Define the system of equations;
xn+1=xn−yn
yn+1=2xn+4yn
so if we take A=(1−124)
and define
(xn+1yn+1)=(1−124)(xnyn)
Then given that the eigenvalues and a corresponding eigenvector to the eigenvalue can be defined as
λ1=3
μ1=(1−2)
λ2=2
μ2=(1−1)
(I already checked this)
Is it then true that the general solution can be defined as
xn=3nc1+2nc2
yn=−2∗3nc1−2nc2
Reinout
Perhaps I've misunderstood, but if c1 and c2 are initial values (n=0) for x and y then (for n = 1):
These don't look the same to me!