Must $langle x mid x rangle$ be real on a complex inner product space?

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A complex inner product operator must satisfy four properties, including positivity which says, according to my textbook:



If $x neq 0$ then $langle x mid x rangle > 0$.



Does this mean that $langle x mid x rangle$ must be real?




linear-algebra complex-numbers inner-product-space




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    up vote
    1
    down vote

    favorite












    A complex inner product operator must satisfy four properties, including positivity which says, according to my textbook:



    If $x neq 0$ then $langle x mid x rangle > 0$.



    Does this mean that $langle x mid x rangle$ must be real?




    linear-algebra complex-numbers inner-product-space




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      A complex inner product operator must satisfy four properties, including positivity which says, according to my textbook:



      If $x neq 0$ then $langle x mid x rangle > 0$.



      Does this mean that $langle x mid x rangle$ must be real?




      linear-algebra complex-numbers inner-product-space




      share|cite|improve this question













      A complex inner product operator must satisfy four properties, including positivity which says, according to my textbook:



      If $x neq 0$ then $langle x mid x rangle > 0$.



      Does this mean that $langle x mid x rangle$ must be real?





      linear-algebra complex-numbers inner-product-space





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 1 at 12:43









      Arnaud D.

      14.5k52141




      14.5k52141









      asked Jul 31 at 19:07









      clay

      625312




      625312




















          3 Answers
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          Yes, $langle x, x rangle$ is always real. This comes from conjugate symmetry, as $langle x, x rangle = overlinelangle x, x rangle$.






          share|cite|improve this answer




























            up vote
            4
            down vote













            It is implied by the skew symmetry of the inner product $$langle x,yrangle=overlinelangle y,xrangle$$ applied to $x=y$



            $$langle x,xrangle=overlinelangle x,xrangle$$






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              up vote
              2
              down vote













              I think you mean $langle x,x rangle$, and indeed it must be real.






              share|cite|improve this answer





















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                4
                down vote



                accepted










                Yes, $langle x, x rangle$ is always real. This comes from conjugate symmetry, as $langle x, x rangle = overlinelangle x, x rangle$.






                share|cite|improve this answer

























                  up vote
                  4
                  down vote



                  accepted










                  Yes, $langle x, x rangle$ is always real. This comes from conjugate symmetry, as $langle x, x rangle = overlinelangle x, x rangle$.






                  share|cite|improve this answer























                    up vote
                    4
                    down vote



                    accepted







                    up vote
                    4
                    down vote



                    accepted






                    Yes, $langle x, x rangle$ is always real. This comes from conjugate symmetry, as $langle x, x rangle = overlinelangle x, x rangle$.






                    share|cite|improve this answer













                    Yes, $langle x, x rangle$ is always real. This comes from conjugate symmetry, as $langle x, x rangle = overlinelangle x, x rangle$.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 31 at 19:09









                    Daniel Mroz

                    851314




                    851314




















                        up vote
                        4
                        down vote













                        It is implied by the skew symmetry of the inner product $$langle x,yrangle=overlinelangle y,xrangle$$ applied to $x=y$



                        $$langle x,xrangle=overlinelangle x,xrangle$$






                        share|cite|improve this answer

























                          up vote
                          4
                          down vote













                          It is implied by the skew symmetry of the inner product $$langle x,yrangle=overlinelangle y,xrangle$$ applied to $x=y$



                          $$langle x,xrangle=overlinelangle x,xrangle$$






                          share|cite|improve this answer























                            up vote
                            4
                            down vote










                            up vote
                            4
                            down vote









                            It is implied by the skew symmetry of the inner product $$langle x,yrangle=overlinelangle y,xrangle$$ applied to $x=y$



                            $$langle x,xrangle=overlinelangle x,xrangle$$






                            share|cite|improve this answer













                            It is implied by the skew symmetry of the inner product $$langle x,yrangle=overlinelangle y,xrangle$$ applied to $x=y$



                            $$langle x,xrangle=overlinelangle x,xrangle$$







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered Jul 31 at 19:14









                            JessicaMcRae

                            1264




                            1264




















                                up vote
                                2
                                down vote













                                I think you mean $langle x,x rangle$, and indeed it must be real.






                                share|cite|improve this answer

























                                  up vote
                                  2
                                  down vote













                                  I think you mean $langle x,x rangle$, and indeed it must be real.






                                  share|cite|improve this answer























                                    up vote
                                    2
                                    down vote










                                    up vote
                                    2
                                    down vote









                                    I think you mean $langle x,x rangle$, and indeed it must be real.






                                    share|cite|improve this answer













                                    I think you mean $langle x,x rangle$, and indeed it must be real.







                                    share|cite|improve this answer













                                    share|cite|improve this answer



                                    share|cite|improve this answer











                                    answered Jul 31 at 19:09









                                    Kenny Lau

                                    17.7k2156




                                    17.7k2156






















                                         

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