Generalized Eigenspaces Associated to Different Values 
Let $V$ be some vector space, and $T \in \mathcal{L}(V)$. If $a \neq b$, then $G(a,T) \cap G(b,T) = \{0\}$, where $G(b,T)$ denotes the generalized eigenspace. 

I am having a lot of trouble with this problem; I've been thinking about it for the past few days with no progress. Here is what I've come up with:
If $v \in G(\alpha,T) \cap G(\beta,T)$, then there $i,j \in \Bbb{N}$ such that $(T-aI)^i v =0$ and $(T-bI)^j v = 0$. WLOG, let $i$ and $j$ be the smallest such integers and suppose $i < j$. Then 
$$w_1 := (T-aI)^{i-1}v \neq 0$$
and
$$w_2 := (T-aI)^{j-1} \neq 0$$
Note that $(T-aI)w_1 =0$ and therefore $Tw_1 = aw_1$; similarly, $Tw_2 = bw_2$...
I've played with this equations in  various ways. My strategy = is to get $(\mbox{something non-zero}) \cdot v = 0$, which will obviously force $v=0$; but this is proving extremely difficult. My hope is that I can get $(a-b)v=0$, but I don't see how to do it. 
I could use some help...
 A: Following your set up, let $v\in G(a,T)\cap G(b,T)$ and let $i$ be the smallest number such that $(T-aI)^iv=0$. Let $w=(T-aI)^{i-1}v\neq 0$. Then $(T-aI)w=0$ and so $Tw=av$. As such for $\lambda \in \mathbb{F}$ and $n\in \mathbb{N}$, $(T-\lambda I)^nw=(a-\lambda)^n w$. Now let $j\in\mathbb{N}$ such that $(T-bI)^jv=0$. Now, $(T-aI)^{i-1}(T-bI)^jv=0$. However, both of these are polynomials of $T$ and therefore commute (if you are using Linear Algebra Done Right by Axler as I suspect you are this is 5.20 on page 144). As such, $0=(T-bI)^j(T-aI)^{i-1}v=(T-bI)^jw=(a-b)^jw$. However, that implies that $a=b$ which is a contradiction. As such, no such $v$ exists. (If you are using Axler this proof is modelled off of Axler's proof of 8.13).
A: The proof I know of this follows from a bunch of general theory about finitely-generated modules over principal ideal domains. If that means something to you, definitely check it out. Otherwise, I will try to boil out the important points. You do need to know something about polynomial divisibility though, there is no getting around that.
Let $T: V \to V$ be a linear transformation of the finite-dimensional vector space $V$, and $k[x]$ the set of polynomials over the same field $k$. Given a polynomial $p \in k[x]$, we can create the linear map $p(T)$ in the usual way.
Given a polynomial $p \in k[x]$ we define $V_p := \{v \in V \mid p(T)v = 0\}$, the subspace of $V$ of vectors killed by $p$. For example, if $p(x) = (x - \lambda)$, then $V_p$ is the $\lambda$-eigenspace of $T$. If $p(x) = (x - \lambda)^n$ for $n$ large enough, then $V_p$ will be the generalised eigenspace for $\lambda$.
So we can re-state your problem: if $p(x) = (x - \lambda)^n$ and $q(x) = (x - \mu)^m$ for some $n, m \geq 1$ and $\lambda \neq \mu$, then why is $V_p \cap V_q = 0$? The polynomials $p, q$ are coprime, i.e. they share no factors. For any two coprime polynomials $p, q$, there exist two other polynomials $r, s$ such that $rp + sq = 1$. Plugging $T$ into this identity, we get $r(T)p(T) + s(T)q(T) = \mathrm{id}_V$. Now suppose $v \in V_p \cap V_q$, then by the above identity we have
$$  v = r(T)p(T)v + s(T)q(T)v = 0$$
The only non-obvious part in the above is the assertion that the polynomials $r, s$ exist. You might have seen a proof of this before in the case of integers, where the extended Euclidean algorithm is used to create the integers $r, s$. The proof for polynomials is exactly the same, and the algorithm still works. But perhaps you can learn about that from somewhere else.
