>>13633673nvm, I slightly fucked up
>>13633647>>13633520>>13633484Assuming you know that the eigenvalues are all less than 1 in absolute value or equal to 1, then we know the limit exists.
A^4 = (A+I)/2, call this [1]
Suppose A^n = aA^3+bA^2+cA+dI, call this [2]
Multiply [2] by [1], simplify any powers >= 4, now call this [3]
Now look at the coefficients. Their limit must be the same as the limit of the coefficients in [3], so set the coefficients equal.
Now you have a system of 4 equations. Solve it.
For your question, you get:
a = (a+b)/2
b = (b+c)/2
c = (c+d+a/2)/2
d = (d+a/2)/2
So
a = b = c = 2d
So the limit will be a scalar multiple of A^3+A^2+A+I/2.
Indeed, it's 2/7 times this above matrix.
To find this scalar, recall that Trace(AB) = Trace(BA).
So when you diagonalize and get A = PDP^(-1), Tr(A^n) = Tr(PD^n P^(-1)) = Tr(D^n), but in the limit, Tr(D^n) = a sum of 0's and 1's. There are as many zeros as there are eigenvalues who absolute value is less then 1, and the rest are ones.
We know a root of the characteristic polynomial is 1, and the rest have absolute value less than 1, so the trace of the limit is 1.
So take the trace of A^3+A^2+A+I/2 (which is 7/2) and divide by it.
So a good strategy to start is checking the multiplicity of 1 as a root of the characteristic polynomial.
