Does squaring all unitary matrices give the identity matrix?












1














All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.



Is this true for all unitary matrices?






matrices







share|cite|improve this question



























    1














    All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.



    Is this true for all unitary matrices?






    matrices







    share|cite|improve this question

























      1












      1








      1







      All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.



      Is this true for all unitary matrices?






      matrices







      share|cite|improve this question













      All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.



      Is this true for all unitary matrices?






      matrices




      matrices






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Oct 12 '15 at 16:43









      Alan Wolfe

      514424




      514424






















          3 Answers
          3






          active

          oldest

          votes


















          2














          Certainly not, consider the matrix
          $$J := pmatrix{0&-1\1&0}.$$
          Since $J^* J = I$, it is unitary, but $J^2 = -I$.






          share|cite|improve this answer





















          • I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
            – Alan Wolfe
            Oct 12 '15 at 16:49












          • Me too. Aren't they the same?
            – Ian Miller
            Oct 12 '15 at 16:51










          • $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
            – Empy2
            Oct 12 '15 at 16:52










          • Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
            – Alan Wolfe
            Oct 12 '15 at 16:54










          • Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
            – Travis
            Oct 12 '15 at 17:01



















          2














          The examples you have seen must have been Hermitian; in fact
          any two of the properties unitary, involution, Hermitian:
          $$
          A^*A=I, quad A^2 = I, quad A^* = A
          $$
          imply the third.






          share|cite|improve this answer





























            0














            For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.






            share|cite|improve this answer





















              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%2f1476635%2fdoes-squaring-all-unitary-matrices-give-the-identity-matrix%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              Certainly not, consider the matrix
              $$J := pmatrix{0&-1\1&0}.$$
              Since $J^* J = I$, it is unitary, but $J^2 = -I$.






              share|cite|improve this answer





















              • I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
                – Alan Wolfe
                Oct 12 '15 at 16:49












              • Me too. Aren't they the same?
                – Ian Miller
                Oct 12 '15 at 16:51










              • $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
                – Empy2
                Oct 12 '15 at 16:52










              • Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
                – Alan Wolfe
                Oct 12 '15 at 16:54










              • Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
                – Travis
                Oct 12 '15 at 17:01
















              2














              Certainly not, consider the matrix
              $$J := pmatrix{0&-1\1&0}.$$
              Since $J^* J = I$, it is unitary, but $J^2 = -I$.






              share|cite|improve this answer





















              • I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
                – Alan Wolfe
                Oct 12 '15 at 16:49












              • Me too. Aren't they the same?
                – Ian Miller
                Oct 12 '15 at 16:51










              • $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
                – Empy2
                Oct 12 '15 at 16:52










              • Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
                – Alan Wolfe
                Oct 12 '15 at 16:54










              • Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
                – Travis
                Oct 12 '15 at 17:01














              2












              2








              2






              Certainly not, consider the matrix
              $$J := pmatrix{0&-1\1&0}.$$
              Since $J^* J = I$, it is unitary, but $J^2 = -I$.






              share|cite|improve this answer












              Certainly not, consider the matrix
              $$J := pmatrix{0&-1\1&0}.$$
              Since $J^* J = I$, it is unitary, but $J^2 = -I$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Oct 12 '15 at 16:46









              Travis

              59.3k767145




              59.3k767145












              • I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
                – Alan Wolfe
                Oct 12 '15 at 16:49












              • Me too. Aren't they the same?
                – Ian Miller
                Oct 12 '15 at 16:51










              • $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
                – Empy2
                Oct 12 '15 at 16:52










              • Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
                – Alan Wolfe
                Oct 12 '15 at 16:54










              • Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
                – Travis
                Oct 12 '15 at 17:01


















              • I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
                – Alan Wolfe
                Oct 12 '15 at 16:49












              • Me too. Aren't they the same?
                – Ian Miller
                Oct 12 '15 at 16:51










              • $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
                – Empy2
                Oct 12 '15 at 16:52










              • Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
                – Alan Wolfe
                Oct 12 '15 at 16:54










              • Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
                – Travis
                Oct 12 '15 at 17:01
















              I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
              – Alan Wolfe
              Oct 12 '15 at 16:49






              I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
              – Alan Wolfe
              Oct 12 '15 at 16:49














              Me too. Aren't they the same?
              – Ian Miller
              Oct 12 '15 at 16:51




              Me too. Aren't they the same?
              – Ian Miller
              Oct 12 '15 at 16:51












              $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
              – Empy2
              Oct 12 '15 at 16:52




              $J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
              – Empy2
              Oct 12 '15 at 16:52












              Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
              – Alan Wolfe
              Oct 12 '15 at 16:54




              Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
              – Alan Wolfe
              Oct 12 '15 at 16:54












              Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
              – Travis
              Oct 12 '15 at 17:01




              Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
              – Travis
              Oct 12 '15 at 17:01











              2














              The examples you have seen must have been Hermitian; in fact
              any two of the properties unitary, involution, Hermitian:
              $$
              A^*A=I, quad A^2 = I, quad A^* = A
              $$
              imply the third.






              share|cite|improve this answer


























                2














                The examples you have seen must have been Hermitian; in fact
                any two of the properties unitary, involution, Hermitian:
                $$
                A^*A=I, quad A^2 = I, quad A^* = A
                $$
                imply the third.






                share|cite|improve this answer
























                  2












                  2








                  2






                  The examples you have seen must have been Hermitian; in fact
                  any two of the properties unitary, involution, Hermitian:
                  $$
                  A^*A=I, quad A^2 = I, quad A^* = A
                  $$
                  imply the third.






                  share|cite|improve this answer












                  The examples you have seen must have been Hermitian; in fact
                  any two of the properties unitary, involution, Hermitian:
                  $$
                  A^*A=I, quad A^2 = I, quad A^* = A
                  $$
                  imply the third.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 20 '15 at 19:23









                  Bob Terrell

                  1,665710




                  1,665710























                      0














                      For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.






                      share|cite|improve this answer


























                        0














                        For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.






                          share|cite|improve this answer












                          For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 at 10:32









                          mahavir

                          4481320




                          4481320






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f1476635%2fdoes-squaring-all-unitary-matrices-give-the-identity-matrix%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

                              Berounka

                              Image
                              Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                              Read more

                              Sphinx de Gizeh

                              Image
                              Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                              Read more

                              Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                              Image
                              3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                              Read more