# Limits of matrices

• May 18th 2008, 08:13 AM
pearlyc
Limits of matrices
Calculate lim n->infinity A^n for large n where

$\displaystyle \left[ \begin{array}{cccc} 1/2 & 0 & 0 \\ 1/2 & 1 & 3/4 \\ 0 & 0 & 1/4\end{array}\right]$
• May 18th 2008, 08:18 AM
Moo
Hello,

Quote:

Originally Posted by pearlyc
Calculate lim n->infinity A^n for large n where

$\displaystyle \left[ \begin{array}{cccc} 1/2 & 0 & 0 \\ 1/2 & 1 & 3/4 \\ 0 & 0 & 1/4\end{array}\right]$

Show by induction that :

$\displaystyle A^n=\begin{pmatrix} \frac{1}{2^n} & 0 & 0 \\ (1-\frac{1}{2^n}) & 1 & (1-\frac{1}{4^n}) \\ 0 & 0 & \frac{1}{4^n} \end{pmatrix}$

Then, find its limit ;)
• May 18th 2008, 08:32 AM
flyingsquirrel
Hi

Another solution has been given in this thread.
• May 18th 2008, 08:34 AM
pearlyc
omg, how do you do that ><?
• May 18th 2008, 08:59 AM
Isomorphism
The alternate but more mechanical way of doing this is:

Diagonalize the matrix: That is find an invertible P such that $\displaystyle P^{-1}AP = D$. From this we see that $\displaystyle P^{-1}A^n P = D^n$.

Here the eigenvalues are $\displaystyle 1,\frac12, \frac14$

With $\displaystyle P = \begin{pmatrix} 0 & 1 & 0\\ 1 & -1 & 0 \\ 0 & -1 & 3 \end{pmatrix}$

$\displaystyle P^{-1}AP = D = \begin{pmatrix} 1 & 0 & 0\\ 0 & \frac12 & 0 \\ 0 & 0 & \frac14 \end{pmatrix}$

$\displaystyle P^{-1}A^n P = D^n = \begin{pmatrix} 1 & 0 & 0\\ 0 & \left(\frac12\right)^n & 0 \\ 0 & 0 & \left(\frac14\right)^n \end{pmatrix}$

So
$\displaystyle \lim_{n \to \infty}P^{-1}A^n P = \lim_{n \to \infty} \begin{pmatrix} 1 & 0 & 0\\ 0 & \left(\frac12\right)^n & 0 \\ 0 & 0 & \left(\frac14\right)^n \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

$\displaystyle \lim_{n \to \infty}A^n = P \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} P^{-1}$