I am stuck on the following problem:
Suppose $W$ be a vector space of dimension $m \geq 2.$ Let $F \colon W \to W$ be a linear transformation such that $F^{n+1}=0$ and $F^n \ne 0$ for some $n \ge 1.$ Then I have to check which of the following statement(s) is/are true?
1.$\text{trace}(F) \ne 0 $
2.$\text{Rank}(F^n) \le \text{Nullity}(F^n)$
3.$n=m$
4.$F$ is diagonalizable.
I do not know how to progress with the problem. Any kind of hints with some explanation will be helpful.
Everything here can be solved with the use of Jordan normal form of a matrix.
So, if we want a non-zero nilpotent matrix $F$, we must write it as $$F=T^{-1}JT, $$ $T$ - some matrix with $det T\ne 0$. $J$ consists of Jordan cells, each of them has the view $$J_l=\begin{pmatrix}\lambda&1&0&\dots\\0&\lambda&1&\dots\\\vdots&\ddots&\ddots&\ddots\\0&\dots&0&\lambda \end{pmatrix},$$ $\lambda$ is an eigenvalue.
It's easy to compute the powers of Jordan blocks and hence the powers of matrices. In particular, $(J_l)^k$ has $\lambda^k$ on its diagonal. Thus, in order to this block to be nilpotent, one must take $\lambda=0$. If we now consider the whole matrix $J$, it's easy to see that $tr(F)=0$, since it's invariant under changes of basis.
As Jordan normal form is unique (up to some changes, cf. wiki), the diagonal form of the matrix is its Jordan form. Since all eigenvalues are zero, diagonalisability would mean that the whole matrix is zero. Contradiction.
Next, it's quite easy to see that if the order of a Jordan block $J_l$ with zero eigenvalue is equal to $r$, then $(J_l)^r=0$; it's easy to conclude on (3).
Finally, $F^{n+1}=0$ and $F^{n}\ne 0$ mean that the maxim order of Jordan cells in Jordan decomposition of $F$ is equal to $n+1$. Let's say that we have $s$ cells of the order $n+1$ and the sum of dimensions of all other cells is $t$. Apparently, $$m=s(n+1)+t.$$ It's easy to see that $\text{rank}(F^n)=s$, $\text{nullity}(F^n) = ns+t$, which allows to conclude on (2).
$2:True:$ $FF^n=0\\\implies\text{Im}(F^n)\subset\text{Kernel}(F)\subset\text{Kernel}(F^n)\\\implies\text{Rank}(F^n)\le\text{Nullity}(F^n)$
$1,3,4: False:$ $F:\mathbb R^3\to\mathbb R^3$ given by $F=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$
To start with (and indeed to answer some of the points) try writing down the simplest (who knows, maybe a $2\times 2$) nonzero matrix you can think of which squares to 0.
For the second point, think about what happens to a vector $v$ as we map $v\mapsto Fv \mapsto F^n F v$.
