showHtmlDiv9a4() {\n  var htmlShow9a4 = document.getElementById(\"html-show9a4\");\n  if (htmlShow9a4.style.display === \"none\") {\n    htmlShow9a4.style.display = \"block\";\n  } else {\n    htmlShow9a4.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a4()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a4\" style=\"display: none;\">\nPrimero determinamos la matriz de giro (\\(R\\)):<br \/>\nPara \\(\\theta = 90^\\circ\\), la matriz de giro $2 \\times 2$ es:<br \/>\n\\[<br \/>\nR = \\begin{bmatrix} \\cos 90^\\circ &#038; -\\sin 90^\\circ \\\\ \\sin 90^\\circ &#038; \\cos 90^\\circ \\end{bmatrix} = \\begin{bmatrix} 0 &#038; -1 \\\\ 1 &#038; 0 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>A continuaci\u00f3n, el vector de traslaci\u00f3n (\\(t\\))<br \/>\nEl desplazamiento es \\((t_x, t_y) = (4, -1)\\).<\/p>\n<p>\\[<br \/>\nt = \\begin{bmatrix} 4 \\\\ -1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Hacemos la Matriz de Transformaci\u00f3n Homog\u00e9nea (\\(M\\))<br \/>\nCombinamos $R$ y $t$ en una \u00fanica matriz $3 \\times 3$:<\/p>\n<p>\\[<br \/>\nM = \\begin{bmatrix}<br \/>\n0 &#038; -1 &#038; 4 \\\\<br \/>\n1 &#038; 0 &#038; -1 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Ya tenemos la matriz, ahora aplicamos la transformaci\u00f3n con el Punto de Partida en Coordenadas Homog\u00e9neas \\(v_{1, hom} = \\begin{bmatrix} 3 \\\\ 2 \\\\ 1 \\end{bmatrix}\\):<\/p>\n<p>\\[<br \/>\nv_{2, hom} = M v_{1, hom} = \\begin{bmatrix}<br \/>\n0 &#038; -1 &#038; 4 \\\\<br \/>\n1 &#038; 0 &#038; -1 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix} \\begin{bmatrix} 3 \\\\ 2 \\\\ 1 \\end{bmatrix}= \\begin{bmatrix} 2 \\\\ 2 \\\\ 1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Punto Final (Vector Transformado)<\/p>\n<p>\\[<br \/>\nv_{2, hom} = \\begin{bmatrix} 2 \\\\ 2 \\\\ 1 \\end{bmatrix} \\implies v_2 = \\begin{bmatrix} 2 \\\\ 2 \\end{bmatrix}<br \/>\n\\]<\/p>\n<figure class=\"wp-block-image aligncenter\">\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/uploads.jesussoto.es\/rigid_motion01.png\" alt=\"\" width=\"452\" height=\"425\" class=\"alignnone size-full wp-image-657\" \/><br \/>\n<\/figure>\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejercicio:<\/strong> Se aplica un <em>movimiento r\u00edgido<\/em> (giro m\u00e1s traslaci\u00f3n) a un punto \\(P(1,4)\\) del plano, transform\u00e1ndolo en un punto \\(P'(5,4)\\). Se sabe que el movimiento consiste en una <em>rotaci\u00f3n de \u00e1ngulo<\/em> $\\theta$ en sentido antihorario, seguida de una <em>traslaci\u00f3n <\/em>por el vector $t=[1,3]$. \u00bfCu\u00e1l es el \u00e1ngulo de giro \\(\\theta\\) en grados?<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv9a5() {\n  var htmlShow9a5 = document.getElementById(\"html-show9a5\");\n  if (htmlShow9a5.style.display === \"none\") {\n    htmlShow9a5.style.display = \"block\";\n  } else {\n    htmlShow9a5.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a5()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a5\" style=\"display: none;\">\nRecordemos que la transformaci\u00f3n af\u00edn se define mediante la f\u00f3rmula:<\/p>\n<p>\\[<br \/>\nP&#8217; = R(\\theta) P + t<br \/>\n\\]<\/p>\n<p>Donde \\(R(\\theta) = \\begin{bmatrix} \\cos \\theta &#038; -\\sin \\theta \\\\ \\sin \\theta &#038; \\cos \\theta \\end{bmatrix}\\).