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

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.

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

• Every two polynomials on the same operator commute, since the operator commutes with all its powers. – DonAntonio May 21 '19 at 18:51

## 2 Answers

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.

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).