TITLE: linearly independent generalized eigenvectors
QUESTION [5 upvotes]: I'm self-studying Axler's Linear Algebra Done Right and I am not understanding one step of the proof of 8.13 (Linearly independent generalized eigenvectors). It is the same step as the one that "yields" 1.4.65 in the proof of Lemma 1.4.63 in this book, which also leaves it unexplained.
We are multiplying both sides of the equation $\sum a_i v_i = 0$, where each $v_i$ is a generalized eigenvector, by $(T-\lambda_j)^k\prod_{i \ne j} (T - \lambda_i)^n$, where each $\lambda_i$ is the eigenvalue corresponding to $v_i$, and somehow getting:
$a_j (T-\lambda_j)^k\prod_{i \ne j} (T - \lambda_i)^n v_j = 0$
(i.e., all terms of $\sum a_i v_i$ disappear except for the one contained $v_j$)
I understand that each $v_i$ is an element of $null(T-\lambda_i)^n$ and so its term would disappear if the operator were applied directly to the $v_i$, but only the final $(T-\lambda_i)^n$ applies directly to each $v_i$. Are these transformations commutative for some reason?
I'm likely missing something trivial here and would appreciate your insights. Thanks!

REPLY [6 votes]: Just to bring this question officially to an end, I will make my comment an answer: yes, of course these transformations are commutative, because they are polynomials in $T$.