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Given a strictly diagonally dominant matrix $A$ i.e., $|a_{ii}| > \sum_{j=1,i \neq j}^{n} |a_{ij}|$ for $i = 1 \cdots n$. I need to show that there is atleast one column k which is dominant i.e to show $|a_{kk}| > \sum_{j=1,k \neq j}^{n} |a_{jk}|$ for at least one value of $k = 1 \cdots n$

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  • $\begingroup$ What if it isn't for any $k$? $\endgroup$ – scineram Nov 3 '11 at 6:48
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Each diagonal entry is greater than the sum of the remaining entries in its row. Thus the sum of the diagonal entries is greater than the sum of all remaining entries. If each diagonal entry were less than or equal to the sum of the remaining entries in its row, the sum of the diagonal entries would have to be less than or equal to the sum of the remaining entries. (I omitted the absolute values for clarity.)

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  • $\begingroup$ I agree that the stated example is strictly diagonally dominant matrix. But,we can also see that the col2 and col3 is also dominant. So this wont be a counter example for my question $\endgroup$ – Learner Nov 3 '11 at 8:41
  • $\begingroup$ @Learner: That comment was in very bad style. You changed the question such that it's no longer a counterexample, and then not only failed to mark that change to make it apparent that my counterexample was correct when posted, but even created the appearance by your comment that it wasn't. Would you want anyone to deal with your contributions like this? $\endgroup$ – joriki Nov 3 '11 at 8:43
  • $\begingroup$ @Learner: I've rolled back to an earlier version of my answer where I had coincidentally misinterpreted your original question to mean what you've now changed it to. $\endgroup$ – joriki Nov 3 '11 at 8:46
  • $\begingroup$ Im extremely sorry. I actually mistyped the index. will make sure the question is correctly stated from next time onwards. $\endgroup$ – Learner Nov 3 '11 at 8:50
  • $\begingroup$ @Learner: Mistakes happen; it's just important to resolve them appropriately. $\endgroup$ – joriki Nov 3 '11 at 9:08

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