<\/p>\n<p>Paso 1: Aislar la Rotaci\u00f3n<\/p>\n<p>Primero, despejamos la componente de rotaci\u00f3n restando el vector de traslaci\u00f3n \\(t\\) al vector final \\(P&#8217;\\):<\/p>\n<p>\\[<br \/>\nR(\\theta) P = P&#8217; &#8211; t<br \/>\n\\]<\/p>\n<p>Calculamos el vector de giro puro \\(P_{giro} = P&#8217; &#8211; t\\):<\/p>\n<p>\\[<br \/>\nP_{giro} = \\begin{bmatrix} 5 \\\\ 4 \\end{bmatrix} &#8211; \\begin{bmatrix} 1 \\\\ 3 \\end{bmatrix} =\\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>La ecuaci\u00f3n ahora se simplifica a:<br \/>\n\\[<br \/>\n\\begin{bmatrix} \\cos \\theta &#038; -\\sin \\theta \\\\ \\sin \\theta &#038; \\cos \\theta \\end{bmatrix} \\begin{bmatrix} 1 \\\\ 4 \\end{bmatrix} = \\begin{bmatrix} 4 \\\\ 1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Paso 2: Plantear y Resolver el Sistema<\/p>\n<p>Definimos \\(c = \\cos \\theta\\) y \\(s = \\sin \\theta\\). La multiplicaci\u00f3n matricial resulta en el siguiente sistema de ecuaciones lineales con inc\u00f3gnitas \\(c\\) y \\(s\\):<\/p>\n<p>\\[<br \/>\n\\begin{align*} 1 \\cdot c &#8211; 4 \\cdot s &#038;= 1 \\quad &#038; (E_1) \\\\ 4 \\cdot c + 1 \\cdot s &#038;= 4 \\quad &#038; (E_2) \\end{align*}<br \/>\n\\]<\/p>\n<p>Resolvemos el sistema :<\/p>\n<p>\\[<br \/>\n\\begin{align*}<br \/>\nc &#038;= \\frac{8}{17}\\\\<br \/>\ns &#038;= -\\frac{15}{17}<br \/>\n\\end{align*}<br \/>\n\\]<\/p>\n<p>Ya est\u00e1mos en condiciones de determinar la matriz de giro y el \u00e1ngulo:<\/p>\n<p>Matriz de Rotaci\u00f3n Pura \\(R\\)<br \/>\n\\[<br \/>\nR = \\begin{bmatrix} \\frac{8}{17} &#038; \\frac{15}{17} \\\\ -\\frac{15}{17} &#038; \\frac{8}{17} \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>\u00c1ngulo de Giro \\(\\theta\\)<br \/>\nComo \\(\\cos \\theta > 0\\) y \\(\\sin \\theta < 0\\), el \u00e1ngulo \\(\\theta\\) se encuentra en el <em>cuarto cuadrante<\/em>.<\/p>\n<p>\\[<br \/>\n\\theta = \\arctan \\left( \\frac{\\sin \\theta}{\\cos \\theta} \\right) = \\arctan \\left( \\frac{-15\/17}{8\/17} \\right) = \\arctan \\left( -\\frac{15}{8} \\right)\\approx -61.9275^\\circ<br \/>\n\\]<\/p>\n<p>El giro es de aproximadamente \\(61.9275^\\circ\\) en sentido horario (negativo). Para una rotaci\u00f3n en sentido antihorario (\u00e1ngulo positivo), podemos dar el equivalente entre \\(0^\\circ\\) y \\(360^\\circ\\):<\/p>\n<p>\\[<br \/>\n\\alpha \\approx 360^\\circ &#8211; 61.9275^\\circ \\approx 298.0725^\\circ.<br \/>\n\\]<\/p>\n<figure class=\"wp-block-image aligncenter\">\n<figure class=\"wp-block-image aligncenter\">\n  <img decoding=\"async\" src=\"https:\/\/uploads.jesussoto.es\/Movimiento_rigido01.png\"\n       alt=\"Movimiento r\u00edgido P(1,4) \u2192 P_rot(4,1) \u2192 P'(5,4)\"\n       width=\"100%\"\n       style=\"max-width:800px;\"><\/p>\n<p><em>Rotaci\u00f3n \u221261.9275\u00b0 (horaria) + traslaci\u00f3n (1,3).<\/em><\/p>\n<\/figure>\n<\/figure>\n<\/div>\n<hr \/>\n<p>El escalado, en coordenadas homog\u00e9neas, lo hacemos mediante<\/p>\n<p>\\[<br \/>\nM_{escalado} = \\begin{bmatrix}<br \/>\ns_x &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; s_y &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Con esto ya podemos componer giros, traslaciones y escalados:<\/p>\n<p>\\[<br \/>\n\\begin{align}<br \/>\nM &#038;= M_{traslaci\u00f3n} \\cdot M_{giro} \\cdot M_{escalado}\\\\ &#038;= \\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; t_x \\\\<br \/>\n0 &#038; 1 &#038; t_y \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}\\cdot \\begin{bmatrix}<br \/>\nc &#038; -s &#038; 0 \\\\<br \/>\ns &#038; c &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}\\cdot \\begin{bmatrix}<br \/>\ns_x &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; s_y &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}\\\\<br \/>\n&#038;=\\begin{bmatrix}<br \/>\ns_x c &#038; -s_y s &#038; t_x \\\\<br \/>\ns_x s &#038; s_y c &#038; t_y \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\end{align}<br \/>\n\\]<\/p>\n<p>Donde:<\/p>\n<ul>\n<li>\\(c = \\cos \\theta\\)<\/li>\n<li>\\(s = \\sin \\theta\\)<\/li>\n<li>\\((s_x, s_y)\\) son los factores de escala.