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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.

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2 Answers 2

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 $A$.

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1  
I already knew that, but that does not work for me. And M does not need to be diagonal. The requirement is diagonally dominant. –  kotoll Oct 20 '12 at 3:13
2  
On the same page... if for all $\lambda \in $ spectrum$(A)$ we have $|\lambda| > 1$ then the Jordan form of $A$ is diagonally dominant. –  user2468 Oct 20 '12 at 4:50
    
@Kotoll A diagonal matrix is diagonally dominant. You didn't specify that you weren't interested in this approach until an edit after my post. –  Daryl Oct 20 '12 at 11:41
    
@Daryl : You are right, i should have mentioned at the beginning. –  kotoll Oct 20 '12 at 17:05

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 $$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}^{-1} = \begin{bmatrix}1 & -1 \\ 1 & -1\end{bmatrix}. $$

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Your answer seems logical, but the assumption that you did at the beginning may not be true for my question. So some eigenvalues of $A$ might me zero. –  kotoll Oct 20 '12 at 17:12

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