Does representation irreducibility ensure non-zero determinant?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



Thank you very much!



Cheers,
Collin



P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






matrices group-theory finite-groups determinant







share|cite|improve this question

























    up vote
    2
    down vote

    favorite












    If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



    Thank you very much!



    Cheers,
    Collin



    P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






    matrices group-theory finite-groups determinant







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



      Thank you very much!



      Cheers,
      Collin



      P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






      matrices group-theory finite-groups determinant







      share|cite|improve this question













      If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



      Thank you very much!



      Cheers,
      Collin



      P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!






      matrices group-theory finite-groups determinant




      matrices group-theory finite-groups determinant






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      Collin Ren

      565




      565




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          5
          down vote



          accepted










          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer




















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour 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',
          convertImagesToLinks: true,
          noModals: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2918323%2fdoes-representation-irreducibility-ensure-non-zero-determinant%23new-answer', 'question_page');

          );

          Post as a guest

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          5
          down vote



          accepted










          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer




















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago














          up vote
          5
          down vote



          accepted










          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer




















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.






          share|cite|improve this answer












          The inverse of $M(g)$ is $M(g^-1)$.



          And of course, invertible matrices have non-zero determinant.



          Note that this is true for all representations, not just irreducible ones.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Daniel Mroz

          1,364414




          1,364414











          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago
















          • Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
            – Collin Ren
            1 hour ago















          Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
          – Collin Ren
          1 hour ago




          Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
          – Collin Ren
          1 hour ago

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2918323%2fdoes-representation-irreducibility-ensure-non-zero-determinant%23new-answer', 'question_page');

          );

          Post as a guest
































































          Comments

          Post a Comment

          Popular posts from this blog

          What does second last employer means? [closed]

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
          Read more

          Installing NextGIS Connect into QGIS 3?

          Image
          Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
          Read more

          One-line joke

          A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
          Read more