Prove or disprove: If a matrix has the property $0 \neq |a_{ii}| \geq \sum_{\substack{j=1 \\ j \neq i}} |a_{ij}| $ then Gaussian Elimination (without pivoting) will preserve this property.
I assume this to be true since I have seen other theorems state that "Gaussian elimination without pivoting preserves the diagonal dominance of a matrix" without much other qualification. I am not sure how the inequality to 0 would change this. However, most of those used the strong inequality not the weak, so perhaps that would change things as well. Could someone help me with a proof to verify this? Else provide a counterexample?
I have been trying to come up with one to no avail. I would assume showing that the first row iteration of Gaussian preserves the condition would mostly complete the proof so I have been trying to come at it with that in mind.
Best Answer
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e., $$ \left|a_{2,2}-a_{1,2}{a_{2,1}\over a_{1,1}}\right|\ge\sum_{i=3}^n\left|a_{2,i}-a_{1,i}{a_{2,1}\over a_{1,1}}\right|, $$ which is equivalent to $$ |a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|\ge\sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|. $$ But this is true: \begin{eqnarray*} \sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&\le& |a_{1,1}|\sum_{i=3}^n|a_{2,i}|+|a_{2,1}|\sum_{i=3}^n|a_{1,i}| \\ &\le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \\ &=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\\ &\le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}| \end{eqnarray*}