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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
![{\displaystyle {\begin{aligned}A^{n}&=(TDT^{-1})^{n}\\&=\underbrace {TDT^{-1}TDT^{-1}\cdots TDT^{-1}} _{n{\text{ times}}}\\&=TD^{n}T^{-1}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/44a43d4019c7ad6dd7fb6910469ca4621e81c4f8)
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 :
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/93b4f579bffc75f4f516ad6688b61e6f8af41794)
And so the eigenvalues are 1/2, -1/3 and 1.
We compute an eigenvector for each eigenvalue using row reduction on the system .
For λ = 1 we have the system
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ad33ed1709a0cdbd77b1c90604dc9c9a86cfbc62)
And so this eigenspace is spanned by the eigenvector [2, 3, 3].
For λ = 1/2 we have the system
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a82b0c8b8a525b3cb75b94951499316f3f1420d0)
And so this eigenspace is spanned by the eigenvector [1, -1, 0].
For λ = -1/3 we have the system
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/305a9e6947306a0e776cf500bd027c5551460855)
And so this eigenspace is spanned by the eigenvector [2, 3, -5].
Using this basis of eigenvectors, we have that
![{\displaystyle \displaystyle A=T\cdot D\cdot T^{-1}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f237359a35d3d94734784ef5cf0207703879da33)
Where
![{\displaystyle 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]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/211695c7f7719b7940dbc343cd49ba291c1c6932)
We need now to compute the inverse of the matrix T. We obtain
![{\displaystyle 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]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ffee27a2a32c262b08adb8a9ff7b0f4efd17de7a)
And so
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3a2b58ce1f118475e7a8821e4b9d2c04ac413f25)
So we can compute x100
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/807c0aa740c68f0fbc395e83d053fd8ccd6a8bfa)
As n becomes very large, xn approaches
![{\displaystyle \lim _{n\to \infty }x_{n}=\left[{\begin{array}{c}4\\6\\6\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/e48d29887ca1063c49a89e75dab6a5d909b2020d)
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