Diagonally Dominant Matrix Preserved after Gaussian Elimination (with a modification)












1












$begingroup$


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.






linear-algebra matrices







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  • 1




    $begingroup$
    What is pivoting exactly?
    $endgroup$
    – Git Gud
    Jan 26 '14 at 14:54






  • 1




    $begingroup$
    After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:10










  • $begingroup$
    I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:40










  • $begingroup$
    Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
    $endgroup$
    – Git Gud
    Jan 26 '14 at 15:41
















1












$begingroup$


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.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is pivoting exactly?
    $endgroup$
    – Git Gud
    Jan 26 '14 at 14:54






  • 1




    $begingroup$
    After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:10










  • $begingroup$
    I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:40










  • $begingroup$
    Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
    $endgroup$
    – Git Gud
    Jan 26 '14 at 15:41














1












1








1


1



$begingroup$


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.






linear-algebra matrices







share|cite|improve this question











$endgroup$




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.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 '14 at 14:54









Git Gud

28.7k1050100




28.7k1050100










asked Jan 26 '14 at 14:51









James SnyderJames Snyder

1479




1479








  • 1




    $begingroup$
    What is pivoting exactly?
    $endgroup$
    – Git Gud
    Jan 26 '14 at 14:54






  • 1




    $begingroup$
    After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:10










  • $begingroup$
    I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:40










  • $begingroup$
    Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
    $endgroup$
    – Git Gud
    Jan 26 '14 at 15:41














  • 1




    $begingroup$
    What is pivoting exactly?
    $endgroup$
    – Git Gud
    Jan 26 '14 at 14:54






  • 1




    $begingroup$
    After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:10










  • $begingroup$
    I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
    $endgroup$
    – James Snyder
    Jan 26 '14 at 15:40










  • $begingroup$
    Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
    $endgroup$
    – Git Gud
    Jan 26 '14 at 15:41








1




1




$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54




$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54




1




1




$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10




$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10












$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40




$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40












$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41




$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41










1 Answer
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$begingroup$

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.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
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*}






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    1 Answer
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    1 Answer
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    0












    $begingroup$

    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.,
    $$
    left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
    $$
    which is equivalent to
    $$
    |a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
    $$
    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*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.,
      $$
      left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
      $$
      which is equivalent to
      $$
      |a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
      $$
      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*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.,
        $$
        left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
        $$
        which is equivalent to
        $$
        |a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
        $$
        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*}






        share|cite|improve this answer









        $endgroup$



        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.,
        $$
        left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
        $$
        which is equivalent to
        $$
        |a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
        $$
        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*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 22 '14 at 20:39









        user32828user32828

        14210




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