Is it guaranteed that the inverse of a diagonal dominant matrix, whose elements are all nonnegative, is also diagonal dominant?
In my specific problem, I have a diagonal dominant complex matrix $\mathbf{A}$. Then, I calculate the following:
$$ \left(\left| \mathbf{A} \right|^2\right)^{-1}$$
where the operator $\left| \cdot \right|^2$ is the element-wise absolute value squared, which guarantees the elements of the matrix to be inverted ($\left| \mathbf{A} \right|^2$) to be nonnegative.
My objective is to state confidently that $\left(\left| \mathbf{A} \right|^2\right)^{-1}$ is also diagonal dominant, with the assumption that the values in the main diagonal of
$\left| \mathbf{A} \right|^2$ are at least 10 to 100 times bigger than the off diagonal values. This assumption holds for my specific application, which is irrelevant for the discussion.
On MATLAB, it seems reasonable. Would someone present a counter example?
Best Answer
$$A=\begin{pmatrix}2&1&1\\0&1&0\\0&0&1\end{pmatrix}\;,\;\;A^{-1}=\begin{pmatrix}\frac12&\!\!-\frac12&\!\!-\frac12\\0&\;\;1&\;\;0\\0&\;\;0&\;\;1\end{pmatrix}$$
So $\,A\;$ is diag. dominant with all its entries non-negative, but its inverse is not even diag. dominant since in the first row
$$\left|\frac12\right|=\frac12\rlap{\,\;/}\ge \left|-\frac12\right|+\left|-\frac12\right|=1$$