(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# T-cyclic Operator - linear algebra

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