{"id":2418,"date":"2017-08-15T07:52:17","date_gmt":"2017-08-15T12:52:17","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1003\/?p=2418"},"modified":"2020-03-18T18:06:26","modified_gmt":"2020-03-18T23:06:26","slug":"cl5-02-ilustracion-del-metodo","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/matg1049\/cl5-02-ilustracion-del-metodo\/","title":{"rendered":"cl5-02. Ilustraci\u00f3n del m\u00e9todo"},"content":{"rendered":"

<\/code><\/p>\n

\n
Ejemplo.<\/strong> Sea T:{P}_{2}\\to {P}_{2}<\/span> una T.L. tal que T(a+bx+c{{x}^{2}})=(a+b+5c)+(3b)x+(3a-c){{x}^{2}}<\/span> Hallar sus eigenvalores y eigenvectores.<\/pre>\n
Soluci\u00f3n:<\/strong><\/p>\n
Necesitamos utilizar una base para hallar una matriz de transformaci\u00f3n. Puesto que no se nos indica alguna base en particular, se utilizar\u00e1 la base can\u00f3nica B=\\{1,x,{{x}^{2}}\\}<\/span>.<\/p>\n
Transformando cada vector de la base:<\/p>\n\\begin{array}{rcl} T(1)&=&1+3{{x}^{2}} \\\\ T(x)&=&1+3x \\\\ T({{x}^{2}})&=&5-{{x}^{2}} \\end{array}<\/span>\n
Y hallando las coordenadas de tales transformadas respecto a la base (la misma) del espacio de llegada, se tiene:<\/p>\n\\begin{array}{rcl} & {{\\left[ T(1) \\right]}_{B}}&=&(1,0,3) \\\\ & {{\\left[ T(x) \\right]}_{B}}&=&(1,3,0) \\\\ & {{\\left[ T({{x}^{2}}) \\right]}_{B}}&=&(5,0,-1) \\end{array}<\/span>\n
Por lo cual, la matriz asociada a T es:<\/p>\nA=\\left[ \\begin{array}{rrr} 1 & 1 & 5  \\\\ 0 & 3 & 0 \\\\ 3 & 0 & -1 \\end{array} \\right]<\/span>\n
\n
Ahora, hay que resolver la ecuaci\u00f3n \\det (A-\\lambda I)=0<\/span>, hallando el siguiente determinante:<\/p>\n\\det (A-\\lambda I)=\\det \\left( \\begin{array}{rrr} 1-\\lambda  & 1 & 5 \\\\ 0 & 3-\\lambda  & 0 \\\\ 3 & 0 & -1-\\lambda \\end{array} \\right)<\/span>\n\\begin{array}{rcl} \\det (A-\\lambda I)&=(3-\\lambda )\\left[ \\left( 1-\\lambda  \\right)\\left( -1-\\lambda  \\right)-15 \\right] \\\\ & =(3-\\lambda )\\left[ {{\\lambda }^{2}}-16 \\right] \\\\ & =(3-\\lambda )(\\lambda -4)(\\lambda +4)=0 \\end{array}<\/span>\n
Cuyas soluciones son: \\lambda =3\\vee \\lambda =4\\vee \\lambda =-4<\/span>, es decir los valores propios. Por convenci\u00f3n, los valores propios se ordenan de mayor a menor<\/strong>, por lo cual se tiene:<\/p>\n\\begin{array}{rcl} {{\\lambda }_{1}}&=&4 \\\\ {{\\lambda }_{2}}&=&3 \\\\ {{\\lambda }_{3}}&=&-4 \\end{array}<\/span>\n
\n
Luego, para encontrar los vectores propios debemos hallar bases para el n\u00facleo Nu(A-\\lambda I)<\/span> luego de reemplazar los valores propios.<\/p>\n
Para<\/strong> {{\\lambda }_{1}}=4<\/span>:<\/p>\nNu(A-{{\\lambda }_{1}}I)\\Rightarrow \\left( \\begin{array}{rrr|r} 1-{{\\lambda }_{1}} & 1 & 5 & 0 \\\\ 0 & 3-{{\\lambda }_{1}} & 0 & 0 \\\\ 3 & 0 & -1-{{\\lambda }_{1}} & 0 \\end{array} \\right)<\/span>\n
Resolviendo:<\/p>\n\\left( \\begin{array}{rrr|r} -3 & 1 & 5 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 3 & 0 & -5 & 0 \\end{array} \\right)\\sim ...\\left( \\begin{array}{rrr|r} -3 & 1 & 5 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\end{array} \\right)<\/span>\n
es decir, {{\\alpha }_{2}}=0<\/span> y -3{{\\alpha }_{1}}+5{{\\alpha }_{3}}=0<\/span>. Por lo cual:<\/p>\nNu(A-{{\\lambda }_{1}}I)=\\left\\{ \\left( {{\\alpha }_{1}},{{\\alpha }_{2}},{{\\alpha }_{3}} \\right)\\in {{\\mathsf{\\mathbb{R}}}^{3}}\/{{\\alpha }_{2}}=0\\text{ }\\wedge \\text{ }-3{{\\alpha }_{1}}+5{{\\alpha }_{3}}=0 \\right\\}<\/span>\n
Una base de este n\u00facleo es:
\n{{B}_{\\lambda 1}}=\\left\\{ \\left( \\begin{array}{r} 5\/3 \\\\ 0 \\\\ 1 \\end{array} \\right) \\right\\}<\/span><\/p>\n
Por lo cual, el primer vector caracter\u00edstico es:<\/p>\n{{v}_{1}}=\\left( \\begin{array}{r} 5 \\\\ 0 \\\\ 3 \\end{array} \\right)<\/span>\n
