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

MATH152 April 2013

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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.

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?

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

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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 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.

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