I'm trying to solve some problems in Hoffman and Kunze and I'm kind of stuck on this one. This is 6.5.3 on Hoffman and Kunze.
Here is the question:
Let $T$ be a a linear operator on an $n$-dimensional space, and suppose that $T$ has $n$ distinct characteristic values. Prove that any linear operator $U$ which commutes with $T$ is a polynomial in $T$.
My work so far: Since $T$ has $n$ distinct characteristic values and since the space it acts on is also $n$ dimensional, $T$ must be diagonalisable. Now using the fact $UT=TU$ I can show that $U$ must also be diagonalizable. So these two operators must be simultaneously diagonalizable. Now I'm stuck with where this is going. I'm thinking of simultaneous diagonalizability because this is section 6.5 of Hoffman and Kunze and it deals with simultaneous diagonalizability. Am I on the right track with this?. Can anybody help?
Thanks so much for your time and your answers.