1. The problem statement, all variables and given/known data Let T be a linear operator on a vector space V, and suppose that V is a T-cyclic subspace of itself. Prove that if U is a linear operator on V, then UT=TU if and only if U=g(T) for some polynomial g(t). 2. Relevant equations They suggest supposing that V is generated by v. and choose g(t) such that g(T)(v) = U(v). 3. The attempt at a solution I could get some coherent stuff for one way.. => assume U=g(T), prove UT=TU. first, {v, T(v), T^2(v), ... , T^n-1(v)} is a basis of V (thus V is n-dimensional). U=g(T) let y=U(x) choosem g(t) such that y = g(T)(v). then UT(x)=U(T(x)) = U(T(a_0v + a_1T(v)+...+a_n-1T^n-1(v)) = U(a_0T(v) + a_1T^2(v) + ... +a_n-1T^n(v)) = a_0UT(v) + a_1UT^2(v) + ... +a_n-1UT^n(v) then TU(x)= T(U(x)) (where x is a linear comb of basis vectors, x=a_0v + a_1T(v)+...+a_n-1T^n-1(v)) thats as far as i got.. =( <= the other direction is more difficult for me.. i'm not really sure where to get the proper expression for g(t) from at all.
if U = g(T), then UT = TU is obvious, you don't need to worry about v's or x's or anything, it has nothing to do with V being T-cyclic, you don't use the hint, etc. For the other direction, use the hint. Since V is T-cyclic, there does indeed exist some v such that v generates V. U(v) is an element of V, and since v generates V, U(v) is indeed equal to g(T)(v) for some polynomial g. Let g be this polynomial. Now just show that for any x in V, g(T)(x) = U(x). Remember that in this direction, you may assume UT = TU.
Hi, sorry for reviving this old thread, but I'm doing the same question.. I know we can assume UT = TU for one direction, but I can't make out what that's supposed to mean.. The two operators commute, so am I supposed to work something from the fact that they commute? ( some property that makes them commute..) thank you
I tried a couple of promising things, but I'm just blindly stabbing at it . How would I go about coming up with a plan for this? I need to know something about UT = TU... :S thanks