Let $T:V\to V$ be a linear transformation. Suppose that $T^m = 0$ for some positive integer $m$. Show that $T^n = 0$, where $n = \dim_F V$. 
Let $T:V\to V$ be a linear transformation. Suppose that $T^m = 0$ for some positive integer $m$. Show that $T^n = 0$, where $n = \dim_F V$. [Taken from here]

I don't know how to prove it, and I don't understand what is the meaning of $T^n = 0$. It means $T^n(v) = 0$?
Can anyone find the solutions of the exercise?
 A: The minimal polynomial divides $t^m$, hence has the form $t^k$ for some $k$. But the characteristic polynomial has degree $n$ and divides a power of the minimal polynomial, so it has to be $t^n$. By Cayley-Hamilton, $T^n =0$.
A: HINT:
The decreasing sequence of subspaces
$$V\supset T(V) \supset T^2(V)\supset \ldots T^n(V)$$
stabilizes at some step $k\le n$ since they are all subspaces of $V$ of dimension $n$,
that is
$$T^k(V) = T^{k+1}(V) = \ldots $$
If moreover $T^m(V)=(0)$ for some $m$ then we have
$$T^k(V) = \ldots = T^n(V)=0$$
Note that is works also over a non-commutative field.
A: Many people mentioned the matrix representation to handle this question .But the lesson what I take didn't reach the chapter of matrix representation ,so I choose a method without matrix .
$ImT$ is the image of T, and $Nullity(T) <= Nullify(T^2) <= ... <= Nullify(T^N)$. According to the rank-Nullify  theorem ,$rank(T) >= rank(T^2) >= ... >= rank(T^n)$.
Therefore ,$T^n(V) \subseteq ... T^2(V) \subseteq T(V) \subseteq V$. When It is not equal , the dimension will decrease at least one from $T^i(V)$ to $T^{i+1}(V)$.When it hit an equality ,it will reach a stabilization and $T^{n+1}(V) = T^{n+2}(V) = ... = T^{n+ m}(V)$.
Hence When $dim V = n$, $T^n(V) = 0$.
Derivation:
$$n = (rank(T) - rank(T^2)) + (rank(T^2) - rank(T^3)) + \\
... + (rank(T^m) - rank(T^m+1)) +\\
(rank(T^n) - rank(T^n+1)) + (rank(T^n+1) - rank(T^n+2)) >= m$$
$T^n(V) = 0$ when $dim(V) = n$.
A: Hint:. $T^m=0\iff $ the matrix for $T$ rel the standard basis is similar to a matrix which is upper triangular with zeros on the diagonal (strictly upper triangular).
To prove that, you could use the Jordan normal form, and the fact, relatively easy to prove, that all the eigenvalues are zero.
By noting that the characteristic polynomial is $x^n$, and applying Cayley-Hamilton, one can get that a strictly upper triangular $n\times n$ matrix satisfies $T^n=0$.
