linear iteration

[kevin4maths]

A: (5 7)
....(-2 -4)
eigenvalues are -2 and 3
eigenlines are y=-x and y=-(2/7)x

Calculate the second and third points of the iteration sequence with
recurrence relation Xn+1 = Axn (n = 0, 1, 2, . . .), for each of the
following initial points.
(i) (−1, 1)
(ii) (1, 1)

i have worked it already (i) (-1,1)---(2, -2)----(-4, 4)
(ii) (1, 1)---(12, -6)---(18, 0)

Question: Describe the long-term behaviour of each of the iteration sequences (i) and (ii)

i would really appreciate any help.

 

[HallsofIvy]

(i) (1, -1)-> (-2, 2)-> (4, -4)->
Every step both x and y are multiplied by -2 so "long term behaviour" is that x and y alternate is sign and are powers of 2. This is simple because (1, -1) is in the direction of "y= -x" and so is an eigenvector with eigenvalue -2. You could write this as A^n(1, -1)= ((-2)^n, -(-2)^n).

(ii) (1, 1)-> (12, -6)-> (18, 0)->
This is more complicated because (1, 1) is not an eigenvector. But we can find a and b so that (1, 1)= a(1, -1)+ b(7, -2) (I got (7, -2) by taking x= 7, in y= (-2/7)x). That gives a+ 7b=1, -a- 2b= 1 so that, adding both equations, 5b= 2 so b= 2/5. Then a+ 7b= a+14/5= 1 so that a= 1- 14/5= -9/5. Then A^n(1, 1)= (-9/5)A^n(1, -1)+ (2/5)A^n(7, -2). A, as said above, multiplies (1, -1), repeatedly by -2 and it multiplies (7, -2) by 3. You could write this as A^n(1, 1)= (-9/5)((-2)^n, -(-2)^n)+ (2/5)(7(3)^n, -2(3)^n)= ((-9/5)(-2)^n,+ 7(3)^n, -(9/5)(-2)^n- 2(3)^n).

 

[kevin4maths]

thank you