Proof on unitary diagonalisable matrices












1












$begingroup$


My task is the following:



$Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



My approach was:



$M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    My task is the following:



    $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



    My approach was:



    $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



    But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      My task is the following:



      $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



      My approach was:



      $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



      But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.










      share|cite|improve this question









      $endgroup$




      My task is the following:



      $Let,, A,U in mathbb{C}^{ntimes n} ,,and,,assume ,,that ,, U ,,is ,,unitary ,, and ,, that ,,U^*AU ,,is,, a,, diagonal,, matrix.\Show ,, that ,,A ,,is,,normal. $



      My approach was:



      $M:= U^*AU Rightarrow A = UMU^*. Now ,, I ,, want ,, to ,, show ,, that ,, A ,, is ,, normal ,,so,,I ,, have ,, to ,, show:\A^*A=AA^*$



      But when I now insert $UMU^*$ for A I don't get far. I don't know how to proceed.







      linear-algebra proof-writing diagonalization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 16:57









      Fo Young Areal LoFo Young Areal Lo

      345




      345






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}

          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01











          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%2f3066428%2fproof-on-unitary-diagonalisable-matrices%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









          2












          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}

          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01
















          2












          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}

          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01














          2












          2








          2





          $begingroup$

          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}

          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.






          share|cite|improve this answer











          $endgroup$



          Following your notation:
          begin{eqnarray}
          A=UMU^*Rightarrow A^*&=&UM^*U^*,\
          AA^*=UMU^*UM^*U^*&=&UMM^*U^*,\
          A^*A=UM^*U^*UMU^*&=&UM^*MU^*.
          end{eqnarray}

          Then $A$ is normal, as diagonal matrices $M$ and $M^*$ commute.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 17:14

























          answered Jan 8 at 17:12









          jobejobe

          994615




          994615












          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01


















          • $begingroup$
            Why is A* = U M* U* shouldn't it be A* = U* M* U
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:46












          • $begingroup$
            If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:51










          • $begingroup$
            $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
            $endgroup$
            – jobe
            Jan 8 at 17:53










          • $begingroup$
            Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 17:55












          • $begingroup$
            But I looked it up, seems like I got something wrong. Thank you
            $endgroup$
            – Fo Young Areal Lo
            Jan 8 at 18:01
















          $begingroup$
          Why is A* = U M* U* shouldn't it be A* = U* M* U
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:46






          $begingroup$
          Why is A* = U M* U* shouldn't it be A* = U* M* U
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:46














          $begingroup$
          If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:51




          $begingroup$
          If $X$ and $Y$ are matrices, then $(XY)^*=Y^*X^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:51












          $begingroup$
          $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:53




          $begingroup$
          $A^*=(UMU^*)^*=(U^*)^*M^*U^*=UM^*U^*$.
          $endgroup$
          – jobe
          Jan 8 at 17:53












          $begingroup$
          Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:55






          $begingroup$
          Ok so my thought was $A = UMU^* ,, then ,, A* = (UMU^*)^* = U^* M^* U^{**} = U^* M^* U$
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 17:55














          $begingroup$
          But I looked it up, seems like I got something wrong. Thank you
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 18:01




          $begingroup$
          But I looked it up, seems like I got something wrong. Thank you
          $endgroup$
          – Fo Young Areal Lo
          Jan 8 at 18:01


















          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%2f3066428%2fproof-on-unitary-diagonalisable-matrices%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

          Human spaceflight

          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

          張江高科駅