If $f^{2}-3f+2id=0$, how to show that $E=\ker(f-id)\bigoplus \ker(f-2id)$? Suppose $E$ a vector space and f an endomorphism.  $f^{2}-3f+2id=0$, show that $E=\ker(f-id)\bigoplus \ker(f-2id)$.
My first approach was to take $y \in E$ and show that it is decomposed in such way:
$$y=x_{1}+x_{2}\:with\:x_{1}\in\:\ker(f-id)\:and\:x_{2}\in\:\ker(f-2id)$$
But didn't get far with this.
I attempted another approach that consisted in showing first that $\ker(f-id)\cap \ker(f-2id)= \left \{ 0 \right \}$, and then showing that the sum of dimensions of $\ker(f-2id)$ and $\ker(f-id)$ equals to the dimension of E. But I don't understand how I can find the dimension of $\ker(f-id)$, $\ker(f-2id)$ and of $E$ itself.
Any help will be appreciated.
 A: Suppose $v\in E$ can be written as $v=x+y$, with $x\in\ker(f-\mathit{id})$ and $y\in\ker(f-2\mathit{id})$, so $f(x)=x$ and $f(y)=2y$.
Then $f(v)=f(x)+f(y)=x+2y$. Hence
$$
y=(x+2y)-(x+y)=f(v)-v
$$
so $x=v-y=2v-f(v)$. This tells us what $x$ and $y$ should be.
Now prove that, for all $v\in V$,
$$
2v-f(v)\in\ker(f-\mathit{id}),
\qquad
f(v)-v\in\ker(f-2\mathit{id})
$$
so
$$
v=(2v-f(v))+(f(v)-v)\in\ker(f-\mathit{id})+\ker(f-2\mathit{id})
$$
(Hint: $(f-\mathit{id})((f-2\mathit{id})=0=(f-2\mathit{id})(f-\mathit{id})$.)
Finally, try proving
$$
\ker(f-\mathit{id})\cap\ker(f-2\mathit{id})=\{0\}
$$

You cannot determine the dimensions of the kernels: the only conclusion you can draw is that their sum equals the dimension of $E$. Indeed, let $\{e_1,\dots,e_n\}$ be a basis for $E$ and, for $1\le k\le n$, define a linear map $f_k\colon E\to E$ by
$$
f_k(e_i)=\begin{cases}
e_i & \text{if $1\le i\le k$} \\[4px]
2e_i & \text{if $k<i\le n$}
\end{cases}
\quad(i=1,2,\dots,n)
$$
Then $f_k$ is a linear map satisfying $f_k^2-3f_k+2\mathit{id}=0$ and $\dim\ker(f-\mathit{id})=k$.
A: Hint: Does this remind you of the primary decomposition theorem? The proof of it's first part ($V = ker(f_1^{n_1})\bigoplus$$ker(f_2^{n_2})\bigoplus...\bigoplus$$ker(f_k^{n_k})$) doesn't use the fact that the polynomial under consideration is minimal. It only assumes that $f_1$,$f_2$,...,$f_i$ are co-prime and that f=$f_1^{n_1}*f_2^{n_2}*...*f_k^{n_k}$ annihilates T. So, you can use it's proof.
