Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 05 (b)

MATH152 April 2013

• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QA 9 • QA 10 • QA 11 • QA 12 • QA 13 • QA 14 • QA 15 • QA 16 • QA 17 • QA 18 • QA 19 • QA 20 • QA 21 • QA 22 • QA 23 • QA 24 • QA 25 • QA 26 • QA 27 • QA 28 • QA 29 • QA 30 • QB 1(a) • QB 1(b) • QB 1(c) • QB 1(d) • QB 2(a) • QB 2(b) • QB 2(c) • QB 2(d) • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 5(a) • QB 5(b) • QB 5(c) • QB 6(a) • QB 6(b) • QB 6(c) •

Other MATH152 Exams

• April 2016 • April 2015 • April 2014 • April 2022 • April 2011 • April 2010 • April 2012 • April 2013 • April 2017 •

[hide] Question B 05 (b)
We are given the following transition matrix for a random walk $P={\begin{bmatrix}0&1/2\\1&1/2\end{bmatrix}}$ Find $P^{20}{\begin{bmatrix}1\\0\end{bmatrix}}$

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.

[show] Hint 1
Computing 20th powers of a general matrix by hand is possible but should be avoided at all cost. However, for some matrices is it easy to compute powers quickly. What are those, and how do we transform P into that?

[show] Hint 2
Diagonalize P. To do this, find the eigenvalues and find eigenvectors for this probability matrix.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] 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 begin by computing the eigenvalues. ${\begin{aligned}\det(P-\lambda I)&=\det \left({\begin{bmatrix}0&1/2\\1&1/2\end{bmatrix}}-{\begin{bmatrix}\lambda &0\\0&\lambda \end{bmatrix}}\right)\\&=\det \left({\begin{bmatrix}-\lambda &1/2\\1&1/2-\lambda \end{bmatrix}}\right)\\&=(-\lambda )(1/2-\lambda )-1/2\\&=\lambda ^{2}-{\frac {1}{2}}\lambda -{\frac {1}{2}}\\&={\frac {1}{2}}(2\lambda ^{2}-\lambda -1)\\&={\frac {1}{2}}(2\lambda +1)(\lambda -1)\\\end{aligned}}$ and thus the roots (and hence eigenvalues) are $\displaystyle \lambda =-1/2,1$ . To compute the eigenvectors, we look at the nullspace of $\displaystyle P-\lambda I$ for each eigenvalue. When $\displaystyle \lambda =-1/2$ , we have ${\begin{bmatrix}0&1/2\\1&1/2\end{bmatrix}}-{\begin{bmatrix}-1/2&0\\0&-1/2\end{bmatrix}}={\begin{bmatrix}1/2&1/2\\1&1\end{bmatrix}}$ and a vector (hence an eigenvector) in the kernel of this matrix is given by ${\begin{bmatrix}1\\-1\end{bmatrix}}$ Similarly, for $\lambda =1$ , we have ${\begin{bmatrix}0&1/2\\1&1/2\end{bmatrix}}-{\begin{bmatrix}1&0\\0&1\end{bmatrix}}={\begin{bmatrix}-1&1/2\\1&-1/2\end{bmatrix}}$ and a vector (hence an eigenvector) in the kernel of this matrix is given by ${\begin{bmatrix}1\\2\end{bmatrix}}$ Adjoin these eigenvectors to make a matrix $M={\begin{bmatrix}1&1\\-1&2\end{bmatrix}}$ Then, our theory of diagonalizability gives us that $D=M^{-1}PM\qquad {\text{or, equivalently,}}\qquad P=MDM^{-1}$ where $D={\begin{bmatrix}-1/2&0\\0&1\end{bmatrix}}$ is the diagonal matrix consisting of the eigenvalues. Next, notice that $P^{20}=(MDM^{-1})^{20}=\underbrace {(MDM^{-1})(MDM^{-1})...(MDM^{-1})} _{20{\text{ times}}}=MD^{20}M^{-1}$ Finally, our work will pay off: Taking the 20th power of a diagonal matrix is easy and convenient. Having to inverting a 2x2 matrix is a small price to pay for this convenience. ${\begin{aligned}P^{20}=MD^{20}M^{-1}&={\begin{bmatrix}1&1\\-1&2\end{bmatrix}}{\begin{bmatrix}-1/2&0\\0&1\end{bmatrix}}^{20}{\begin{bmatrix}1&1\\-1&2\end{bmatrix}}^{-1}\\&={\begin{bmatrix}1&1\\-1&2\end{bmatrix}}{\begin{bmatrix}(-1/2)^{20}&0\\0&(1)^{20}\end{bmatrix}}\left({\frac {1}{3}}{\begin{bmatrix}2&-1\\1&1\end{bmatrix}}\right)\\&={\frac {1}{3}}{\begin{bmatrix}1/(2^{20})&1\\-1/(2^{20})&2\end{bmatrix}}{\begin{bmatrix}2&-1\\1&1\end{bmatrix}}\\&={\frac {1}{3}}{\begin{bmatrix}1/(2^{19})+1&1-1/(2^{20})\\2-1/(2^{19})&1/(2^{20})+2\end{bmatrix}}\end{aligned}}$ Applying the vector ${\begin{bmatrix}1\\0\end{bmatrix}}$ to this matrix gives $P^{20}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}{\frac {1/(2^{19})+1}{3}}\\{\frac {-1/(2^{19})+2}{3}}\end{bmatrix}}$ and this completes the problem.