TITLE: Characteristic polynomial with coefficients c0=, c1=cn=1. Prove: $V = Ker(T) \oplus T(V) $
QUESTION [5 upvotes]: Question from final exam: 
$V$ is a vector space , $\dim V = n$, and $T:V\rightarrow V$ is a linear transformation. 
We assume that the characteristic polynomial of the linear transformation
$$p_T(x) = \sum_{i=0}^nc_ix^i $$ 
has coefficients of $c_0 = 0$, $c_1=c_n= 1$.
I need to prove that: 
$$V = Ker(T) \oplus T(V)$$
Please any hints, I don't have a clue how to approach this question.

REPLY [7 votes]: You know that $\chi_T=X+c_2X^2+\cdots+c_{n-1}X^{n-1}+X^n$.  
We show that $\ker T\cap \operatorname{im} T=0$. Indeed, suppose that $Tv=0$, and $v=Tw$ for some $w$. You want to show $v=0$. Alternatively, you want to show that if $T^2w=0$; then $w\in\ker T$. Using Hamilton Cayley, you know $$T+c_2T^2+\cdots+c_{n-1}T^{n-1}+T^n=0$$
Plugging in $w$, can you see what happens?

SPOILERS

 Plugging in $w$, you get $$Tw+c_2T^2w+\dots c_{n-1}T^{n-1}w+T^nw=0$$
 But you know $Tv=TTw=0$, so every term after $Tw$ vanishes, and you get $Tw=0$ as desired. Thus, $\ker T\cap \operatorname{im}T=0$. Since we always have $\dim \ker T+\dim{\rm im}\; T=\dim V$; we're done.