TITLE: $p(T) = 0$ with $\deg p \le n^2$
QUESTION [2 upvotes]: Let $V$ be a vector space with $\dim V = n$, and let $T \in \mathcal{L}(V)$. I need to prove that there is a non-zero polynomial $p(x)$ of degree no more than $n^2$ such that $p(T) = 0$. 
Any help or hints would be splendid.

REPLY [3 votes]: This result is more elementary than Cayley-Hamilton, and it is not required for the proof.   
Note that
$\dim \mathscr L(V) = n^2;  \tag 1$
note further that the set
$\mathscr S = \{T^k, \; 0 \le k \le n^2 \} \tag 2$
satisfies
$\vert \mathscr S \vert = n^2 + 1; \tag 2$
thus a linear dependence must hold amongst the elements of $\mathscr S$, that is, there exist $n^2 + 1$ scalars $p_0, p_1, p_2, \ldots, p_{n^2}$, not all zero, such that
$\displaystyle \sum_0^{n^2} p_iT^i = 0; \tag 3$
$T$ then satisfies the polynomial
$p(x) = \displaystyle \sum_0^{n^2} p_i x^i, \tag 4$
and
$\deg p(x) \le n^2. \tag 5$
$OE\Delta$.
{"set_name": "stack_exchange", "score": 2, "question_id": 3370320}