{"id":223,"date":"2019-09-10T17:16:13","date_gmt":"2019-09-10T17:16:13","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=223"},"modified":"2020-06-18T16:44:25","modified_gmt":"2020-06-18T16:44:25","slug":"eigenvalues-and-eigenvectors","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/linear-algebra\/eigenvalues-and-eigenvectors\/","title":{"rendered":"Eigenvalues and Eigenvectors"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<title>Eigenvalues and Eigenvectors &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nWe review here the basics of computing eigenvalues and eigenvectors.\nEigenvalues and eigenvectors play a prominent role in the study of\nordinary differential equations and in many applications in the\nphysical sciences.  Expect to see them come up in a variety of\ncontexts!\n\n<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Definitions<\/h4>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"alignright\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"175\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/eigen1.gif?resize=300%2C175&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-347\"\/><\/figure><\/div>\n\n\n\n<p> Let $A$ be an $n \\times n$ matrix.  The number $\\lambda$ is an <b> eigenvalue<\/b> of $A$ if there exists a non-zero vector ${\\bf v}$ such that $$ A{\\bf v} = \\lambda {\\bf v}. $$ In this case, vector ${\\bf v}$ is called an <b>eigenvector<\/b> of $A$ corresponding to $\\lambda$. <br><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Computing Eigenvalues and Eigenvectors<\/h4>\n\n\n\n<p>\n\t\tWe can rewrite the condition $A{\\bf v} = \\lambda {\\bf v}$ as\n\t\t$$\n\t\t(A- \\lambda I){\\bf v} = {\\bf 0}.\n\t\t$$\n\t\twhere $I$ is the $n \\times n$ identity matrix.  Now, in order for a\n\t\t<i>non-zero<\/i> vector ${\\bf v}$ to satisfy this equation, $A &#8211;\n\t\t\\lambda I$ must <i>not<\/i> be invertible.\n<\/p>\n\n\n\n<p> Otherwise, if $A &#8211; \\lambda I$ has an inverse, \\begin{eqnarray*} (A &#8211; \\lambda I)^{-1}(A &#8211; \\lambda I){\\bf v} &amp; = &amp; (A &#8211; \\lambda  I)^{-1}{\\bf 0} \\\\ {\\bf v} &amp; = &amp; {\\bf 0}.  \\end{eqnarray*} But we are looking for a non-zero vector ${\\bf v}$. That is, the determinant of $A &#8211; \\lambda I$ must equal 0.  We call $p(\\lambda )= \\det (A &#8211; \\lambda I)$ the <strong>characteristic<br> polynomial<\/strong> of $A$.  The eigenvalues of $A$ are simply the roots of the characteristic polynomial of $A$.  <\/p>\n\n\n\n<p> <\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet $A = \\left[ \\begin{array}{rr}\n\t2 &amp; -4\\\\\n\t-1 &amp; -1 \n\\end{array}\\right] $.  Then $$\\begin{array}{rcl}\n\tp(\\lambda) &amp; = &amp; \\det \\left[\\begin{array}{cc}\n\t\t2-\\lambda &amp; -4 \\\\\n\t\t-1 &amp; -1-\\lambda\n\t\\end{array}\\right]\\\\\n\t&amp; = &amp; (2-\\lambda)(-1-\\lambda)-(-4)(-1)\\\\\n\t&amp; = &amp; \\lambda^{2} -\\lambda -6\\\\\n\t&amp; = &amp; (\\lambda -3)(\\lambda +2).\n\\end{array}$$\n\nThus, $\\lambda_1 =3$ and $\\lambda_2 = -2$ are the eigenvalues of $A$.\n\n<\/p>\n\n\n\n<p>\nTo find eigenvectors ${\\bf v} = \\left[ \\begin{array}{c}\n\tv_1\\\\\n\tv_2\\\\\n\t\\vdots\\\\\n\tv_n\n\\end{array}\\right] $ corresponding to an eigenvalue $\\lambda$, we\nsimply solve the system of linear equations given by\n$$\n(A-\\lambda I) {\\bf v} = {\\bf 0}.