I am reading a paper and there is a steep in a proof that I cannot understand. The proof says:
Suppose $v_\lambda$ is a generalized eigenvector corresponding to an eigenvalue $\lambda$ of the map $T$, then there is a $n \, \in \mathbb{N}$ such that $(T-\lambda\,I)^n(v_\lambda)=0$
and they arrive to:
$$Tr\{(T-\lambda\,I)^n(v_\lambda)\}=(1-\lambda)^n\,Tr\;v_\lambda=0$$
Due to $T$ is trace-preserving, its trace is 1, but I am not able to see why $v_\lambda$ is treated as a matrix ( at least it is what I understand) and no as a vector. So I would like to know if someone has a reference to understand this or if someone can explain this to me. Thanks in advance.