If $m$ is the smallest positive integer such that $T^m = 0_v$, then $m \leq \dim(V)$ Let $V$ be a vector space and $T \in L(v)$. Prove that
If $m$ is the smallest positive integer such that $T^m = 0_v$, then $m \leq \dim(V)$
I have no idea how to prove this.
 A: Suppose that the smallest such $m$ is greater than $k = \dim V$. Pick a vector $v$ such that $T^{m-1} v \ne 0$. Then $v, Tv, T^2v, \ldots, T^{m-1} v$ are all nonzero. (If any are zero, then all subsequent ones are!). 
I claim that these are  linearly independent. 
Suppose $\sum c_i T^i v = 0$. 
Apply $T^{m-1}$ to both sides to conclude that $c_0 = 0$.
Then apply $T^{m-2}$ to both sides to conclude that $c_1 = 0$.
Continue in this way. 
Conclusion: the $m$ vectors are independent, so $\dim V \ge m > k = \dim V$, a contradiction. 
Improved version, avoiding contradiction: 
Since $T^{m-1} \ne 0$, there's some vector $v$ with $T^{m-1} v \ne 0$. 
Then $v, Tv, T^2v, ..., T^{m-1} v$ are all nonzero. (If any are zero, then all subsequent ones are!). 
I claim that these are  linearly independent. 
Suppose 
$$
\sum_{i=0}^{m-1} c_i T^i v = 0.$$ 
Applying $T^{m-1}$ to both sides gives
$$
\sum_{i=0}^{m-1} c_i T^{i+m-1} v = 0 \text{, and since $T^r = 0$ for $r \ge m$,} \\
\sum_{i=0}^{0} c_i T^{i+m-1} v = 0 \text{, i.e}\\
c_0 T^{m-1} v = 0, 
$$ 
hence $c_0 = 0$. 
Applying smaller and smaller powers of $T$ to both sides, we can conclude that $c_1, c_2$, etc., are all zero. This lets us conclude that the $m$ vectors are all linearly independent. 
Since the number of elements of a linearly independent set is no greater than the dimension, we have that $m \le \dim V$. 
A: Most of the time there isn't any positive integer $m$ such that $T^m=0$, in which case talking about the smallest such $m$ makes no sense. I will assume however that one is supposed to be in the (exceptional) circumstance that such $m$ exists (such $T$ are called nilpotent), and $m$ is then chosen to be as small as possible.
This means there is a vector $v$ such that $T^m\cdot v=0$ (obviously) but $T^{m-1}\cdot v\neq0$ (since if there were no such $v$ then $m$ would not be minimal). This means that $v\notin\ker(T^{m-1})$ and so $v$ witnesses that $\ker(T^{m-1})$ is strictly contained in $\ker(T^m)=V$. The idea is to continue the sequence of subspaces $\ker(T^m),\ker(T^{m-1}),\ker(T^{m-2}),\ldots\ker(T),\ker(T^0)=\{0\}$ and to show that each one is strictly contained in its predecessor; this will give $m$ steps where the dimension drops, so that the initial dimension $\dim V$ must have been at least$~m$. That each space is contained in its predecessor is obvious from the definition of kernels; as witnesses for the fact that the inclusion is strict, one can use images of the vector$~v$.
A: Let $\;v\in V\;$ be such that $\;T^{m-1}\neq 0\;$ (why such $\;v\;$ exists?), then
$$\{v\,,\,Tv\,,\,T^2v\,,\ldots\,,T^{m-1}v\}\;\;\text{is linearly independent, since:}$$
$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0\implies 0=T(0)=a_0Tv+a_1T^2v+\ldots+a_{m-1}T^mv$$
But $\;T^mv=0\;$  , so in fact we have
$$a_0Tv+\ldots+a_{m-2}T^{m-1}v=0$$
Apply $\;T\;$ again and get a shorter trivial combination. After some times (how many?), you get that $\;a_0T^{m-1}v=0\implies a_0=0\;$ . 
Inductively, $\;a_0=a_1=\ldots a_{m-1}=0\;$ and we're done
Finally, remember that in an $\;n$-dimensional space there can't be more than $\;n\;$ vectors linearly independent.
A: Since $T^{m-1}v\ne0$, there must be some vector $v$ for which $T^{m-1}v\ne0$, and that entails that $v\ne0,\ Tv\ne0,\ T^2 v\ne0,\ldots,T^{m-1} v\ne0$.