\n
Para<\/strong> {{\\lambda }_{2}}=3<\/span>:<\/p>\nNu(A-{{\\lambda }_{2}}I)\\Rightarrow \\left( \\begin{array}{rrr|r} -2 & 1 & 5 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 3 & 0 & -4 & 0 \\end{array} \\right)\\sim ...\\left( \\begin{array}{rrr|r} -2 & 1 & 5 & 0 \\\\ 0 & 3 & 7 & 0 \\\\ 0 & 0 & 0 & 0 \\end{array} \\right)<\/span>\n
Es decir,<\/p>\nNu(A-{{\\lambda }_{2}}I)=\\left\\{ \\left( {{\\alpha }_{1}},{{\\alpha }_{2}},{{\\alpha }_{3}} \\right)\\in {{\\mathsf{\\mathbb{R}}}^{3}}\/{{\\alpha }_{2}}=-(7\/3){{\\alpha }_{3}}\\text{ }\\wedge \\text{ }{{\\alpha }_{1}}=(4\/3){{\\alpha }_{3}} \\right\\}<\/span>\n
Por lo cual, el respectivo vector caracter\u00edstico es:<\/p>\n{{v}_{2}}=\\left( \\begin{array}{r} 4 \\\\ -7 \\\\ 3 \\end{array} \\right)<\/span>\n
\n
Para<\/strong> {{\\lambda }_{3}}=-4<\/span>:<\/p>\nNu(A-{{\\lambda }_{3}}I)\\Rightarrow \\left( \\begin{array}{rrr|r} 5 & 1 & 5 & 0 \\\\ 0 & 7 & 0 & 0 \\\\ 3 & 0 & 3 & 0 \\end{array} \\right)\\sim ...\\left( \\begin{array}{rrr|r} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\end{array} \\right)<\/span>\n
Es decir,<\/p>\nNu(A-{{\\lambda }_{3}}I)=\\left\\{ \\left( {{\\alpha }_{1}},{{\\alpha }_{2}},{{\\alpha }_{3}} \\right)\\in {{\\mathsf{\\mathbb{R}}}^{3}}\/{{\\alpha }_{2}}=0\\text{ }\\wedge \\text{ }{{\\alpha }_{1}}+{{\\alpha }_{3}}=0 \\right\\}<\/span>\n
Por lo cual, el respectivo vector caracter\u00edstico es:<\/p>\n{{v}_{3}}=\\left( \\begin{array}{r} -1 \\\\ 0 \\\\ 1 \\end{array} \\right)<\/span>\n
\n
Finalmente, los vectores propios de la matriz de transformaci\u00f3n son coordenadas respecto a la base B (can\u00f3nica en este caso), de los vectores propios de la transformaci\u00f3n, por lo cual los resultados quedan as\u00ed:<\/p>\n
Para {{\\lambda }_{1}}=4<\/span>, el eigenvector es: {{v}_{1}}=5+3{{x}^{2}}<\/span>.
\nPara {{\\lambda }_{2}}=3<\/span>, el eigenvector es: {{v}_{2}}=4-7x+3{{x}^{2}}<\/span>.
\nPara {{\\lambda }_{3}}=-4<\/span>, el eigenvector es: {{v}_{3}}=-1+{{x}^{2}}<\/span>.<\/p>\n
Se cumplir\u00e1 en cada caso que:
\nT({{v}_{1}})={{\\lambda }_{1}}{{v}_{1}}<\/span>T({{v}_{2}})={{\\lambda }_{2}}{{v}_{2}}<\/span>T({{v}_{3}})={{\\lambda }_{3}}{{v}_{3}}<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"
Ejemplo. Sea una T.L. tal que Hallar sus eigenvalores y eigenvectores. Soluci\u00f3n: Necesitamos utilizar una base para hallar una matriz de transformaci\u00f3n. Puesto que no se nos indica alguna base en particular, se utilizar\u00e1 la base can\u00f3nica . Transformando cada vector de la base: Y hallando las coordenadas de tales transformadas respecto a la base … Sigue leyendo cl5-02. Ilustraci\u00f3n del m\u00e9todo<\/span><\/a><\/p>\n","protected":false},"author":9991,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"categories":[1414634],"tags":[],"class_list":["post-2418","post","type-post","status-publish","format-standard","hentry","category-temas-2da-evaluacion"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/posts\/2418","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/users\/9991"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/comments?post=2418"}],"version-history":[{"count":1,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/posts\/2418\/revisions"}],"predecessor-version":[{"id":7918,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/posts\/2418\/revisions\/7918"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/media?parent=2418"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/categories?post=2418"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/matg1049\/wp-json\/wp\/v2\/tags?post=2418"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}