You are given that $\\var{vector(1,-b,-a)}$ is an eigenvector of $M$.
\nFind the corresponding eigenvalue: [[0]]
\nAlso $\\var{vector(1,1-b,-a)}$ is an eigenvector of $M$.
\nFind the corresponding eigenvalue: [[1]]
\n", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "-lambda^3+{coefflamsq}*lambda^2+{coefflam}*lambda+{det}", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "all", "marks": 1, "vsetrangepoints": 5}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "eigen2", "minValue": "eigen2", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"answer": "{-b}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}, {"answer": "{b}/{g}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "Enter numbers as fractions or integers and not as decimals.
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all,fractionNumbers", "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "Find the characteristic polynomial of $M$ and hence another eigenvalue for $M$.
\nEnter the characteristic polynomial in the form $P_M(\\lambda) = -\\lambda^3+a\\lambda^2+b\\lambda+c$.
\nWrite the letter $\\lambda$ as lambda.
Characteristic polynomial: $P_M(\\lambda) = \\;$[[0]]
\nHence find another eigenvalue of $M$: [[1]]
\nFind a corresponding eigenvector for this eigenvalue. Scale your vector such that the first component is $1$. You have to find the other two components:
\nEigenvector = $\\Bigg($ $1$[[2]][[3]] $\\Bigg)$
\nInput both components as fractions or integers and not as decimals.
", "showCorrectAnswer": true, "marks": 0}], "statement": "Let $M=\\var{M}$. Answer the following questions.
", "tags": ["characteristic equation", "characteristic polynomial", "checked2015", "determinant", "eigenvalues", "eigenvectors", "linear algebra", "linear equations", "MAS1602", "matrices", "matrix algebra"], "rulesets": {}, "preamble": {"css": ".vector-input {\n display: inline-block;\n vertical-align: middle;\n text-align: center;\n}\n.vector-input .component {\n float: center;\n clear: both;\n display: block !important;\n text-align: center;\n}\n.vector-input .part {\n margin-top: 0;\n margin-bottom: 0;\n}", "js": ""}, "type": "question", "metadata": {"notes": "Created 19/09/2014
", "licence": "Creative Commons Attribution 4.0 International", "description": "Given a 3 x 3 matrix, and two eigenvectors find their corresponding eigenvalues. Also fnd the characteristic polynomial and using this find the third eigenvalue and a normalised eigenvector $(x=1)$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "For the eigenvector $\\var{vector(1,-b,-a)}$ we have:
\n$\\var{M}\\var{vector(1,-b,-a)}=\\var{vector(eigen1,-eigen1*b,-eigen1*a)}=\\var{eigen1}\\var{vector(1,-b,-a)}$.
\nHence the corresponding eigenvalue is $\\var{eigen1}$.
\nSimilarly, for the eigenvector $\\var{vector(1,1-b,-a)}$ we have:
\n$\\var{M}\\var{vector(1,1-b,-a)}=\\var{vector(eigen3,eigen3*(1-b),-eigen3*a)}=\\var{eigen3}\\var{vector(1,1-b,-a)}$.
\nHence the corresponding eigenvalue for this eigenvector is $\\var{eigen3}$.
\nThe characteristic polynomial is given by $P_M(\\lambda)=\\operatorname{det}(M-\\lambda I_3)$. The roots of this are the eigenvalues.
\nWe find that in this case:
\n\\[\\begin{align}\\operatorname{det}(M-\\lambda I_3)&=\\operatorname{det}\\begin{pmatrix}\\var{m[0][0]}-\\lambda &\\var{m[0][1] }&\\var{m[0][2]}\\\\ \\var{m[1][0]}&\\var{m[1][1] }-\\lambda &\\var{m[1][2]}\\\\ \\var{m[2][0]}&\\var{m[2][1] } &\\var{m[2][2]}-\\lambda \\end{pmatrix}\\\\&=\\simplify{-lambda^3+{coefflamsq}*lambda^2+{coefflam}*lambda+{det}}\\end{align}\\]
\nNow we know that $\\simplify{lambda-{eigen1}}$ and $\\simplify{lambda-{eigen3}}$ are both factors of the characteristic polynomial.
\nHence we have:
\n$\\simplify{-lambda^3+{coefflamsq}*lambda^2+{coefflam}*lambda+{det}=-(lambda-{eigen1})(lambda-{eigen3})(lambda-r)}$ for some number $r$.
\nSince the constant term in the characteristic polynomial is the product of the three eigenvalues, $r=\\simplify[!basic]{{det}/({eigen1}*{eigen3})={eigen2}}$ and this is the third eigenvalue.
\nTo find an eigenvector corresponding to this eigenvalue we solve the equation:
\n$\\simplify{M*vector(x,y,z)={M}*vector(x,y,z)={eigen2}*vector(x,y,z)}$
\nThis gives the equations:
\n\\[\\begin{align} \\simplify{{m[0][0]}*x+{m[0][1]}*y+{m[0][2]}*z}&=\\simplify{{eigen2}*x}\\\\ \\simplify{{m[1][0]}*x+{m[1][1]}*y+{m[1][2]}*z}&=\\simplify{{eigen2}*y}\\\\ \\simplify{{m[2][0]}*x+{m[2][1]}*y+{m[2][2]}*z}&=\\simplify{{eigen2}*z}\\end{align}\\]
\nThe question asked you to find an eigenvector with $x=1$ and if we substitute this into the equations we find (only need to use two of the equations) that $y=\\var{-b}$ and $z=\\simplify[all,fractionNumbers]{{b}/{g}}$.
\nHence the eigenvector we want is $\\simplify[all,fractionNumbers]{vector(1,{-b},{b}/{g})}$.
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Ugur Efem", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18261/"}]}