Equation of matrices Let $V$, a 3d vector space above $F$. Let $T:V\rightarrow V$, linear transformation and $E$, an "ordered" basis such that:  
$$[ T ]_E = \left( \matrix{
   0 & 0 & a  \cr 
   1 & 0 & b  \cr 
   0 & 1 & c 
 }  \right)$$ 
where $a,b,c \in F$.

Show there's $v\in V$ such that:
  $$v \in \ker\left(T^3 + (-c)T^2 + (-b)T + (-a)\operatorname{Id}_v\right)$$
  Moreover, Show that $$T^3 + (-c)T^2 + (-b)T + (-a)\operatorname{Id}_v = 0$$

It seems to me a bit unreasonable to calculate $T^3, T^2$ (Which I did actually)
What's the catch of this exercise? What approach should I take?
Thanks.

This seems to be a case of Cayley's theorem ; the result that every matrix is a root of its characteristic polynomial . Compute the characteristic polynomial and then, by Cayley's theorem, if you evaluate $T$ in it (with $T^n=T \circ T\circ\cdots\circ T$
n times ) you will get $0$ (meaning the matrix with all-zero entries). See, e.g.: http://mathworld.wolfram.com/Cayley-HamiltonTheorem.html
