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

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

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