An inequality of matrix norm












3












$begingroup$


Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
$$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
$$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






linear-algebra matrices functional-analysis norm







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
    $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



    Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
    $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



    I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






    linear-algebra matrices functional-analysis norm







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
      $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



      Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
      $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



      I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






      linear-algebra matrices functional-analysis norm







      share|cite|improve this question











      $endgroup$




      Let $V,W$ be complex inner product spaces. Suppose $T: V to W$ is a linear map, then we define
      $$|T|:=sup{|Tv|_{W}:|v|_{V}=1}$$ where $|v_{V}|:=sqrt{langle v,vrangle}$ and $|Tv|_{W}:=sqrt{langle Tv,Tvrangle}$.



      Question: Suppose $U_1,ldots,U_k$ and $V_1,ldots,V_k$ are $n {times} n$ unitary matrices. Show that
      $$|U_1cdots U_k-V_1cdots V_k| leq sum_{i=1}^{k}|U_i-V_i|$$



      I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!






      linear-algebra matrices functional-analysis norm




      linear-algebra matrices functional-analysis norm






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Bernard

      123k741116




      123k741116










      asked 2 hours ago









      bbwbbw

      51739




      51739






















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            56 mins ago











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149889%2fan-inequality-of-matrix-norm%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            56 mins ago
















          6












          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            56 mins ago














          6












          6








          6





          $begingroup$

          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)






          share|cite|improve this answer











          $endgroup$



          For $k = 1$, the inequality becomes an identity. So, start with the special case $k = 2$:
          $$
          begin{array}{ll}
          & ||U_1 U_2 - V_1 V_2||\ \
          = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\ \
          = &
          || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \ \
          leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\
          end{array}
          $$
          (The last inequality is the triangle inequality.) Now, since $U_2$ is unitary, we have $||U_{2}|| = 1$, so
          $$
          || ( U_1 - V_1) U_2 || leq || U_1 - V_1 ||.
          $$
          A similar bound obtains for $||V_1 (U_2 - V_2) ||$.



          This should give you enough "building blocks".:)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 56 mins ago

























          answered 1 hour ago









          avsavs

          3,424513




          3,424513












          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            56 mins ago


















          • $begingroup$
            Thank you so much!
            $endgroup$
            – bbw
            1 hour ago






          • 1




            $begingroup$
            You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
            $endgroup$
            – avs
            56 mins ago
















          $begingroup$
          Thank you so much!
          $endgroup$
          – bbw
          1 hour ago




          $begingroup$
          Thank you so much!
          $endgroup$
          – bbw
          1 hour ago




          1




          1




          $begingroup$
          You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
          $endgroup$
          – avs
          56 mins ago




          $begingroup$
          You are welcome. I edited to remove the (unjustified) assumption that the matrices commute.:)
          $endgroup$
          – avs
          56 mins ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3149889%2fan-inequality-of-matrix-norm%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          What other Star Trek series did the main TNG cast show up in?

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
          Read more

          Berlina muro

          Image
          La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
          Read more

          Berlina aerponto

          Image
          Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa
          Read more