I am currently reviewing Jordan Normal Form.
Say we have $T,$ a linear operator, on a finite-dimensional vector space $V.$ So if we consider an eigenvector $v$ with eigenvalue $\lambda,$ then our eigenspace is given by the null space of $(T - \lambda).$ And the generalized eigenvector is a vector $x$ such that $(T - \lambda)^d \cdot x = 0.$
Although our vector space is really just a group in disguise, I couldn't help be note some similarity between this and the radical of an ideal $I$ of some ring $R.$ Where we define the radical to be: $$\sqrt{I} = \{a \in R \mid \exists n \in \mathbb{N}: a^n \in I\}.$$
And we usually talk about the radical of some ideal of polynomials. However, in the case of the eigenspace, this acts as the vanishing set for the operator $(T - \lambda).$ So I am not sure these are actually related, but if someone could give me more insight, I would be very grateful.
When you have an operator $T$ on your $k$-vector space $V$, it's the same thing as having a $k[X]$-module structure on $V$, where $X$ acts as $T$ (can you see why ?)
Then for given $x$, you have the ideal $Anh(x)=\{P\in k[X]\mid P\cdot x = 0\}$ (without the module notation, this is $\{P\in k[X]\mid P(T)(x) = 0\}$, but it's the same thing).
Now $x$ is an eigenvector of $T$ with eigenvalue $\lambda$ if and only if $X-\lambda \in Anh(x)$. Similarly, $x$ is a generalized eigenvector if and only if $(X-\lambda)^d\in Anh(x)$. Now the connection is that if you take $\sqrt{Anh(x)}$ then you get that $X-\lambda \in\sqrt{Anh(x)}$ if $x$ is a generalized-or-not eigenvector of $T$ with eigenvalue $\lambda$.
So there is indeed a connection between the two notions. More generally if you look at a commutative ring $R$ and an $R$-module $M$, and $x\in M$, then you can be interested in $Anh(x)$ and $\sqrt{Anh(x)}$.