For a linear operator,T, show the existence of a polynomial, p, such that p(T) is a nilpotent operator I am faced with the following (homework) problem.
Given an operator $T : C^{n} \rightarrow C^{n}$, show that there exists a polynomial $p(z) \in C$ of degree at most $n - 1$ such that $p(T)$ is a nilpotent operator.
Unfortunately, I am unsure about how to approach this problem. Can anyone give me some clues?
 A: Not to put too fine a point on it, but except for trivialities, this cannot be the case as stated.  Counterexample:   consider the linear map $T:C^2 \to C^2$ such that 
$T(z_1, z_2)^T = (\lambda_1 z_1, \lambda_2 z_2)^T, \tag{1}$
where $\lambda_1, \lambda_2 \in C$, $\lambda_1 \ne \lambda_2$, and $\lambda_1 \ne 0 \ne \lambda_2$.  In matrix form, we have
$T = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}.\tag{2}$
Any polynomial  $p(z) \in C[z]$, $\deg p(z)) \le 1$,  may be written in the form $p(z) = az + b$ for $a,b \in C$; thus $p(T) = aT + b$.  Then, from (2),
$aT + b = \begin{bmatrix} a \lambda_1 + b & 0 \\ 0 & a \lambda_2 + b \end{bmatrix}. \tag{3}$
If $aT + b$ were nilpotent, then we would have $(aT + b)^n = 0$ for some ingeter $n \ge 2$.  Then (3) yields
$0 = (aT + b)^n = \begin{bmatrix} (a \lambda_1 + b)^n & 0 \\ 0 & (a \lambda_2 + b)^n \end{bmatrix}, \tag{3}$
forcing 
$a \lambda_1 + b = a \lambda_2 + b = 0. \tag{4}$
Now if we allow $a = 0$, (4) becomes $b = 0$, so $aT + b = 0$, which is not nilpotent in any non-trivial sense.  If $a \ne 0$, then (4) forces $\lambda_1 = \lambda_2$, a contradiction.  
This simple counterexample shows that such polynomials with $p(T)$ nilpotent cannot exist in general without further conditions on $T$.
I hope this helps sort things out.  Cheers, 
and as always,
Fiat Lux!!!
