Science:Math Exam Resources/Courses/MATH221/December 2011/Question 07

MATH221 December 2011

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Question 07
Let $A=\left[{\begin{array}{ccc}1/2&0&1/3\\0&1/2&1/2\\1/2&1/2&1/6\end{array}}\right]\qquad {\text{and let}}\qquad x_{0}=\left[{\begin{array}{c}5\\5\\6\end{array}}\right]$ Let xn = Anx0. Then find x100 = A100x0. What happens to xn as n becomes very large? Note: there is a typo in the exam, it is written xn = Anxn instead of xn = Anx0

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
Don't calculate the 100th power of A directly. Instead, diagonalize A = TDT -1 and then take the 100th power of that.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We are not going to multiply A with itself a hundred times, at least not directly. Instead, we are going to diagonalize A first, because the 100th power of a diagonal matrix is much easier to calculate. Indeed, if D is the diagonal matrix corresponding to A (i.e. is a matrix with the eigenvalues of A on the diagonal), and if T is the transformation matrix of eigenvectors of A, then ${\begin{aligned}A^{n}&=(TDT^{-1})^{n}\\&=\underbrace {TDT^{-1}TDT^{-1}\cdots TDT^{-1}} _{n{\text{ times}}}\\&=TD^{n}T^{-1}\end{aligned}}$ So let's find D, T and T -1. We need the eigenvalues and eigenvectors of A, so we calculate the roots of the characteristic polynomial $\det(A-\lambda I)$: ${\begin{aligned}\det(A-\lambda I)&=\left|{\begin{array}{ccc}1/2-\lambda &0&1/3\\0&1/2-\lambda &1/2\\1/2&1/2&1/6-\lambda \end{array}}\right|\\&=({\frac {1}{2}}-\lambda )({\frac {1}{2}}-\lambda )({\frac {1}{6}}-\lambda )+0+0-{\frac {1}{3}}({\frac {1}{2}}-\lambda ){\frac {1}{2}}-0-({\frac {1}{2}}-\lambda ){\frac {1}{2}}{\frac {1}{2}}\\&=({\frac {1}{2}}-\lambda )({\frac {1}{12}}-{\frac {1}{2}}\lambda -{\frac {1}{6}}\lambda +\lambda ^{2}-{\frac {1}{6}}-{\frac {1}{4}})\\&=({\frac {1}{2}}-\lambda )(\lambda ^{2}-{\frac {2}{3}}\lambda -{\frac {1}{3}})\\&=({\frac {1}{2}}-\lambda )(\lambda +{\frac {1}{3}})(\lambda -1)\end{aligned}}$ And so the eigenvalues are 1/2, -1/3 and 1. We compute an eigenvector for each eigenvalue using row reduction on the system $A-\lambda I$. For λ = 1 we have the system ${\begin{aligned}A-I&=\left[{\begin{array}{ccc}-1/2&0&1/3\\0&-1/2&1/2\\1/2&1/2&-5/6\end{array}}\right]\\&\sim \left[{\begin{array}{ccc}1&0&-2/3\\0&1&-1\\0&0&0\end{array}}\right]\\\end{aligned}}$ And so this eigenspace is spanned by the eigenvector [2, 3, 3]. For λ = 1/2 we have the system ${\begin{aligned}A-(1/2)I&=\left[{\begin{array}{ccc}0&0&1/3\\0&0&1/2\\1/2&1/2&-1/3\end{array}}\right]\\&\sim \left[{\begin{array}{ccc}1&1&0\\0&0&1\\0&0&0\end{array}}\right]\\\end{aligned}}$ And so this eigenspace is spanned by the eigenvector [1, -1, 0]. For λ = -1/3 we have the system ${\begin{aligned}A+(1/3)I&=\left[{\begin{array}{ccc}5/6&0&1/3\\0&5/6&1/2\\1/2&1/2&1/2\end{array}}\right]\\&\sim \left[{\begin{array}{ccc}1&1&1\\5&0&2\\0&5&3\end{array}}\right]\\&\sim \left[{\begin{array}{ccc}1&1&1\\0&5&3\\0&0&0\end{array}}\right]\\\end{aligned}}$ And so this eigenspace is spanned by the eigenvector [2, 3, -5]. Using this basis of eigenvectors, we have that $\displaystyle A=T\cdot D\cdot T^{-1}$ Where $D=\left[{\begin{array}{ccc}1&0&0\\0&1/2&0\\0&0&-1/3\end{array}}\right]\quad T=\left[{\begin{array}{ccc}2&1&2\\3&-1&3\\3&0&-5\end{array}}\right]$ We need now to compute the inverse of the matrix T. We obtain $T^{-1}=\left[{\begin{array}{ccc}1/8&1/8&1/8\\3/5&-2/5&0\\3/40&3/40&-1/8\end{array}}\right]$ And so ${\begin{aligned}A^{n}&=(T\cdot D\cdot T^{-1})^{n}\\&=T\cdot D^{n}\cdot T^{-1}\\&=\left[{\begin{array}{ccc}2&1&2\\3&-1&3\\3&0&-5\end{array}}\right]\cdot \left[{\begin{array}{ccc}1^{n}&0&0\\0&(1/2)^{n}&0\\0&0&(-1/3)^{n}\end{array}}\right]\cdot \left[{\begin{array}{ccc}1/8&1/8&1/8\\3/5&-2/5&0\\3/40&3/40&-1/8\end{array}}\right]\\&=\left[{\begin{array}{ccc}2&2^{-n}&2(-1/3)^{n}\\3&-2^{-n}&(-1)^{n}3^{1-n}\\3&0&-5(-1/3)^{n}\end{array}}\right]\cdot \left[{\begin{array}{ccc}1/8&1/8&1/8\\3/5&-2/5&0\\3/40&3/40&-1/8\end{array}}\right]\\&=\left[{\begin{array}{ccc}1/4+(3/5)(1/2)^{n}+(3/20)(-1/3)^{n}&1/4-(2/5)(1/2)^{n}+(3/20)(-1/3)^{n}&1/4-(1/4)(-1/3)^{n}\\3/8-(3/5)(1/2)^{n}+(9/40)(-1/3)^{n}&3/8+(2/5)(1/2)^{n}+(9/40)(-1/3)^{n}&3/8-(3/8)(-1/3)^{n}\\3/8-(3/8)(-1/3)^{n}&3/8-(3/8)(-1/3)^{n}&3/8+(5/8)(-1/3)^{n}\end{array}}\right]\end{aligned}}$ So we can compute x100 ${\begin{aligned}x_{100}&=A^{100}x_{0}\\&=\left[{\begin{array}{ccc}1/4+(3/5)2^{-100}+(3/20)3^{-100}&1/4-(2/5)2^{-100}+(3/20)3^{-100}&1/4-(1/4)3^{-100}\\3/8-(3/5)2^{-100}+(9/40)3^{-100}&3/8+(2/5)2^{-100}+(9/40)3^{-100}&3/8-(3/8)3^{-100}\\3/8-(3/8)3^{-100}&3/8-(3/8)3^{-100}&3/8+(5/8)3^{-100}\end{array}}\right]\left[{\begin{array}{c}5\\5\\6\end{array}}\right]\\&=\left[{\begin{array}{c}4+(1/2)^{100}\\6-(1/2)^{100}\\6\end{array}}\right]\\\end{aligned}}$ As n becomes very large, xn approaches $\lim _{n\to \infty }x_{n}=\left[{\begin{array}{c}4\\6\\6\end{array}}\right]$

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