Suppose $A$ is a square matrix $d\times d$ and that it's a linear opetaror over the albgeraicly closed field $\mathbb{C}$. In particular, $A$ has $n$ different eigen values, each one with multiplicity $m_i$. Therefore $A$ is similar to a Jordan matrix in a basis $\mathcal{J}:=(w_{i,j}; i=1, ... , n, j=1, ..., m_i)$ formed by the generalized eigen vectors of $A$. Now, why would the next statement be true?
$$A w_{i,j}=\lambda_i w_{i,j} + w_{i,j+1} \text{ where } w_{i,m_i+1}=0.$$
I think that the convention $w_{i,m_i+1}=0$ is because since $A$ is a Jordan form, it's block diagonal with $\lambda_i, \; i=1,..., n$ in its diagonal and 1's over them except for the rows where we change eigen value. Also, by thinking of $A$ as a Jordan form, and multiplying it for a vector $[v]_{J}=(v_1, v_2, ... , v_d)^T$ written in de $\mathcal{J}$ base, I can see how \begin{align} Av = \begin{pmatrix} \lambda_1 v_1 + v_2\\ \vdots\\ \lambda_1 v_{m_1} + 0 v_{m_1+1}\\ \lambda_2 v_{m_1+1} + v_{m_1+2}\\ \vdots \end{pmatrix} \end{align} So, my intuition is that to calculate the $k-th$ row, you multiply the $k-th$ entry on your vector $v$ by the asociated eigen value $\lambda$ and then add it with the next entry. However I'm having trouble seeing why this, in the case of the vectors of the base, would be as if the $k-th$ row of $w_{i,j}$ is precisely $w_{i,j}$ and then the next one is that of $w_{i,j+1}$, which in principle should be a different vector altogether. I also tried thinking $w_{i,j}$ as $0's$ and a $1$ in the $k-th$ row... however with that I get that it should be $A w_{i,j}=-\lambda_i w_{i,j} + w_{i,j-1}$ instead.
Any guidance in the right direction is highly appreciated. As well as any reference where I can find more about the Jordan form and its properties.