<\/li>\n<li>\\((t_x, t_y)\\) es el vector de traslaci\u00f3n.<\/li>\n<\/ul>\n<blockquote><p><strong>Ejemplo:<\/strong> Sea el punto P(1,2) al que realizamos un escalado s(2,2), un giro\\(90^\\circ\\) antihorario y una translaci\u00f3n por vector \\(t = (5, 3)\\). \u00bfCu\u00e1l el el punto final?<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv9a8() {\n  var htmlShow9a8 = document.getElementById(\"html-show9a8\");\n  if (htmlShow9a8.style.display === \"none\") {\n    htmlShow9a8.style.display = \"block\";\n  } else {\n    htmlShow9a8.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv9a8()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show9a8\" style=\"display: none;\">\nPara trabajar m\u00e1s c\u00f3modos lo haremos en coordenadas homog\u00e9neas.<\/p>\n<p>En coordenadas homog\u00e9neas, un punto \\(P = (x, y)\\) se representa como \\(\\tilde{P} = (x, y, 1)^T\\). Las transformaciones lineales (rotaci\u00f3n, escalado) y afines (traslaci\u00f3n) se representan mediante matrices \\(3 \\times 3\\).<\/p>\n<p>Para el ejemplo con punto inicial \\(P = (1, 2)\\), las transformaciones en coordenadas homog\u00e9neas son:<\/p>\n<p>1. Escalado por \\(k=2\\):<br \/>\n\\[<br \/>\nS = \\begin{bmatrix}<br \/>\n2 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 2 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>2. Giro \\(90^\\circ\\) antihorario:<br \/>\n\\[<br \/>\nR = \\begin{bmatrix}<br \/>\n0 &#038; -1 &#038; 0 \\\\<br \/>\n1 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>3. Traslaci\u00f3n por vector \\(t = (5, 3)\\):<br \/>\n\\[<br \/>\nT = \\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; 3 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>El punto final homog\u00e9neo \\(\\tilde{P}&#8217;\\) se obtiene aplicando las transformaciones en orden (primero escalado, luego rotaci\u00f3n, luego traslaci\u00f3n):<\/p>\n<p>\\[<br \/>\n\\tilde{P}&#8217; = T \\cdot R \\cdot S \\cdot \\tilde{P} =<br \/>\n\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; 3 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\cdot<br \/>\n\\begin{bmatrix}<br \/>\n0 &#038; -1 &#038; 0 \\\\<br \/>\n1 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\cdot<br \/>\n\\begin{bmatrix}<br \/>\n2 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 2 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\cdot<br \/>\n\\begin{bmatrix}<br \/>\n1 \\\\ 2 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Calculando paso a paso:<\/p>\n<p>\\[<br \/>\nS \\cdot \\tilde{P} =<br \/>\n\\begin{bmatrix}<br \/>\n2 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 2 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\begin{bmatrix}<br \/>\n1 \\\\ 2 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n=<br \/>\n\\begin{bmatrix}<br \/>\n2 \\\\ 4 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>\\[<br \/>\nR \\cdot (S \\cdot \\tilde{P}) =<br \/>\n\\begin{bmatrix}<br \/>\n0 &#038; -1 &#038; 0 \\\\<br \/>\n1 &#038; 0 &#038; 0 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\begin{bmatrix}<br \/>\n2 \\\\ 4 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n=<br \/>\n\\begin{bmatrix}<br \/>\n-4 \\\\ 2 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>\\[<br \/>\nT \\cdot (R \\cdot S \\cdot \\tilde{P}) =<br \/>\n\\begin{bmatrix}<br \/>\n1 &#038; 0 &#038; 5 \\\\<br \/>\n0 &#038; 1 &#038; 3 \\\\<br \/>\n0 &#038; 0 &#038; 1<br \/>\n\\end{bmatrix}<br \/>\n\\begin{bmatrix}<br \/>\n-4 \\\\ 2 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n=<br \/>\n\\begin{bmatrix}<br \/>\n1 \\\\ 5 \\\\ 1<br \/>\n\\end{bmatrix}<br \/>\n\\]<\/p>\n<p>Por lo tanto,<\/p>\n<p>\\[<br \/>\n\\tilde{P}&#8217; = \\begin{bmatrix} 1 \\\\ 5 \\\\ 1 \\end{bmatrix}<br \/>\n\\]<\/p>\n<p>que corresponde al punto \\(P&#8217; = (1, 5)\\) en coordenadas cartesianas.<\/p>\n<div style=\"text-align: center;\">\n<img decoding=\"async\" src=\"https:\/\/uploads.jesussoto.es\/affine_transform01.png\" \n     alt=\"Transformaci\u00f3n af\u00edn P(1,2) \u2192 P'(1,5)\" \n     width=\"100%\" \n     style=\"max-width:800px;\"><\/p>\n<p><em>Escalado k=2 \u2192 Giro 90\u00b0 \u2192 Traslaci\u00f3n (5,3)<\/em><\/p>\n<\/div>\n<\/div>\n<hr \/>\n<h3>Simetr\u00edas<\/h3>\n<p>Si \\(|A|=-1\\), ser\u00e1<br \/>\n\\[A=\\begin{bmatrix}a&#038; b\\\\ b&#038; -a\\end{bmatrix},\\]<br \/>\ny habr\u00e1n dos vectores \\(\\vec{v}_1\\) y \\(\\vec{v}_2\\), tales que \\(f(\\vec{v}_1)=\\vec{v}_1\\) y \\(f(\\vec{v}_2)=-\\vec{v}_2\\), siendo \\(\\vec{v}_1\\bullet\\vec{v}_2=0\\). De este modo, la aplicaci\u00f3n ortogonal es una simetr\u00eda respecto de la recta \\(r=\\{\\lambda\\vec{v}_1;\\lambda\\in\\mathbb{R}\\}\\)<\/p>\n<h2>Clasificaci\u00f3n de las aplicaciones ortogonales en \\(\\mathbb{R}^3\\)<\/h2>\n<p>En esta caso pueden darse cuatro matrices:<\/p>\n<p>1) La matriz identidad<\/p>\n<p>2) La matriz \\[\\begin{bmatrix}1&#038; 0 &#038; 0\\\\ 0 &#038; 1 &#038; 0\\\\ 0&#038; 0 &#038; -1\\end{bmatrix},\\] y la aplicaci\u00f3n ortogonal es una simetr\u00eda respecto del plano generado por los vectores tales que \\(f(\\vec{v}_1)=\\vec{v}_1\\) y \\(f(\\vec{v}_2)=\\vec{v}_2\\).<\/p>\n<p>3) La matriz \\[\\begin{bmatrix}1&#038; 0 &#038; 0\\\\ 0 &#038; \\cos(\\alpha) &#038; -\\sin(\\alpha)\\\\ 0&#038; \\sin(\\alpha) &#038; \\cos(\\alpha)\\end{bmatrix},\\] y la aplicaci\u00f3n ortogonal es un giro con eje en la recta generada por el \u00fanico vector tal que \\(f(\\vec{v}_1)=\\vec{v}_1\\), y de \u00e1ngulo \\(\\alpha\\) con<br \/>\n\\[Tr(A)=1+2\\cos(\\alpha)\\Rightarrow \\alpha=\\arccos\\left(\\tfrac{1}{2}(Tr(A)-1)\\right)\\]<\/p>\n<p>4) La matriz \\[\\begin{bmatrix}-1&#038; 0 &#038; 0\\\\ 0 &#038; \\cos(\\alpha) &#038; -\\sin(\\alpha)\\\\ 0&#038; \\sin(\\alpha) &#038; \\cos(\\alpha)\\end{bmatrix},\\] y la aplicaci\u00f3n ortogonal es una simetr\u00eda, respecto del plano generado por el ortogonal por vector tal que \\(f(\\vec{v}_1)=-\\vec{v}_1\\), compuesta con un giro de recta \\(r=\\{\\lambda\\vec{v}_1;\\lambda\\in\\mathbb{R}\\}\\) y de \u00e1ngulo \\(\\alpha\\) con<br \/>\n\\[Tr(A)=-1+2\\cos(\\alpha)\\Rightarrow \\alpha=\\arccos\\left(\\tfrac{1}{2}(Tr(A)+1)\\right)\\] <\/p>\n<p>Un caso particular de este \u00faltimo es cuando \\(\\alpha=\\pi\\), en cuyo caso \\[\\begin{bmatrix}-1&#038; 0 &#038; 0\\\\ 0 &#038; -1 &#038; 0\\\\ 0&#038; 0 &#038; -1\\end{bmatrix},\\] que resulta una simetr\u00eda central con centro en el origen.<\/p>\n<p>La justificaci\u00f3n de lo anterior reside en el siguiente apartado de este tema, donde abordaremos los autovalores y autovectores de un endomorfismo, y, por ende, de una matriz.