# $A =\begin{pmatrix} 2 & -\frac{3}{2}\\ 1 & - \frac{1}{2} \end{pmatrix}$ and $B =\begin{pmatrix} 3 & -3\\ 2 & - 2\end{pmatrix}$, show $\lim A^n = B$.

Put $A =\left(\begin{matrix} 2 & -\frac{3}{2}\\ 1 & - \frac{1}{2} \end{matrix}\right)$ and $B =\left(\begin{matrix} 3 & -3\\ 2 & - 2\end{matrix}\right)$,

Show that $\lim_{n \to \infty} A^n = B$.

First I try to write $A$ as $PDP^{-1}$, $D$ is a diagonal matrix, which makes $A^n = PD^nP^{-1}$, but I am not sure about how to determine the matrices $P$ and $D$. Also, I am not sure about whether this is a right approach to this problem.

Could you help me to figure it out? Thank you!

• Hint: Eigenvalues make up $D$ and $P$ is the corresponding eigenvectors, this approach works perfectly. As a check, you should get $D = \left( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \\ \end{array} \right)$. – Moo Mar 28 '16 at 5:00
• Oh I got it! Thank you very much!!!! – Nhay Mar 28 '16 at 5:10