Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 04 (c)/Solution 1

For $P={\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}}$, we solve $det(P-\lambda I)=0$ to find the eigenvalues. So, we have

$det(P-\lambda I)=det{\begin{pmatrix}0.9-\lambda &0.2\\0.1&0.8-\lambda \end{pmatrix}}=\lambda ^{2}-1.7\lambda +0.7=(\lambda -1)(\lambda -0.7)=0.$

We then get that $\lambda _{1}=0.7$ and $\lambda _{2}=1.$

We now find the corresponding eigenvector $v_{1}={\begin{pmatrix}a_{1}\\b_{1}\end{pmatrix}}$ to the eigenvalue $\lambda _{1}=0.7$:

$(P-\lambda _{1}I)v_{1}={\begin{pmatrix}0.9-0.7&0.2\\0.1&0.8-0.7\end{pmatrix}}v_{2}={\begin{pmatrix}0.2&0.2\\0.1&0.1\end{pmatrix}}{\begin{pmatrix}a_{1}\\b_{1}\end{pmatrix}}={\begin{pmatrix}0.2a_{1}+0.2b_{1}\\0.1a_{1}+0.1b_{1}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}.$

This implies that $0.1a_{1}+0.1b_{1}=0,$ and therefore $b_{1}=-a_{1}.$ Now letting $a_{1}=1,$ we have $b_{1}=-1,$ and $v_{1}={\begin{pmatrix}1\\-1\end{pmatrix}}.$

Let $v_{2}={\begin{pmatrix}a_{2}\\b_{2}\end{pmatrix}}$ be an eigenvector corresponding to the eigenvalue $\lambda _{2}=1$. Similarly, we have

$(P-\lambda _{2}I)v_{2}={\begin{pmatrix}0.9-1&0.2\\0.1&0.8-1\end{pmatrix}}v_{2}={\begin{pmatrix}-0.1&0.2\\0.1&-0.2\end{pmatrix}}{\begin{pmatrix}a_{2}\\b_{2}\end{pmatrix}}={\begin{pmatrix}-0.1a_{2}+0.2b_{2}\\0.2a_{2}-0.1b_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}.$

We then get that $-0.1a_{1}+0.2b_{1}=0,$ and therefore $a_{1}=2b_{1}.$ Now letting $b_{1}=1,$ we have $a_{1}=2,$ and $v_{2}={\begin{pmatrix}2\\1\end{pmatrix}}.$

Therefore, we have two eigenvalues and the basis of eigenvector as

$\color {blue}{\lambda _{1}=0.7,\lambda _{2}=1,\left\{{\begin{pmatrix}1\\-1\end{pmatrix}},{\begin{pmatrix}2\\1\end{pmatrix}}\right\}.}$