Diagonally dominant matrix with matrix similarity Applying similarity transform to a matrix $A$ gives:
$$M=P^{-1}AP$$
$M$ and $A$ have same eigenvalues. What is the way to to find $P$ such that $M$ is diagonally dominant case of $A$? $M$ is diagonally dominant if
$$|{m_{ii}}| \ge \sum\limits_{j \ne i} | {m_{ij}}|\quad {\rm{for \quad  all}}\quad i,{\mkern 1mu} $$
Note: I want $P$ to be something other than eigenvectors of $A$
EDIT:
Some eigenvalues of $A$ might be zero.
 A: If $A$ is diagonalisable, then you can set $P$ to be the matrix of eigenvectors. Then $M$ will be a diagonal matrix with entries equal to the eigenvalues of $A$.
A: Taking Jennifer Dylan's reasoning a tiny bit further:
Assume that $A$ is invertible, i.e. there are no zero eigenvalues. Then $A$ is similar to $$\operatorname{diag}(J_{\lambda_1}(k_1),\dots,J_{\lambda_r}(k_r))$$ where each $J_{\lambda_i}(k_i)$ is a Jordan block of size $k_i \times k_i$ and $\lambda_i \ne 0$. For any $\varepsilon>0$, the block
$$J_{\lambda_i}(k_i) = \begin{bmatrix}
\lambda_i \\
1 & \ddots \\
  & \ddots & \ddots \\
  & & 1 & \lambda_i
\end{bmatrix}$$
is similar to
$$\widetilde{J}_{\lambda_i}(k_i) = \begin{bmatrix}
\lambda_i \\
\varepsilon & \ddots \\
  & \ddots & \ddots \\
  & & \varepsilon & \lambda_i
\end{bmatrix},$$
which you can easily verify by computing $P J_{\lambda_i}(k_i) P^{-1}$ where $$P=\operatorname{diag}(1,\varepsilon,\dots,\varepsilon^{k_i-1}).$$
For each $i$ simply pick $\varepsilon$ to be less than $|\lambda_i|$, and you've got a (strictly) diagonally dominant matrix. Note that if $A$ is singular then it cannot be similar to a strictly diagonally dominant matrix.
Remaining question: what do we do with Jordan blocks for $\lambda=0$?
In the $2 \times 2$ case we have
$$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}^{-1} = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix}. $$
