Relationship between Nilpotent Matrix and Matrix with all zero diagonal factors. solving Linear Algebra HW, I suddenly became curious about the relationship between Nilpotent Matrix and matrix with all zero diagonal factors such that $A_{11} = A_{22} = \cdots = A_{nn} = 0$
Does Nilpotent Matrix implies the matrix with all zero diagonal factors? What about converse? I guess it might be iff relationship. Could you develop more?
Thank you very much. 
 A: We have this general result:

For a linear transformation $f$ such that  $\mathrm{tr}(f)=0$  there's a basis $\mathcal B$ in which the matrix of $f$ is  with all zero diagonal factors.

The proof of this result is in these points:


*

*$f$ isn't an homothetie so there's $x$ such that $x$ and $f(x)$ are linearly independant

*We consider $F$ a complement space of $\mathrm{span}(x)$ containing $f(x)$ and we define the projection $p$ on $F$ with kernel $\mathrm{span}(x)$ so the linear transformation $v:=p\circ f_{|F}$ of $F$ verify $\mathrm{tr}(v)=0$

*We prove the result by induction using the last point.

A: Let $A$ be a $n\times n$ matrix for some $n\times n$, over some algebraically closed field. The following holds:
$$A \text{ is nilpotent }\iff A\text {'s only eigenvalue is }0.$$

Question 1: Does $A$ being nilpotent imply its diagonal entries are all $0$  ?

According to the above characterization of nilpotency, absolutely not. 
Take for instance the matrix $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}$. It's easy to check that $\begin{bmatrix} -3 & -1\\ 9 & 3\end{bmatrix}^2=0_{2\times 2}$.

Question 2: Is any matrix with only $0$'s on the diagonal entries necessarily nilpotent?

Again, no. There are matrices in these conditions which don't even have $0$ has an eigenvalue. For instance $\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$.

More can be said, but it depends on your knowledge whether it is worth saying or not. As far as I know it all comes down to a matrix's Jordan Normal Form:
Any nilpotent matrix $n\times n$ is similar to some block diagonal matrix $$ {\begin{bmatrix}
\color{blue}{J_1} &    0   &\dots    &\dots   & 0\\
  0  & \color{blue}{J_2} &  0  & \dots  &0\\
  \vdots  &    \ddots   & \ddots & \ddots  &\vdots\\
   \vdots &  \ddots      &  \ddots   &\ddots   & 0\\
   0 &   \dots    &  \dots   &  0 &  \color{blue}{ J_k}\\
 \end{bmatrix}}_{n\times n},$$ for some $k\in \Bbb N$. Where, for each $i\in \{1\ldots ,k\},\,J_i$ is a $m_i\times m_i$ matrix, for some $m_i\in \Bbb N$, that looks like 
$$\begin{bmatrix}0& 1 &&& \\
& 0 & 1 &\huge 0& \\
& & \ddots & \ddots &\\
&\huge 0 &&  0 &1 \\
&&&  & 0 \\ 
\end{bmatrix}_{m_i\times m_i}.$$
A: By diagonal factors, do you mean diagonal elements of the matrix? If so, it's wrong.
See Wikipedia for more examples, but here is one:
$$A = \left( \begin{array}{cc}
0 & 1 \\
0 & 0 \end{array} \right)$$
$$P = \left( \begin{array}{cc}
3 & 1 \\
5 & 2 \end{array} \right)$$
Then let $B=P^{-1} A P$,
$$B = \left( \begin{array}{cc}
10 & 4 \\
-25 & -10 \end{array} \right)$$
$A$ and $B$ are nilpotent, and $A^2=B^2=0$.
The converse is also wrong, the following matrix $C$ has zero diagonal entries, but it's not nilpotent, and actually $C^2=I$.
$$C = \left( \begin{array}{cc}
0 & 1 \\
1 & 0 \end{array} \right)$$
