$A^tto 0$ when its row sum is strictly less than one?












0












$begingroup$


$A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.



Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.






matrices limits eigenvalues-eigenvectors matrix-norms







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    0












    $begingroup$


    $A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.



    Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.






    matrices limits eigenvalues-eigenvectors matrix-norms







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      $A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.



      Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.






      matrices limits eigenvalues-eigenvectors matrix-norms







      share|cite|improve this question











      $endgroup$




      $A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.



      Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.






      matrices limits eigenvalues-eigenvectors matrix-norms




      matrices limits eigenvalues-eigenvectors matrix-norms






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 19:03









      Martin Sleziak

      44.8k9118272




      44.8k9118272










      asked Dec 6 '18 at 16:17









      MarkovMarkov

      17.3k1058178




      17.3k1058178






















          1 Answer
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          $begingroup$

          The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.



          Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.



          This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
            $endgroup$
            – Markov
            Dec 6 '18 at 16:31












          • $begingroup$
            @Wow, I think the extra information is redundant.
            $endgroup$
            – lhf
            Dec 6 '18 at 16:37











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.



          Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.



          This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
            $endgroup$
            – Markov
            Dec 6 '18 at 16:31












          • $begingroup$
            @Wow, I think the extra information is redundant.
            $endgroup$
            – lhf
            Dec 6 '18 at 16:37
















          2












          $begingroup$

          The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.



          Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.



          This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
            $endgroup$
            – Markov
            Dec 6 '18 at 16:31












          • $begingroup$
            @Wow, I think the extra information is redundant.
            $endgroup$
            – lhf
            Dec 6 '18 at 16:37














          2












          2








          2





          $begingroup$

          The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.



          Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.



          This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.






          share|cite|improve this answer











          $endgroup$



          The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.



          Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.



          This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 16:39

























          answered Dec 6 '18 at 16:25









          lhflhf

          165k10171395




          165k10171395












          • $begingroup$
            where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
            $endgroup$
            – Markov
            Dec 6 '18 at 16:31












          • $begingroup$
            @Wow, I think the extra information is redundant.
            $endgroup$
            – lhf
            Dec 6 '18 at 16:37


















          • $begingroup$
            where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
            $endgroup$
            – Markov
            Dec 6 '18 at 16:31












          • $begingroup$
            @Wow, I think the extra information is redundant.
            $endgroup$
            – lhf
            Dec 6 '18 at 16:37
















          $begingroup$
          where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
          $endgroup$
          – Markov
          Dec 6 '18 at 16:31






          $begingroup$
          where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
          $endgroup$
          – Markov
          Dec 6 '18 at 16:31














          $begingroup$
          @Wow, I think the extra information is redundant.
          $endgroup$
          – lhf
          Dec 6 '18 at 16:37




          $begingroup$
          @Wow, I think the extra information is redundant.
          $endgroup$
          – lhf
          Dec 6 '18 at 16:37


















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