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 $v_{1}={\begin{pmatrix}0\\1\end{pmatrix}}$ and $v_{2}={\begin{pmatrix}2\\1\end{pmatrix}}$, respectively.

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

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

$A^{k}=PD^{k}P^{-1}={\begin{pmatrix}0&2\\1&1\end{pmatrix}}{\begin{pmatrix}1&0\\0&5^{k}\end{pmatrix}}{\begin{pmatrix}-{\frac {1}{2}}&1\\{\frac {1}{2}}&0\end{pmatrix}}={\begin{pmatrix}5^{k}&0\\{\frac {1}{2}}(5^{k}-1)&1\end{pmatrix}}.$

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