Find the eigenvalues of $A$.
\nLet $a_1$ be the least eigenvalue of $A,\\;\\;\\; a_1=\\;\\;$[[0]]
\nLet $a_2$ be the greatest eigenvalue of $A,\\;\\; a_2=\\;\\;$[[1]]
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{s*(mnA-a11)}", "minValue": "{s*(mnA-a11)}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{s*(mxA-a11)}", "minValue": "{s*(mxA-a11)}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "Find eigenvectors for $A$.
\nLet $(1,y_1)^T$ be an eigenvector corresponding to $a_1,\\;\\;\\;\\;y_1=\\;\\;$[[0]]
\nLet $(1,y_2)^T$ be an eigenvector corresponding to $a_2,\\;\\;\\;\\;y_2=\\;\\;$[[1]]
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{mnB}", "minValue": "{mnB}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{mxB}", "minValue": "{mxB}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "\n \n \nFind the eigenvalues of $B$.
\n \n \n \nLet $b_1$ be the least eigenvalue of $B,\\;\\;\\; b_1=\\;\\;$[[0]]
\n \n \n \nLet $b_2$ be the greatest eigenvalue of $B,\\;\\; b_2=\\;\\;$[[1]]
\n \n \n ", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{x1}", "minValue": "{x1}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{x2}", "minValue": "{x2}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "Find eigenvectors for $B$.
\nLet $(x_1,1)^T$ be an eigenvector corresponding to $b_1,\\;\\;\\;\\;x_1=\\;\\;$[[0]]
\nLet $(x_2,1)^T$ be an eigenvector corresponding to $b_2,\\;\\;\\;\\;x_2=\\;\\;$[[1]]
", "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{bn11}", "minValue": "{bn11}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{bn12}", "minValue": "{bn12}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{bn21}", "minValue": "{bn21}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}, {"showCorrectAnswer": true, "allowFractions": false, "scripts": {}, "type": "numberentry", "maxValue": "{bn22}", "minValue": "{bn22}", "correctAnswerFraction": false, "marks": 1, "showPrecisionHint": false}], "type": "gapfill", "prompt": "\nFind $B^{\\var{n}}$ using the last two parts of this question:
\n| $B^{\\var{n}} = \\Bigg($ | \n[[0]] | \n[[1]] | \n$\\Bigg)$ | \n
| [[2]] | \n[[3]] | \n
Input your answers as integers.
\n ", "showCorrectAnswer": true, "marks": 0}], "statement": "\n \n \nFind the eigenvalues and eigenvectors for the matrices $A$ and $B$ where:
\\[ A=\\begin{pmatrix} \\var{a11}&\\var{a12}\\\\ \\var{a21}&\\var{a22} \\end{pmatrix},\\;\\;\\;\\;\\;\n \n B=\\begin{pmatrix} \\var{b11}&\\var{b12}\\\\ \\var{b21}&\\var{b22} \\end{pmatrix}\n \n \\]
10/07/2012:
\nAdded tags.
\nIn the Advice section it is not explained how to find the trace and the determinant of the matrix - Should this be included?
\nQuestion appears to be working correctly.
\n24/12/2012:
\nChecked calculations, OK. Added tested1 tag.
", "licence": "Creative Commons Attribution 4.0 International", "description": "$A,\\;B$ $2 \\times 2$ matrices. Find eigenvalues and eigenvectors of both. Hence or otherwise, find $B^n$ for largish $n$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "a)
\n\\[A - \\lambda I_2 = \\begin{pmatrix} \\var{a11}-\\lambda & \\var{a12}\\\\ \\var{a21} & \\var{a22}-\\lambda \\end{pmatrix}\\]
Hence the characteristic polynomial $p(\\lambda)$ is: \\[\\begin{eqnarray*} \\mathrm{det}\\left(A-\\lambda I_2 \\right)&=&\\simplify[zeroTerm]{({a11}-lambda)({a22}-lambda)-{a12}*{a21}}\\\\ &=& \\simplify[std]{lambda^2-{trA}*lambda+{dA}}\\\\ &=&\\simplify[std]{(lambda-{a})(lambda-{b})} \\end{eqnarray*} \\]
We see that on solving $p(\\lambda)=0$ we get the eigenvalues:
\\[\\lambda_1=\\var{mnA},\\;\\;\\;\\lambda_2=\\var{mxA}\\]
Note: We could have found the characteristic polynomial by noting that for a 2 × 2 matrix $A$ then the characteristic polynomial is
\\[\\lambda^2-\\mathrm{trace}(A)+\\mathrm{det}(A)\\]
where $\\mathrm{trace}(A) = \\var{trA},\\;\\;\\;\\mathrm{det}(A)=\\var{dA}$
b)
\n1. $\\lambda=\\var{mnA}$
\nWe have the eigenspace is given by all $v=(x,y)^T$ such that $(\\simplify{A-{mnA}}I_2)v=(0,0)^T$ i.e.
