# Why is $(T-\lambda I)^pT(x) = T(T-\lambda I)^p(x)$?

Definition. Let $$T$$ be a linear operator on a vector space $$V$$, and let $$\lambda$$ be an eigenvalue of $$T$$. The generalized eigenspace of $$T$$ corresponding to $$\lambda_1$$ denoted $$K_\lambda$$, is the subset of $$V$$ defined by

$$K_\lambda$$ = $$\{x \in V: (T-\lambda I)^p(x) = 0$$ for some positive integer $$p\}$$.

To show that $$K_\lambda$$ is $$T$$-invariant, consider any $$x \in K_\lambda$$. Choose a positive integer $$p$$ such that $$(T-\lambda)^p = 0$$. Then

$$(T-\lambda I)^pT(x) = T(T-\lambda I)^p(x) = T(0) = 0$$

I wanted to know how we go from $$(T-\lambda I)^pT(x) = T(T-\lambda I)^p(x)$$?

• Induction on $p$? Or write $T$ as $(T-\lambda I)+\lambda I$? – Angina Seng Aug 4 '18 at 15:42
• Take both expressions and expand out $(T-\lambda I)^p$ using the binomial theorem. After simplifying, you will find there are equal. – Mike Earnest Aug 4 '18 at 15:42

Since $T$ commutes with both $T$ and $\lambda\operatorname{Id}$, $T$ commutes with $T-\lambda\operatorname{Id}$ and therefore it comutes with any power of $T-\lambda\operatorname{Id}$.