Matrix inequality with its norm












0












$begingroup$


Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
I am trying to find a suitable upper bound of this matrix.



Is this inequality correct?



$M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.






linear-algebra matrices







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
    I am trying to find a suitable upper bound of this matrix.



    Is this inequality correct?



    $M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.






    linear-algebra matrices







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
      I am trying to find a suitable upper bound of this matrix.



      Is this inequality correct?



      $M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.






      linear-algebra matrices







      share|cite|improve this question









      $endgroup$




      Let M $in mathbb{R}^{ntimes n}$ with its norm given by $|{M}|$.
      I am trying to find a suitable upper bound of this matrix.



      Is this inequality correct?



      $M leq |M|mathbb{I}$, where $mathbb{I} in mathbb{R}^{ntimes n}$ is a matrix of all 1s and the inequality is implied entry wise.






      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










      asked Dec 25 '18 at 3:16









      jjgarrisonjjgarrison

      316




      316






















          1 Answer
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          $begingroup$

          Yes, that inequality is correct.



          Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.



          Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.



          Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.



          Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.






          share|cite|improve this answer









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            $begingroup$

            Yes, that inequality is correct.



            Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.



            Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.



            Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.



            Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.






            share|cite|improve this answer









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              2












              $begingroup$

              Yes, that inequality is correct.



              Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.



              Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.



              Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.



              Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes, that inequality is correct.



                Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.



                Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.



                Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.



                Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.






                share|cite|improve this answer









                $endgroup$



                Yes, that inequality is correct.



                Suppose at least one entry of $M$ is larger than $|M|$, i.e. $M_{i,j} > |M|$ for some $i,j$.



                Let $e_1,e_2, ldots, e_n$ be the Euclidean basis vectors, i.e. the $k$-th entry of $e_k$ is $1$ and all the other entries are $0$.



                Then, $|Me_j| = |M_{i,j}e_i| = |M_{i,j}| ge M_{i,j} > |M| = |M| cdot |e_j|$. This is a contradiction, since the matrix norm satisfies $|Mx| le |M| cdot |x|$ for all vectors $x$.



                Therefore, $M_{i,j} le |M|$ for all $i,j$, as desired.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 25 '18 at 3:28









                JimmyK4542JimmyK4542

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                41.8k248110






























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