How they commute $(T-\mu I)(T-\lambda I)^{p-1}(x)=^?(T-\lambda I)^{p-1}(T-\mu I)(x)$

Question

Im currently reading "Linear Algebra, Fourth Edition,Stephen H. Friedberg, Arnold J. Insel, Lawrence E. Spence"

At page 485 a theorem is located as follows:

Theorem 7.1 Let T be a linear operator on a vector space V, and let $\lambda$ be an eigen value of T. Then a)...

b)For any scalar $\mu\not=\lambda$, the restriction of $T-\mu I$ to $K_\lambda$ is injective.

Question: How to commute following elements which are used in the proof.

$$1)(T-\lambda I)^pT(x)=^?T(T-\lambda I)^p(x)$$ $$2)(T-\mu I)(T-\lambda I)^{p-1}(x)=^?(T-\lambda I)^{p-1}(T-\mu I)(x)$$

Answer 1

Hint By induction it suffices to show that the identities hold for $p = 1$. But, for example, for (1) we have $(T - \lambda I) T = T^2 - \lambda T = T (T - \lambda I)$.

A more general version of this argument shows that for any polynomials $f, g$ $f(T)$ and $g(T)$ commute.

Answer 2

In part 1), write $T=(T-\lambda I)+\lambda I$. It is clear that $(T-\lambda I)^p$ commutes with both summands, and therefore with $T$.

There is a similar trick you can use in part 2).