<\/p>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\">\n<p><strong>Ejercicio:<\/strong>Sea \\(S=\\left\\{[a, b; 2b, a+b]\\in\\mathcal{M}_2(\\mathbb{R})\\right\\}\\). \u00bfCu\u00e1l es la traza de la proyecci\u00f3n de \\([2,3; -1,2]\\) sobre \\(S^\\bot\\)?<\/p>\n<div id=\"menu-a\">\n<ul>\n<li>11\/5<\/li>\n<li>0<\/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><!-- Text cell --><\/p>\n<div class=\"comment\">En este caso no necesitamos calcular el ortogonal, \\(S^\\bot\\), pues la matriz nos la dar\u00eda directamente si conocemos la proyecci\u00f3n sobre \\(S\\), y para ello necesitamos los vectores de una base del subespacio:<\/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\">(%i2) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">b1<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">ev<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">matrix<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">a<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">b<\/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_operator\">\u00b7<\/span><span class=\"code_variable\">b<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">a<\/span><span class=\"code_operator\">+<\/span><span class=\"code_variable\">b<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">a<\/span><span class=\"code_operator\">=<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">b<\/span><span class=\"code_operator\">=<\/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\">b2<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">ev<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">matrix<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">[<\/span><span class=\"code_variable\">a<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">b<\/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_operator\">\u00b7<\/span><span class=\"code_variable\">b<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">a<\/span><span class=\"code_operator\">+<\/span><span class=\"code_variable\">b<\/span><span class=\"code_operator\">]<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">a<\/span><span class=\"code_operator\">=<\/span><span class=\"code_number\">0<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_variable\">b<\/span><span class=\"code_operator\">=<\/span><span class=\"code_number\">1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\begin{bmatrix}1 &amp; 0\\\\0 &amp; 1\\end{bmatrix}\\]<\/p>\n<p>\\[\\begin{bmatrix}0 &amp; 1\\\\2 &amp; 1\\end{bmatrix}\\]<\/p>\n<p><!-- Text cell --><\/p>\n<div class=\"comment\">Ahora hacemos que la base sea ortogonal:<\/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\">(%i4) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">:<\/span><span class=\"code_variable\">b1<\/span><span class=\"code_endofline\">;<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_variable\">u2<\/span><span class=\"code_operator\">:<\/span><span class=\"code_variable\">b2<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_variable\">b2<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">\/<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">\u00b7<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_endofline\">;<\/span><span class=\"code_endofline\"><br \/><\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\begin{bmatrix}1 &amp; 0\\\\0 &amp; 1\\end{bmatrix}\\]<\/p>\n<p>\\[\\begin{bmatrix}-\\frac{1}{2} &amp; 1\\\\2 &amp; \\frac{1}{2}\\end{bmatrix}\\]<\/p>\n<p><!