Science:Math Exam Resources/Courses/MATH221/April 2013/Question 09

MATH221 April 2013

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Question 09
Given the discrete dynamical system xn+1 = Axn with $A=\left[{\begin{array}{cc}0.8&0.5\\0.2&0.5\end{array}}\right]$ Find the eigenvalues and eigenvectors of the system. Then find closed formulae for the components of xn with the initial value $\mathbf {x} _{0}=\left[{\begin{array}{c}4\\1\end{array}}\right]$

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?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
If you want to avoid the fractions in the matrix $\ A$, you can determine the eigenvalues and eigenvectors of the matrix $\ B=10A$. The eigenvector of $\ B$ and $\ A$ are the same, while the eigenvalues of $\ B$ are ten times the eigenvalues of $\ A$.

Hint 2
Once you compute the eigenvalues $\ \lambda _{1},\lambda _{2}$ and eigenvectors $\ v_{1},v_{2}$, you can write $\ x_{0}$ as $\ x_{0}=av_{1}+bv_{2}$ for some $a,b\in \mathbb {R}$. How does that help to compute $\ x_{n}$?

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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. Determine eigenvalues of $\ B=10A$ We will first find the eigenvalues of the matrix $B=\left[{\begin{array}{cc}8&5\\2&5\end{array}}\right]$ to avoid dealing with fractions. The eigenvalues of the matrix A, will simply be ${\frac {1}{10}}$ of this integer matrix B. To find the eigenvalues we calculate as usual: ${\begin{aligned}\det \left(\left[{\begin{array}{cc}8-t&5\\2&5-t\end{array}}\right]\right)&=(8-t)(5-t)-(5)(2)\\&=t^{2}-13t+30\\&=(t-3)(t-10)\\&=0\end{aligned}}$ Determine eigenvectors of $\ B=10A$ For the eigenvectors with eigenvalue 3 we look at $\left[{\begin{array}{cc}8-3&5\\2&5-3\end{array}}\right]=\left[{\begin{array}{cc}5&5\\2&2\end{array}}\right]$ which gives the eigenvector $\left[{\begin{array}{cc}1\\-1\end{array}}\right]$. For the eigenvalue 10, we look at the $\left[{\begin{array}{cc}-2&5\\2&-5\end{array}}\right]$ which gives the eigenvectors $\left[{\begin{array}{cc}5\\2\end{array}}\right]$. Eigenvalues and eigenvectors of $\ A$ For the matrix A this means that A has the eigenvalue ${\frac {3}{10}}$ with eigenvector $v_{1}=\left[{\begin{array}{cc}1\\-1\end{array}}\right]$ eigenvalue $\ 1$ with eigenvector $v_{2}=\left[{\begin{array}{cc}5\\2\end{array}}\right]$ Find $\ a,b\in \mathbb {R}$ such that $\ x_{0}=av_{1}+bv_{2}$ To find the closed formulae for the components of xn we decompose the vector $x_{0}$ into a sum of the eigenvectors of A. To do so, we use the extended matrix: $[v_{1}\ v_{2}\ |\ x_{0}]=\left[{\begin{array}{cc|c}1&5&4\\-1&2&1\\\end{array}}\right]\sim \left[{\begin{array}{cc|c}1&5&4\\0&7&5\\\end{array}}\right]\sim \left[{\begin{array}{cc|c}1&0&3/7\\0&1&5/7\\\end{array}}\right]$ Thus a = 3/7 and b = 5/7 and therefore $\left[{\begin{array}{cc}4\\1\end{array}}\right]=3/7\left[{\begin{array}{cc}1\\-1\end{array}}\right]+5/7\left[{\begin{array}{cc}5\\2\end{array}}\right]$. Computing $\ x_{n}$ We can now compute $x_{n}$ as follows: ${\begin{aligned}x_{n}=A^{n}x_{0}&=A^{n}\left({\frac {3}{7}}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}\left[{\begin{array}{cc}5\\2\end{array}}\right]\right)\\&={\frac {3}{7}}A^{n}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}A^{n}\left[{\begin{array}{cc}5\\2\end{array}}\right]\\&={\frac {3}{7}}{\frac {3^{n}}{10^{n}}}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}1^{n}\left[{\begin{array}{cc}5\\2\end{array}}\right]\end{aligned}}$