Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 05 (b)/Solution 1

We begin by computing the eigenvalues.

$\begin{array}{rl}\det (P-\lambda I)& =\det (\left[\begin{array}{cc}0& 1/2\\ 1& 1/2\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right])\\ & =\det \left(\left[\begin{array}{cc}-\lambda & 1/2\\ 1& 1/2-\lambda \end{array}\right]\right)\\ & =(-\lambda )(1/2-\lambda )-1/2\\ & ={\lambda }^2-\frac{1}{2}\lambda -\frac{1}{2}\\ & =\frac{1}{2}(2{\lambda }^2-\lambda -1)\\ & =\frac{1}{2}(2\lambda +1)(\lambda -1)\\ \end{array}$

and thus the roots (and hence eigenvalues) are $\lambda =-1/2,1$. To compute the eigenvectors, we look at the nullspace of $P-\lambda I$ for each eigenvalue. When $\lambda =-1/2$, we have

$\left[\begin{array}{cc}0& 1/2\\ 1& 1/2\end{array}\right]-\left[\begin{array}{cc}-1/2& 0\\ 0& -1/2\end{array}\right]=\left[\begin{array}{cc}1/2& 1/2\\ 1& 1\end{array}\right]$

and a vector (hence an eigenvector) in the kernel of this matrix is given by $\left[\begin{array}{c}1\\ -1\end{array}\right]$

Similarly, for $\lambda =1$, we have

$\left[\begin{array}{cc}0& 1/2\\ 1& 1/2\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-1& 1/2\\ 1& -1/2\end{array}\right]$

and a vector (hence an eigenvector) in the kernel of this matrix is given by $\left[\begin{array}{c}1\\ 2\end{array}\right]$

Adjoin these eigenvectors to make a matrix

$M=\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]$

Then, our theory of diagonalizability gives us that

$D=M^{-1}PM\text{or, equivalently,}P=MD{M}^{-1}$

where

$D=\left[\begin{array}{cc}-1/2& 0\\ 0& 1\end{array}\right]$

is the diagonal matrix consisting of the eigenvalues. Next, notice that

$P^{20}=(MD{M}^{-1}{)}^{20}=\underset{20\text{ times}}{\underbrace{(MD{M}^{-1})(MD{M}^{-1})...(MD{M}^{-1})}}=M{D}^{20}{M}^{-1}$

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.

$\begin{array}{rl}{P}^{20}=M{D}^{20}{M}^{-1}& =\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]{\left[\begin{array}{cc}-1/2& 0\\ 0& 1\end{array}\right]}^{20}{\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]}^{-1}\\ & =\left[\begin{array}{cc}1& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}(-1/2{)}^{20}& 0\\ 0& (1{)}^{20}\end{array}\right]\left(\frac{1}{3}\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\right)\\ & =\frac{1}{3}\left[\begin{array}{cc}1/(2^{20})& 1\\ -1/(2^{20})& 2\end{array}\right]\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right]\\ & =\frac{1}{3}\left[\begin{array}{cc}1/(2^{19})+1& 1-1/(2^{20})\\ 2-1/(2^{19})& 1/(2^{20})+2\end{array}\right]\end{array}$

Applying the vector $\left[\begin{array}{c}1\\ 0\end{array}\right]$ to this matrix gives

$P^{20}\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}\frac{1/(2^{19})+1}{3}\\ \frac{-1/(2^{19})+2}{3}\end{array}\right]$

and this completes the problem.