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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.

2 Answers 2

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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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    @Daryl : You are right, i should have mentioned at the beginning.2012-10-20
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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.2012-10-20