-- Text cell --><\/p>\n<div class=\"comment\">Y calculamos la proyecci\u00f3n:<\/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\">v<\/span><span class=\"code_operator\">:<\/span><span class=\"code_function\">matrix<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">[<\/span><span class=\"code_number\">2<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">3<\/span><span class=\"code_operator\">]<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_operator\">[<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_number\">1<\/span><span class=\"code_endofline\">,<\/span><span class=\"code_number\">2<\/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_variable\">p<\/span><span class=\"code_operator\">:<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u1<\/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_operator\">)<\/span><span class=\"code_operator\">\/<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">\u00b7<\/span><span class=\"code_variable\">u1<\/span><span class=\"code_operator\">+<\/span><span class=\"code_operator\">(<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u2<\/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_operator\">)<\/span><span class=\"code_operator\">\/<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_function\">transpose<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">u2<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">.<\/span><span class=\"code_variable\">u2<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">)<\/span><span class=\"code_operator\">\u00b7<\/span><span class=\"code_variable\">u2<\/span><span class=\"code_endofline\">;<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\begin{bmatrix}2 &amp; 3\\\\-1 &amp; 2\\end{bmatrix}\\]<\/p>\n<p>\\[\\begin{bmatrix}\\frac{21}{11} &amp; \\frac{2}{11}\\\\\\frac{4}{11} &amp; \\frac{23}{11}\\end{bmatrix}\\]<\/p>\n<p><!-- Text cell --><\/p>\n<div class=\"comment\">La proyeci\u00f3n sobre el ortogonal ser\u00e1<\/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\">(%i8) <\/span><\/td>\n<td style=\"vertical-align: top;padding: 1mm;\"><span class=\"input\"><span class=\"code_variable\">v<\/span><span class=\"code_operator\">\u2212<\/span><span class=\"code_variable\">p<\/span><span class=\"code_endofline\">;<\/span><span class=\"code_endofline\"><br \/><\/span><span class=\"code_function\">mat_trace<\/span><span class=\"code_operator\">(<\/span><span class=\"code_variable\">%<\/span><span class=\"code_operator\">)<\/span><span class=\"code_endofline\">;<\/span><\/span><\/td>\n<\/tr>\n<\/table>\n<p>\\[\\begin{bmatrix}\\frac{1}{11} &amp; \\frac{31}{11}\\\\-\\frac{15}{11} &amp; -\\frac{1}{11}\\end{bmatrix}\\]<\/p>\n<p>\\[0\\]<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Aplicaciones y matrices ortogonales Definimos las aplicaciones ortogonales a las aplicaciones de un espacio vectorial con producto escalar \\((\\mathcal{E},\\bullet)\\) que conservan el producto escalar; es decir, \\(f:\\mathcal{E}\\to \\mathcal{E}\\), es ortogonal si \\[f(\\vec{x})\\bullet&#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-618","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/618","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=618"}],"version-history":[{"count":25,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/618\/revisions"}],"predecessor-version":[{"id":669,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/618\/revisions\/669"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=618"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=618"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=618"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}