2
$\begingroup$

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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.