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

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.