What can you say about $T$ if dim$(V) =$ Rank$(T - \lambda I)$?

Question

I stumbled across this condition and I wanted to know what you could say about this:

Let $T:V \to V$ be a linear transformation, with $V$ having a finite dimension. What can you say about $T$ if dim$(V) =$ Rank$(T - \lambda I)$, where $\lambda$ is some eigenvalue of $T$?

I think this implies that there doesn't exist a Jordan canonical form of $T$, but I was curious if there were any other things you could deduce from this condition.

Answer 1

Given a finite-dimensional $k$-vector space $V$ and a linear operator $T : V \to V$ such that $\dim V = \operatorname{rank}(T - \lambda I)$ for some eigenvalue $\lambda$ of $T,$ we can immediately conclude that $T - \lambda I$ is injective by the Rank-Nullity Theorem; however, we also have that $$(T - \lambda I)(v) = T(v) - \lambda I(v) = \lambda v - \lambda v = 0$$ for every eigenvector $v$ of $T$ corresponding to $\lambda.$ Considering that $T - \lambda I$ is injective, we conclude that $v = 0$ -- a contradiction. Explicitly, an eigenvector must be nonzero.

Ultimately, we conclude that this condition is impossible to satisfy, so any statement you want to make would be vacuously true. For instance, if $\dim V = \operatorname{rank}(T - \lambda I),$ then you are immortal.