\n\\[\\begin{pmatrix} \\var{a11-mnA}&\\var{a12}\\\\ \\var{a21}&\\var{a22-mnA} \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix}\\]
\nThis gives the two equations:
\n\\[ \\begin{eqnarray*} \\simplify[std]{{a11-mnA}x + {a12}y}&=&0\\\\ \\simplify[std]{{a21}x + {a22-mnA}y}&=&0 \\end{eqnarray*} \\]
There is only one equation here as we see that the equations are the same (one is a multiple of the other).
So putting $x=1$ in the first equation we get $y_1=\\var{-s*(a11-mnA)}$
\nHence the eigenvector we want is \\[\\begin{pmatrix} 1 \\\\ \\var{-s*(a11-mnA)} \\end{pmatrix}\\]
\n2. $\\lambda=\\var{mxA}$
\nIn this case we have the equations:
\n\\[ \\begin{eqnarray*} \\simplify[std]{{a11-mxA}x + {a12}y}&=&0\\\\ \\simplify[std]{{a21}x + {a22-mxA}y}&=&0 \\end{eqnarray*} \\]
\nOnce again there is only one equation, so putting $x=1$ in the first equation we get $y_2=\\var{-s*(a11-mxA)}$
\nHence the eigenvector we want is \\[\\begin{pmatrix} 1 \\\\ \\var{-s*(a11-mxA)} \\end{pmatrix}\\]
\nc)
\nThe characteristic polynomial is given by:
\n\\[p(\\lambda)=\\simplify[std]{lambda^2-{b11+b22}*lambda + {dB}}\\]
\nSolving $p(\\lambda)=0$, we find the eigenvalues for $B$ are:
\\[\\lambda_1=\\var{mnB},\\;\\;\\;\\lambda_2=\\var{mxB}\\]
d)
\n1. $\\lambda=\\var{mnB}$
\nThe equations are:
\\[ \\begin{eqnarray*} \\simplify[std]{{b11-mnB}x + {b12}y}&=&0\\\\ \\simplify[std]{{b21}x + {b22-mnB}y}&=&0 \\end{eqnarray*} \\]
Putting $y=1$ in the second equation we get $x_1=\\var{s*(b22-mnB)}$
\nHence the eigenvector we want is \\[\\begin{pmatrix} \\var{s*(b22-mnB)}\\\\1 \\end{pmatrix}\\]
\n2. $\\lambda=\\var{mxB}$
\nThe equations are:
\\[ \\begin{eqnarray*} \\simplify[std]{{b11-mxB}x + {b12}y}&=&0\\\\ \\simplify[std]{{b21}x + {b22-mxB}y}&=&0 \\end{eqnarray*} \\]
Putting $y=1$ in the second equation we get $x_2=\\var{s*(b22-mxB)}$
Hence the eigenvector we want is \\[\\begin{pmatrix} \\var{s*(b22-mxB)}\\\\1 \\end{pmatrix}\\]
\ne)
\nFor the last part we use the diagonalisation of $B$ given by the last two parts.
\nThus if $x_1,\\;\\;x_2,\\;\\;\\lambda_1,\\;\\;\\lambda_2$ are as above for $B$ then we have $B=PDP^{-1} \\Rightarrow B^{\\var{n}}=PD^{\\var{n}}P^{-1}$ where:
\n\\[\\begin{eqnarray*} P &=& \\begin{pmatrix} x_1 & x_2\\\\1&1 \\end{pmatrix} = \\begin{pmatrix} \\var{s*(b22-mnB)} & \\var{s*(b22-mxB)} \\\\1&1 \\end{pmatrix}\\Rightarrow P^{-1}= \\simplify[std]{1/{x1-x2}}\\begin{pmatrix} 1 & \\var{-s*(b22-mxB)} \\\\-1&\\var{s*(b22-mnB)} \\end{pmatrix}\\\\ \\\\ D&=& \\begin{pmatrix} \\lambda_1 & 0\\\\0&\\lambda_2 \\end{pmatrix} = \\begin{pmatrix} \\var{mnB} & 0\\\\0&\\var{mxB} \\end{pmatrix} \\Rightarrow D^{\\var{n}}=\\begin{pmatrix} \\var{mnB^n} & 0\\\\0&\\var{mxB^n} \\end{pmatrix} \\end{eqnarray*} \\]
\nHence \\[\\begin{eqnarray*}B^{\\var{n}}&=&PD^{\\var{n}}P^{-1}\\\\ \\\\ &=&\\simplify[std]{1/{x1-x2}}\\begin{pmatrix} \\var{s*(b22-mnB)} & \\var{s*(b22-mxB)} \\\\1&1 \\end{pmatrix}\\begin{pmatrix} \\var{mnB^n} & 0\\\\0&\\var{mxB^n} \\end{pmatrix}\\begin{pmatrix} 1 & \\var{-s*(b22-mxB)} \\\\-1&\\var{s*(b22-mnB)} \\end{pmatrix}\\\\ \\\\ &=&\\begin{pmatrix} \\var{bn11} & \\var{bn12}\\\\\\var{bn21}&\\var{bn22} \\end{pmatrix} \\end{eqnarray*} \\]
", "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/"}]}