Decomposition of Complex Operator A complex operator $T \in L(V)$ on a finite dimensional space decomposes the space into a sum of invariant, generalized eigenspaces.  I'm having trouble intuitively understanding why these generalized eigenspaces can't overlap.
The book I am reading from (Axler) proves this with an argument based on the sum of the dimensions of the generalized eigenspaces being equal to the dimension of the space.  You know that is true because of the following theorem
Theorem: each eigenvalue appears on the diagonal of an upper triangular matrix for $T$ exactly dim null $(T-\lambda I)^n$ times.
In Axler, this is proven by induction and by recovering the general case from the case where $\lambda = 0$, and for that reason, I don't feel I really understand why it's true.  
This may be slightly vague, but are there any proofs of the theorem above that are constructive or that shed additional light on why the generalized eigenspaces sum to the space?  Decomposing a complex operator seems fairly important/deep, and I am trying to get as much intuition about the generalized eigenspaces as I can.
 A: You can find the following proof in Hoffman's book.
Let $p(x)=\prod_{i=1}^n(x-\lambda_i)^{n_i}$ be the characteristic polynomial of $T$, where $\lambda_i\neq\lambda_j$. Define the polynomials $p_i(x)=\dfrac{p(x)}{(x-\lambda_i)^{n_i}}$. 
Notice that $\operatorname{mdc}(p_1(x),\ldots,p_n(x))=1$. Thus, exist polynomials $q_i(x)\in\mathbb{C}[x]$ such that $\sum_{i=1}^nq_i(x)p_i(x)=1$. Replacing $x$ by $T$ we obtain: $\sum_{i=1}^nq_i(T)p_i(T)=Id$
Thus, every $x=\sum_{i=1}^nq_i(T)p_i(T)x$. 
Now, $(T-\lambda_iId)^{n_i}q_i(T)p_i(T)x=q_i(T)(T-\lambda_iId)^{n_i}p_i(T)=q_i(T)p(T)x$. Remind by  Cayley-Hamilton thm that $p(T)\equiv 0$. Thus, $(T-\lambda_iId)^{n_i}q_i(T)p_i(T)x=0$. 
So every $x\in V$ is a sum of vectors $q_i(T)p_i(T)x\in\ker(T-\lambda_iId)^{n_i}$.
Therefore, $V=\sum_{i=1}^n\ker(T-\lambda_iId)^{n_i}$. We must prove that the overlap of these kernels is $0$, i.e, $\ker(T-\lambda_iId)^{n_i}\cap \ker(T-\lambda_jId)^{n_j}=0$.
Since $\operatorname{mdc}((x-\lambda_i)^{n_i},(x-\lambda_j)^{n_j})=1$ then exist polynomials $r_i(x),r_j(x)\in\mathbb{C}[x]$ such that $r_i(x)(x-\lambda_i)^{n_i}+r_j(x)(x-\lambda_j)^{n_j}=1$. 
Replacing $x$ by $T$, we obtain $r_i(T)(T-\lambda_iId)^{n_i}+r_j(T)(T-\lambda_jId)^{n_j}=Id$. 
So if $x\in \ker(T-\lambda_i)^{n_i}\cap \ker(T-\lambda_j)^{n_j}$ then $0=r_i(T)(T-\lambda_iId)^{n_i}x+r_j(T)(T-\lambda_jId)^{n_j}x=x$. 
So $V=\bigoplus_{i=1}^n\ker(T-\lambda_iId)^{n_i}$.
It is not difficult to show that $\ker(T-\lambda_iId)^{n_i}$ is invariant by $T$. Try to prove it.
A: I would do this in the following steps.


*

*If $f_1$ and $f_2$ are relatively prime polynomials, and $f = f_1 f_2$, prove that $\ker f(T) =  \ker f_1(T) \oplus \ker f_2(T)$.

*Extend the first point by induction to a product $f = f_1 f_2 \dots f_r$.

*Apply this result to the decomposition of the minimal polynomial $f$ of $T$ as a product of powers of linear factors.
Of these points, only the first is at all difficult. It is clear that $\ker f_1(T)$ and $\ker f_2(T)$ are subspaces of $\ker f(T)$. By Bezout's lemma, there are polynomials $\alpha_1$ and $\alpha_2$ satisfying $\alpha_1 f_1 + \alpha_2 f_2 = 1$. Thus for any $v \in V$, we have 
$$v = \alpha_1(T)f_1(T)\cdot v + \alpha_2(T)f_2(T) \cdot v.$$
On the one hand, this equality shows that if $v \in \ker f_1(T) \cap \ker f_2(T)$, then $v = 0$.
On the other, if $v \in \ker f(T)$, then it provides a decomposition of $v$ into the sum of an element of $\ker f_2(T)$ and one of $\ker f_1(T)$, proving what we need.
Also, you should return to this question after you have studied the structure theorem for finitely-generated modules over a principal ideal domain, which is a topic in abstract algebra. (I presume, since you're reading Axler's book, that you haven't studied this yet.) The idea is to consider $V$ as a $\mathbf{C}[X]$-module, where $X \cdot v$ is defined to be $T(v)$. This idea greatly clarifies the structure theory of endomorphisms of $V$.
