Abstract Algebra?

T is a linear operator on the vector space V, find an ordered basis for the T-cyclic subspace generated by the vector z.





V = R^4, T(a, b, c, d) = (a + b, b - c, a + c, a+ d) and z = (1, 0, 0, 0)





Solution:





Let W be the T-cyclic subspace generated by z and β its basis.





W = {z, T(z), T²(z)....}





Therefore,


z = (1, 0, 0, 0)


T(z) = (1, 0, 1, 1)


T²(z) = (1, -1, 2, 2)


T³(z) = (0, -3, 3, 3)


T^4(z) = (-3, -6, 3, 3)


T^5(z) = -(-9, -9, 0, 0)


T^6(z) = (-18, -9, -9, -9)


T^7(z) = (-27, 0, -27, -27)


T^8(z) = (-27, -27, -54, -54)





So obviously they have begun to repeat as multiples of the first 7 elements in W. The back of th text says that β = {(1, 0, 0, 0), (1, 0, 1, 1), (1, -1, 2, 2)} which means that dim(W) = 3 but doesnt that necessarily mean that there should only be three coordinates being "used" in each of the points rather then 4? And how is it that you check that the basis will span W when I dont see a general way of expressing the elements in W?





Thank you

Abstract Algebra?
T^3(z) = 0 z -3 T(z) + 3 T^2(z)





So you can get T^3 form the first 3.





But that means you can get them all from the first 3!





T^4z = T T^3 z = 0 T(z) - 3 T^2(z) + 3 T^3 (z) =





- 3 T^2(z) + 3 ( -3 T(z) + 3 T^2(z)) = -9 T(z) + 6T^2(z)





etc. Each time apply T once more and reexpress T^3(z) using the first formula I showed.





All powers T^n(z) are linear combinations of z, T(z), and T^2(z), which is what you are supposed to prove.





Its a three dimensional subspace of a 4-dimensional space.

local florist