Are $-A, \ A^T, \ A+B$ are strictly diagonally dominant as per rows?
If $A,B\in \mathbb{R}^{n\times n}$ are strictly diagonally dominant as per rows, then check if $-A, \ A^T, \ A+B$ are strictly diagonally dominant as per rows.
We have that:
A square matrix $A$ is strictly diagonally dominant if for all rows the absolute value of the diagonal element in a row is strictly greater than the sum of absolute value of the rest of the elements in that row.
$-A$ is also strictly diagonally dominant because we check the elements with their absolute values,which are equal to these of $A$.
$A^T$ is not necesarily strictly diagonally dominant, because in this case it would mean that for all columns in $A$ the absolute value of the diagonal element in a column is strictly greater than the sum of absolute value of the rest of the elements in that column in $A$, which we don't know.
We have that $\displaystyle{|a_{ii}|>\sum_{i\neq j}|a_{ij}|}$ and $\displaystyle{|b_{ii}|>\sum_{i\neq j}|b_{ij}|}$, but we have that $|a_{ii}+b_{ii}|<|a_{ii}|+|b_{ii}|$ and so we cannot conculde that $A+B$ is also strictly diagonally dominant.
Is everything correct?
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$\begingroup$No, points 1 and 2 are correct, but point 3 is flawed. You have $$|a_{ii}+b_{ii}| \leq |a_{ii}| + |b_{ii}|,$$ rather than $$|a_{ii}+b_{ii}| \leq |a_{ii}| + |b_{ii}|.$$ Regardless, this inequality contains no information about the relationship between the diagonal values of $C=A+B$ and the off diagonal values of $C$.
Instead consider the choice of $B=-A$. If $A$ is strictly diagonally dominant by rows, then so is $B$, but $C=0$ is not strictly diagonally dominant by rows. Similarly, if $B=A$, then $C=2A$ is strictly diagonally dominant by rows.
These two examples show that we cannot conclude that $C=A+B$ is strictly diagonally dominant by rows if our information is limited to the fact that $A$ and $B$ are both strictly diagonally dominant by rows.
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