Exercise 1.6.6

• The eigenvalues of $A$ are: ${\lambda }_1=0.4$, ${\lambda }_2=1$. The eigenvectors are $v_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ and $v_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$
• The eigenvalues of $A^{\infty }$ are: ${\lambda }_1=0$, ${\lambda }_2=1$. The eigenvectors are $v_1=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right]$ and $v_2=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$

The eigenvalues of $A^{100}$ are ${\lambda }_1=0.4^{100}=1.6e-40$, ${\lambda }_2=1^{100}=1$.

Since $A=\mathit{X\Lambda }{X}^{-1}$, $A^{100}=X{\Lambda }^{100}{X}^{-1}$, the ${\Lambda }^{100}$ approach the eigenvalues ${\Lambda }_{\infty }$ of $A^{\infty }$, and $A$ and $A^{\infty }$ both have the same eigenvectors, $A^{\infty }=X{\Lambda }_{\infty }{X}^{-1}$, so $A^{100}$ is close to $A^{\infty }$.