What are the limits as k -> infinity of the following:

[.4 .2; .6 .8]^k[1; 0]

[.4 .2; .6 .8]^k[0; 1]

[.4 .2; .6 .8]^k

I think the first two go to [1; 0] and the last goes to [0 0; 0 0], but am not sure.

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- Nov 17th 2010, 06:51 PMveronicak5678Steady States
What are the limits as k -> infinity of the following:

[.4 .2; .6 .8]^k[1; 0]

[.4 .2; .6 .8]^k[0; 1]

[.4 .2; .6 .8]^k

I think the first two go to [1; 0] and the last goes to [0 0; 0 0], but am not sure. - Nov 17th 2010, 07:11 PMpickslides
- Nov 17th 2010, 07:37 PMveronicak5678
Could you please explain how you got that answer?

- Nov 18th 2010, 11:51 AMpickslides
I used my calculator to find that matrix to a very high power. In this example either $\displaystyle A^{50}$ or $\displaystyle A^{100}$ will work.

- Nov 18th 2010, 03:18 PMAckbeet
The more complete method of solution involves diagonalizing the matrix. Let's say that for a matrix $\displaystyle A$ you could find an invertible matrix $\displaystyle P$ and a diagonal matrix $\displaystyle D$ such that $\displaystyle A=PDP^{-1}.$ Then

$\displaystyle A^{2}=(PDP^{-1})(PDP^{-1})=PDDP^{-1}=PD^{2}P^{-1},$

$\displaystyle A^{3}=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PDDDP^{-1}=PD^{3}P^{-1}.$

More generally,

$\displaystyle A^{k}=PD^{k}P^{-1}.$

But the kth power of a diagonal matrix is just the (diagonal) matrix with the diagonal elements of the original diagonal matrix raised to the kth power.

So you should be able to compute the limit exactly.