In terms of matrices: $\forall v\in V,\phi(\phi(v))=0$ $\phi: V\to V$( a linear operator here)
How to interpret $\forall v\in V,\phi(\phi(v))=0$ in terms of matrices? Can I have some hint?

I suppose $\phi(V)= \begin{bmatrix} \phi(v_1)\\\phi(v_2)\\\dots\\\phi(v_n)\end{bmatrix}$ for $V=\dim n$
 A: $\phi(\phi(v))=0$ is equivalent to saying that this $A^2=0$. What does this mean? This means that $A$ has a nilpotency index of $2$. So what we want is to find all nilpotent index $2$ matrices of size $n\times n$. What matrices satisfy this?
All matrices are similar to a Jordan form and hence there is some $J$ Jordan form satisfying  $A=P^{-1} J A$. This $J^2=0$ also, and since all nilpotent matrices have eigenvalues of $0$, we form some combination of Jordan blocks of the form:
$$\begin{bmatrix}0&1&\\0&0&1\\&&\ddots&\ddots\\0&\dots&&0&1\\0&\dots&&&0\end{bmatrix}_{n\times n}$$
Now another fact we know is the following:
$$J^2=\begin{bmatrix}{J_1}^2\\&{J_2}^2\\&&\ddots\\&&&&{J_n}^2\end{bmatrix}$$
Where $J_i$ are the jordan blocks making up this Jordan form. What this means is, each of $J_i$ must have a nilpotency index of $2$.
What size Jordan blocks have a nilpotency index of $2$?
$\begin{bmatrix}0\end{bmatrix}$ and $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ are the only two such sizes. Why? The three by three case and above will always have a nilpotency index greater than $2$.
This means that all such matrices are equivalent to some permutation of Jordan blocks of size $1$ and $2$ across the diagonal.
