Science:Math Exam Resources/Courses/MATH221/December 2008/Question 05 (b)/Solution 1

From part (a), we know that the eigenvalues for $A$ are ${\lambda }_1=1$ and ${\lambda }_2=5$, with corresponding eigenvectors $v_1=\left(\begin{array}{c}0\\ 1\end{array}\right)$ and $v_2=\left(\begin{array}{c}2\\ 1\end{array}\right)$, respectively.

Let $D=\left(\begin{array}{cc}1& 0\\ 0& 5\end{array}\right)$ denote the matrix of eigenvalues, and $P=\left(\begin{array}{cc}0& 2\\ 1& 1\end{array}\right)$ the matrix whose columns are the corresponding eigenvectors. Then $A$ is diagonalized via $D=P^{-1}AP$, or, $A=PD{P}^{-1}$.

This decomposition allows for easy computations of powers of $A$, as for any integer $k$, we have $A^k=(PD{P}^{-1}{)}^k=P{D}^k{P}^{-1}$, and $D^k=\left(\begin{array}{cc}{1}^k& 0\\ 0& 5^k\end{array}\right)$. Therefore, we compute $P^{-1}=\left(\begin{array}{cc}-\frac{1}{2}& 1\\ \frac{1}{2}& 0\end{array}\right)$, and multiply the three matrices to obtain

$A^k=P{D}^k{P}^{-1}=\left(\begin{array}{cc}0& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 5^k\end{array}\right)\left(\begin{array}{cc}-\frac{1}{2}& 1\\ \frac{1}{2}& 0\end{array}\right)=\left(\begin{array}{cc}{5}^k& 0\\ \frac{1}{2}(5^k-1)& 1\end{array}\right).$

Then for $k=1000$, we have $A^{1000}=\left(\begin{array}{cc}{5}^{1000}& 0\\ \frac{1}{2}(5^{1000}-1)& 1\end{array}\right).$