If $\mathrm{Tr}(A)=0$ then $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$ 
Prove that if $A$ is a square matrix and $\mathrm{Tr}(A)=0$, then there exists an invertible matrix $R$ such that the matrix $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$.

It seems like the formula $A=S\Lambda S^{-1}$, but maybe it does not help.
Thanks so much.
 A: $\newcommand{\trace}[0]{\mathrm{trace}}$A proof by induction on the size $n$ of $A$ appears to be possible, the case $n = 1$ being obvious.
We have an implicit assumption that the characteristic of the underlying field is zero here. (Otherwise $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ is a counterexample in characteristic $2$, which can be extended to all positive characteristics.)
So $A$ is not scalar (unless $A = 0$ and we're done), so there is a vector $v \ne 0$ which is not an eigenvector. In particular $v$ and $A v$ are independent. Consider a basis $\mathcal{B}$ which starts with $v, A v$. The matrix of $A$ with respect to this basis will be of the form
$$
\begin{bmatrix}
0 & a\\
b & C\\
\end{bmatrix},
$$
where $C$ is a square $(n-1) \times (n-1)$ matrix, and $b, c$ are vectors of the appropriate size.
(This seems to me like a quicker proof of Lemma 2 of the paper quoted in a comment.)
(Actually
$$
b =
\begin{bmatrix}
1\\0\\\vdots\\0
\end{bmatrix},
$$
as $A$ maps the first element of the basis $\mathcal{B}$ onto the second element.)
Now clearly $\trace(C) = 0$. By the induction hypothesis, there is an invertible $(n-1) \times (n-1)$ matrix $Z$ such that all elements on the diagonal of $Z^{-1} C Z$ are zero.
Now
$$
\begin{bmatrix}
1 & 0\\
0 & Z\\
\end{bmatrix}^{-1}
\begin{bmatrix}
0 & a\\
b & C\\
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & Z\\
\end{bmatrix}
=
\begin{bmatrix}
0 & a Z\\
Z^{-1} b & Z^{-1} C Z\\
\end{bmatrix},
$$
and we're done.

The proof can be quickly adapted to yield Corollary 4 of the paper mentioned above, that is, you can conjugate any matrix $A$ to a matrix where all diagonal elements are the same.  So this diagonal element has to be $\lambda = \trace(A)/n$. (But please do check Marc van Leeuwen's comment below, to the effect that you can reduce the general case to the case of $\trace(A) = 0$ already treated above.)
In fact, if $A$ is not already scalar, there is a nonzero vector $v$ which is not an eigenvector. So $v, A v$ are linearly independent. Consider a basis $\mathcal{B}$ which starts like $v, A v - \lambda v$. With respect to this basis, $A$ has the form
$$
\begin{bmatrix}
\lambda & a\\
b & C\\
\end{bmatrix},
$$
as $A v = \lambda v + (A v - \lambda v)$.
Now $\trace(C) = (n-1) \lambda$, so induction applies.
