Commutativity of linear operators

Let $$T: V \rightarrow V$$ be a linear operator on a vector space $V$ over a field $F$ and with a basis $$B = (\mathbf{u}_{1}, \cdots, \mathbf{u}_{n}).$$ Let $$f(x), g(x) \in F[x]$$ be polynomials. Then

$$f(T)g(T) = g(T)f(T).$$

Does that mean if i am given a linear operator $T:V \to V$, and another map $T - \lambda I_V$, i can say both maps commute with each other because of the above theorem.

I say : $$f(x) = x,~~~ g(x) = x - \lambda$$ Then $$f(T) = T,~~~ g(T) = T - \lambda I_V$$

and hence they commute. Moreover, if $g(x) = (x-\lambda)^n$, then $h(T) = (T-\lambda I_V)^n$ and subsequently by the above theorem, we have $$T \circ (T- \lambda I_V)^n = (T - \lambda I_V)^n \circ T$$

Can anyone clarify this? I am working on the proof of Jordan normal form.

• Yes, $T$ commutes with powers of $T-\lambda I$. Commented Nov 25, 2017 at 6:13
• Yes, but is my line of logic correct in showing that they commute ?
– nan
Commented Nov 25, 2017 at 6:14
• Have a little confidence! Commented Nov 25, 2017 at 6:17