This is from Artin's "Algebra" (2nd Edition), Proposition 4.7.1 and Lemma 4.7.2..
Let $x$ be a generalized eigenvector of a linear operator $T$, with eigenvalue $\lambda$ and exponent $d$. For $j\geq0$, let $u_{j}=(T-\lambda)^{j}x$, $B=(u_{0},...,u_{d-1})$ and $z$ be a nontrivial linear combination of $B$. Then $z$ is a generalized eigenvector with eigenvalue $\require{cancel} \cancel{\lambda}$ and exponent $\require{cancel} \cancel{d-j}$ . (Edited)
I'm having trouble understanding some of the steps in the proof of the lemma. Firstly, it is claimed that for a linear combination $y=c_{j}u_{j}+\ ... \ +c_{d-1}u_{d-1}$ with $j\leq d-1$ and $c_{j}\neq0$, $$(T-\lambda)^{d-j-1}y=c_{j}u_{d-1}.$$
By the way it is written I believe the author meant simply to ensure that $y$ is a nontrivial linear combination of the elements of $B$ and that the $j$-th element wouldn't be repeated, this might be where I am mistaken. That would yield, $$\begin{equation} \begin{aligned} (T-\lambda)^{d-j-1}y&=c_{j}(T-\lambda)^{d-j-1}u_{j}+\sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d-j-1}u_{i} \\ &=c_{j}(T-\lambda)^{d-1}x \ + \sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j-1}x \\ &=c_{j}u_{d-1} \ + \sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j-1}x. \end{aligned} \end{equation}$$
In this last equation I don't see how the terms on the right add up to $0$ or how they are all $0$ for any $j\leq d-1$. The next step of the proof claims the following, $$(T-\lambda)^{d-j}y=0.$$ Which from the previous equations becomes, $$\begin{equation} \begin{aligned} (T-\lambda)^{d-j}y&=c_{j}(T-\lambda)^{d}x \ + \sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j}x \\ &=0 \ + \sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j}x. \end{aligned} \end{equation}$$
And I have essentially the same question here. What I can think of is, in both cases, splitting the operator, $$\sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j-1}x=\sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{i-j-1}(T-\lambda)^{d}x,$$ and similarly, $$\sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{d+i-j}x=\sum_{i=0;i\neq j}^{d-1}c_{i}(T-\lambda)^{i-j}(T-\lambda)^{d}x.$$
This would lead me to the answer, however I'm unsure about the details regarding invertibility, or even one sided inverses, of the operator $(T-\lambda)$ as some of the powers would be negative.
Any explanations or hints regarding theorems or lemmas from linear algebra that I might be forgetting will be appreciated. Thanks in advance.
You've written the statement of the lemma wrong in part of the statement and in your work.
Note that $y = c_j u_j + \cdots + c_{d-1} u_{d-1}$ with $c_j \ne 0$. In particular, there are no terms involving $u_1, \ldots, u_{j-1}$.
Therefore your sums should be $\sum_{i = j+1}^{d-1}$. Then $i-j > i-j-1 \ge 0$ so there are no negative exponents in the end of your proof attempt.