\n$$\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nThe matrix $A = \\left[ \\begin{array}{rr}\n\t2 &amp; -4\\\\\n\t-1 &amp; -1 \n\\end{array}\\right] $ of the previous example has eigenvalues\n$\\lambda_1 =3$ and $\\lambda_2 = -2$.  Let&#8217;s find the eigenvectors\ncorresponding to $\\lambda_1 =3$.  Let ${\\bf v}= \\left[{v_1 \\atop\nv_2}\\right]$.  Then $(A-3I){\\bf v}={\\bf 0}$ gives us\n$$\n\\left[\\begin{array}{cc}\n\t2-3 &amp; -4\\\\\n\t-1 &amp; -1-3\n\\end{array}\\right]\\left[\\begin{array}{c}\n\tv_1\\\\\n\tv_2\n\\end{array}\\right] = \\left[\\begin{array}{c}\n\t0\\\\\n\t0\n\\end{array}\\right],\n$$\nfrom which we obtain the duplicate equations\n\\begin{eqnarray*}\n\t-v_1-4v_2 &amp; = &amp; 0 \\\\\n\t-v_1-4v_2 &amp; = &amp; 0. \n\\end{eqnarray*}\nIf we let $v_2=t$, then $v_1=-4t$.  All eigenvectors corresponding to\n$\\lambda_1 =3$ are multiples of $\\left[{-4 \\atop 1}\\right] $ and thus\nthe eigenspace corresponding to $\\lambda_1 =3$ is given by the span of\n$\\left[{-4 \\atop 1}\\right] $.  That is, $\\left\\{\\left[{-4 \\atop\n1}\\right]\\right\\}$ is a <b>basis<\/b> of the eigenspace corresponding to\n$\\lambda_1 =3$.\n\n<\/p>\n\n\n\n<p>\nRepeating this process with $\\lambda_2 = -2$, we find that\n\\begin{eqnarray*}\n\t4v_1 -4V_2 &amp; = &amp; 0  \\\\\n\t-v_1  +  v_2 &amp; = &amp; 0\n\\end{eqnarray*}\nIf we let $v_2=t$ then $v_1=t$ as well.  Thus, an eigenvector\ncorresponding to $\\lambda_2 = -2$ is $\\left[{1 \\atop 1}\\right]$ and\nthe eigenspace corresponding to $\\lambda_2 = -2$ is given by the span\nof $\\left[{1 \\atop 1}\\right]$.  $\\left\\{\\left[{1 \\atop\n1}\\right]\\right\\}$ is a basis for the eigenspace corresponding to\n$\\lambda_2 = -2$.\n\n<\/p>\n\n\n\n<p>\nIn the following example, we see a two-dimensional eigenspace.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet $A=\\left[\\begin{array}{ccc}\n\t5 &amp; 8 &amp; 16\\\\\n\t4 &amp; 1 &amp; 8\\\\\n\t-4 &amp; -4 &amp; -11\n\\end{array}\\right]$.  Then $p(\\lambda ) = \\det\\left[\\begin{array}{ccc}\n\t5-\\lambda &amp; 8 &amp; 16\\\\\n\t4 &amp; 1-\\lambda &amp; 8\\\\\n\t-4 &amp; -4 &amp; -11-\\lambda\n\\end{array}\\right] = (\\lambda-1)(\\lambda+3)^{2}$\nafter some algebra!  Thus, $\\lambda_1 = 1$ and $\\lambda_2=-3$ are the\neigenvalues of $A$.  Eigenvectors ${\\bf v} = \\left[\\begin{array}{c}\n\tv_1\\\\\n\tv_2\\\\\n\tv_3\n\\end{array}\\right]$ corresponding to $\\lambda_1=1$ must satisfy\n\n<\/p>\n\n\n\n<center>\n\t$\\begin{array}{rcrcrcr}\n\t\t4v_1 &amp; + &amp; 8v_2 &amp; + &amp; 16v_3 &amp; = &amp; 0\\\\\n\t\t4v_1 &amp;   &amp;      &amp; + &amp;  8v_3 &amp; = &amp; 0\\\\\n\t\t-4v_1 &amp; &#8211; &amp; 4v_2 &amp; &#8211; &amp; 12v_3 &amp; = &amp; 0.\n\t\\end{array}$\n<\/center>\n\n\n\n<p>\nLetting $v_3=t$, we find from the second equation that $v_1=-2t$, and\nthen $v_2=-t$.  All eigenvectors corresponding to $\\lambda_1=1$ are\nmultiples of $\\left[\\begin{array}{c}\n\t-2\\\\\n\t-1\\\\\n\t1\n\\end{array}\\right]$, and so the eigenspace corresponding to\n$\\lambda_1=1$ is given by the span of $\\left[\\begin{array}{c}\n\t-2\\\\\n\t-1\\\\\n\t1\n\\end{array}\\right]$.  $\\left\\{\\left[\\begin{array}{c}\n\t-2\\\\\n\t-1\\\\\n\t1\n\\end{array}\\right]\\right\\}$ is a basis for the eigenspace\ncorresponding to $\\lambda_1=1$.\n\n<\/p>\n\n\n\n<p>\nEigenvectors corresponding to $\\lambda_2=-3$ must satisfy\n\n<\/p>\n\n\n\n<center>\n\t$\\begin{array}{rcrcrcr}\n\t\t8v_1 &amp; + &amp; 8v_2 &amp; + &amp; 16v_3 &amp; = &amp; 0\\\\\n\t\t4v_1 &amp; + &amp; 4v_2 &amp; + &amp;  8v_3 &amp; = &amp; 0\\\\\n\t\t-4v_1 &amp; &#8211; &amp; 4v_2 &amp; &#8211; &amp; 8v_3 &amp; = &amp; 0.\n\t\\end{array}$\n<\/center>\n\n\n\n<p>\nThe equations here are just multiples of each other!\nIf we let $v_3 = t$ and $v_2 = s$, then $v_1 = -s -2t$.  Eigenvectors\ncorresponding to $\\lambda_2=-3$ have the form \n$$\n\\left[\\begin{array}{c}\n\t-1\\\\\n\t1\\\\\n\t0\n\\end{array}\\right]s+\\left[\\begin{array}{c}\n\t-2\\\\\n\t0\\\\\n\t1\n\\end{array}\\right]t.\n$$\nThus, the eigenspace corresponding to \n$\\lambda_2=-3$ is two-dimensional and is spanned by $\\left[\\begin{array}{c}\n\t-1\\\\\n\t1\\\\\n\t0\n\\end{array}\\right]$ and $\\left[\\begin{array}{c}\n\t-2\\\\\n\t0\\\\\n\t1\n\\end{array}\\right]$.  $\\left\\{\\left[\\begin{array}{c}\n\t-1\\\\\n\t1\\\\\n\t0\n\\end{array}\\right],\\left[\\begin{array}{c}\n\t-2\\\\\n\t0\\\\\n\t1\n\\end{array}\\right]\\right\\}$ is a basis for the eigenspace\ncorresponding to $\\lambda_2=-3$.\n\n<\/p>\n\n\n\n<center>\n<h4>Notes<\/h4>\n<\/center>\n\n\n\n<ul class=\"wp-block-list\"><li>Eigenvalues and eigenvectors can be complex-valued as well as real-valued. <br><\/li><li>The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. <br> <\/li><li>The techniques used here are practical for $2 \\times 2$ and $3 \\times 3$ matrices.  Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods. <\/li><\/ul>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h4>Key Concepts<\/h4>\n<\/center>\n\n\n\n<p> Let $A$ be an $n \\times n$ matrix.  The eigenvalues of $A$ are the roots of the characteristic polynomial $$ p(\\lambda )= \\det (A &#8211; \\lambda I). $$ For each eigenvalue $\\lambda$, we find eigenvectors ${\\bf v} =\\left[ \\begin{array}{c} v_1\\\\ v_2\\\\ \\vdots\\\\ v_n \\end{array}\\right] $ by solving the linear system $$ (A &#8211; \\lambda I){\\bf v} = {\\bf 0}. $$ The set of all vectors ${\\bf v}$ satisfying $A{\\bf v}= \\lambda {\\bf v}$ is called the <strong>eigenspace<\/strong> of $A$ corresponding to $\\lambda$. <br><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ3910\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Eigenvalues and Eigenvectors &#8211; HMC Calculus Tutorial We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts! Definitions Let $A$ be an&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":61,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-223","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/223","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=223"}],"version-history":[{"count":5,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/223\/revisions"}],"predecessor-version":[{"id":1233,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/223\/revisions\/1233"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/61"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=223"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}