On the supremum norm of matrices












3












$begingroup$


Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that



$||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.



Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?



I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.



Please help



NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$






linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that



    $||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.



    Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?



    I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.



    Please help



    NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$






    linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that



      $||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.



      Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?



      I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.



      Please help



      NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$






      linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets







      share|cite|improve this question











      $endgroup$




      Let $D=diag (d_{ii}) in M_n(mathbb R)$ be a diagonal matrix and $Ein M_n(mathbb R)$ be such that



      $||E||_infty < min _{ine j} Bigg|dfrac{d_{ii}-d_{jj}}{2}Bigg|$.



      Then how to show that there is an ordering of the eigenvalues of $D+E$ as ${mu_1,...,mu_n}$ such that $|d_{ii}-mu_i|<||E||_infty $ ?



      I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.



      Please help



      NOTE: Here $||E||_infty :=sup_{||x||_infty=1}||Ex||_infty$






      linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets




      linear-algebra matrices eigenvalues-eigenvectors norm gershgorin-sets






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 11 at 6:26







      user521337

















      asked Jan 11 at 6:04









      user521337user521337

      1,2201417




      1,2201417






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.



          We note that
          $$
          |E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
          $$
          Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.



          Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why are we try to show that the disks will not overlap ?
            $endgroup$
            – user521337
            Jan 11 at 8:31










          • $begingroup$
            Because if the disks have no overlap, then each contains one eigenvalue.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 14:55










          • $begingroup$
            Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 17:03












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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.



          We note that
          $$
          |E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
          $$
          Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.



          Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why are we try to show that the disks will not overlap ?
            $endgroup$
            – user521337
            Jan 11 at 8:31










          • $begingroup$
            Because if the disks have no overlap, then each contains one eigenvalue.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 14:55










          • $begingroup$
            Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 17:03
















          2












          $begingroup$

          Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.



          We note that
          $$
          |E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
          $$
          Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.



          Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            why are we try to show that the disks will not overlap ?
            $endgroup$
            – user521337
            Jan 11 at 8:31










          • $begingroup$
            Because if the disks have no overlap, then each contains one eigenvalue.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 14:55










          • $begingroup$
            Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 17:03














          2












          2








          2





          $begingroup$

          Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.



          We note that
          $$
          |E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
          $$
          Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.



          Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.






          share|cite|improve this answer











          $endgroup$



          Hint: For convenience, I'll take $delta := min _{ine j} |{d_{ii}-d_{jj}}|$.



          We note that
          $$
          |E|_{infty} = max_{i=1,dots,n} sum_{j=1}^n |e_{ij}|
          $$
          Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = sum_{j neq i} |e_{ij}| < frac 12 delta - |e_{ii}|$ (note that we must have $|e_{ii}| leq |E|_infty < frac 12 delta$). I claim that this is enough for us to conclude that the disks will not overlap.



          Note that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 14:56

























          answered Jan 11 at 6:37









          OmnomnomnomOmnomnomnom

          129k794188




          129k794188












          • $begingroup$
            why are we try to show that the disks will not overlap ?
            $endgroup$
            – user521337
            Jan 11 at 8:31










          • $begingroup$
            Because if the disks have no overlap, then each contains one eigenvalue.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 14:55










          • $begingroup$
            Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 17:03


















          • $begingroup$
            why are we try to show that the disks will not overlap ?
            $endgroup$
            – user521337
            Jan 11 at 8:31










          • $begingroup$
            Because if the disks have no overlap, then each contains one eigenvalue.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 14:55










          • $begingroup$
            Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
            $endgroup$
            – Omnomnomnom
            Jan 11 at 17:03
















          $begingroup$
          why are we try to show that the disks will not overlap ?
          $endgroup$
          – user521337
          Jan 11 at 8:31




          $begingroup$
          why are we try to show that the disks will not overlap ?
          $endgroup$
          – user521337
          Jan 11 at 8:31












          $begingroup$
          Because if the disks have no overlap, then each contains one eigenvalue.
          $endgroup$
          – Omnomnomnom
          Jan 11 at 14:55




          $begingroup$
          Because if the disks have no overlap, then each contains one eigenvalue.
          $endgroup$
          – Omnomnomnom
          Jan 11 at 14:55












          $begingroup$
          Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
          $endgroup$
          – Omnomnomnom
          Jan 11 at 17:03




          $begingroup$
          Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $|E|_infty$.
          $endgroup$
          – Omnomnomnom
          Jan 11 at 17:03


















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