If you can show that $v,\ Tv,\ T^2v,\ \ldots,T^{m-1}v$ are linearly independent, then you can conclude $\dim V\ge m$ since there are $m$ vectors in that set.  If those vectors are linearly dependent, then you have
$$
c_0 v + c_1 Tv + c_2 T^2 v + \cdots + c_{m-1} T^{m-1} v=0
$$
for some scalars $c_0,\ldots,c_m$, not all $0$.  Consequently
$$
T(c_0 v + c_1 Tv + c_2 T^2 v + \cdots + c_{m-1} T^{m-1} v) = T0= 0,
$$
so
$$
c_0 Tv + c_1 T^2 v + \cdots + c_{m-2} T^{m-1}v + c_{m-1}T^m v=0, 
$$
but the last term on the left above is $0$, so
$$
c_0 Tv + c_1 T^2 v + \cdots + c_{m-2} T^{m-1}v =0.
$$
Then in the same way you can conclude that
$$
c_0 T^2 v + c_2 T^3 v + \cdots + c_{m-3}T^{m-1}v=0,
$$
and then that
$$
c_0 T^3 v+ c_1 T^4 v + \cdots + c_{m-4} T^{m-1}v=0,
$$
and so on, until you have $c_0 T^{m-1}v=0$.  Since $T^{m-1}v\ne 0$, this implies $c_0=0$.  Then you have
$$
c_1Tv+c_2 T^2v+\cdots+c_{m-1}T^{m-1}v=0,
$$
and then do the same thing with $Tv$ that we did with $v$ above, and conclude that $c_1=0$.  Then keep going in the same way to show that $c_0=c_1=\cdots=c_{m-1}=0$.
Finally, conclude that $v,\ Tv,\ T^2v,\ldots,\ T^{m-1}v$ are linearly independent.
A: I am not an expert but I'll try to sketch an idea. Let $\dim(V)=n$, and let $\mathbb{O}=0_{L(V)}$.
Since $T$ is an endomorphism and $T^m=\mathbb{O}$, we must have $\ker(T)\neq\{0_V\}$, because otherwise you would have that $T$ is invertible and so $\mathbb{I}=T^{-m}\circ T^m=T^{-m}\circ \mathbb{O}=\mathbb{O}$ which is absurd. So, take $v\in\ker(T)$. $T^k(v)=(T^{k-1}\circ T)(v)=T^{k-1}(T(v))=T^{k-1}(0_V)=0_V$ so $v$ is also in $\ker(T^k)$ for any $k$. This means that $\ker(T)\subseteq\ker(T^2)\subseteq\ker(T^3)\subseteq\dots \subseteq\ker(T^m)=V$. Moreover, if $\ker(T)\subseteq im(T)$, that means that for $v\neq 0_V,v\in\ker(T)$ there exists $w$ such that $v=T(w)$ so that $T(v)=T^2(w)=0_V$: this implies that $w\in\ker(T^2)$ but $w\notin\ker(T)$, so the above chain is really
\begin{equation}
\ker(T)\subset\ker(T^2)\subset\ker(T^3)\subset\dots \subset\ker(T^m)=V
\end{equation}
(If otherwise $im(T)\subseteq \ker(T)$, what happens?)
Passing to the dimension of these subspaces,
\begin{equation}
\dim(\ker(T))<\dim(\ker(T^2))<\dim(\ker(T^3))<\dots<\dim(\ker(T^m))=n
\end{equation}
That means that there are $m$ positive integers $d_1,\dots,d_m$ such that
\begin{equation}
0<d_1<d_2<d_3<\dots<d_m=n
\end{equation} 
That shows that $m\leq n$.
In fact, the best possible situation is when you have $d_k=k$; in that case $m=n$.
A: Although I wouldn't personally consider this a very good answer (using too heavy artillery for the proof of a relatively simple fact), it may be noted that a proof using the Cayley-Hamilton theorem is also possible.
Since it is (apparently) given that $T$ is nilpotent (there exists $k$ with $T^k=0$), all eigenvalues$~\lambda$ (in an algebraically closed field) of$~T$ satisfy $\lambda^k=0$ and hence $\lambda=0$. The characteristic polynomial then can only be $X^n$ where $n=\dim(V)$, so $T^n=0$ by the Cayley-Hamilton theorem, and $m\leq n=\dim(V)$.
