Is there always another non-trivial hermitian matrix $A$ for $B$ that $e^{iAx_0} = e^{iBx_1}$, $x_0 neq x_1$...












0












$begingroup$


Suppose $A$ and $B$ are hermitian $n times n$ matrix, with $A neq B$. $x_0$ and $x_1$ are real numbers, with $x_0 neq x_1$.



Let $B$ a given matrix, and $x_0$, $x_1$ are given numbers as well, with $x_0 neq x_1$. Would there always be some $A$ such that $e^{iAx_0} = e^{iBx_1}$ that is not simply $Bx_1/x_0$?



What would happen if $n$ goes to infinity?



edit:
What if I also constrain $A$ to be so that $A neq Bx_1/x_0 + 2pi k/x_0$ with $k in mathbb{Z}$?






linear-algebra







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Suppose $A$ and $B$ are hermitian $n times n$ matrix, with $A neq B$. $x_0$ and $x_1$ are real numbers, with $x_0 neq x_1$.



    Let $B$ a given matrix, and $x_0$, $x_1$ are given numbers as well, with $x_0 neq x_1$. Would there always be some $A$ such that $e^{iAx_0} = e^{iBx_1}$ that is not simply $Bx_1/x_0$?



    What would happen if $n$ goes to infinity?



    edit:
    What if I also constrain $A$ to be so that $A neq Bx_1/x_0 + 2pi k/x_0$ with $k in mathbb{Z}$?






    linear-algebra







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose $A$ and $B$ are hermitian $n times n$ matrix, with $A neq B$. $x_0$ and $x_1$ are real numbers, with $x_0 neq x_1$.



      Let $B$ a given matrix, and $x_0$, $x_1$ are given numbers as well, with $x_0 neq x_1$. Would there always be some $A$ such that $e^{iAx_0} = e^{iBx_1}$ that is not simply $Bx_1/x_0$?



      What would happen if $n$ goes to infinity?



      edit:
      What if I also constrain $A$ to be so that $A neq Bx_1/x_0 + 2pi k/x_0$ with $k in mathbb{Z}$?






      linear-algebra







      share|cite|improve this question











      $endgroup$




      Suppose $A$ and $B$ are hermitian $n times n$ matrix, with $A neq B$. $x_0$ and $x_1$ are real numbers, with $x_0 neq x_1$.



      Let $B$ a given matrix, and $x_0$, $x_1$ are given numbers as well, with $x_0 neq x_1$. Would there always be some $A$ such that $e^{iAx_0} = e^{iBx_1}$ that is not simply $Bx_1/x_0$?



      What would happen if $n$ goes to infinity?



      edit:
      What if I also constrain $A$ to be so that $A neq Bx_1/x_0 + 2pi k/x_0$ with $k in mathbb{Z}$?






      linear-algebra




      linear-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 2:14







      Mark Vanderbilt

















      asked Dec 4 '18 at 0:18









      Mark VanderbiltMark Vanderbilt

      284




      284






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The equation is equivalent to $e^{iA} = e^{iBt}$ where $t = x_1/x_0$.



          Since the exponential map is a group homomorphism, the preimage is the set of matrices of the form $Bt + X$ where $e^{iX} = 1$, i.e. $X$ is in the kernel. The kernel includes diagonal matrices with entries of the form $2pi n$ for $ninmathbb Z$ which are Hermitian.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
            $endgroup$
            – jgon
            Dec 4 '18 at 1:19










          • $begingroup$
            Thanks that’s better
            $endgroup$
            – Ben
            Dec 4 '18 at 1:59



















          0












          $begingroup$

          Let $A=(Bx_1+2pi k)/x_0$ (which is Hermitian), for any $kinBbb{Z}$. Then
          $$e^{iAx_0}=e^{i(Bx_1+2pi k)}=e^{iBx_1+2pi i}=e^{iBx_1},$$
          so as long as $x_0ne 0$, then the answer is yes, you can always find $Ane B$ with the property you want.



          Also you lost the $i$ in the equation in the body, but its there in the title, so I assume I'm answering the right question.






          share|cite|improve this answer









          $endgroup$













            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%2f3024920%2fis-there-always-another-non-trivial-hermitian-matrix-a-for-b-that-eiax-0%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The equation is equivalent to $e^{iA} = e^{iBt}$ where $t = x_1/x_0$.



            Since the exponential map is a group homomorphism, the preimage is the set of matrices of the form $Bt + X$ where $e^{iX} = 1$, i.e. $X$ is in the kernel. The kernel includes diagonal matrices with entries of the form $2pi n$ for $ninmathbb Z$ which are Hermitian.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
              $endgroup$
              – jgon
              Dec 4 '18 at 1:19










            • $begingroup$
              Thanks that’s better
              $endgroup$
              – Ben
              Dec 4 '18 at 1:59
















            1












            $begingroup$

            The equation is equivalent to $e^{iA} = e^{iBt}$ where $t = x_1/x_0$.



            Since the exponential map is a group homomorphism, the preimage is the set of matrices of the form $Bt + X$ where $e^{iX} = 1$, i.e. $X$ is in the kernel. The kernel includes diagonal matrices with entries of the form $2pi n$ for $ninmathbb Z$ which are Hermitian.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
              $endgroup$
              – jgon
              Dec 4 '18 at 1:19










            • $begingroup$
              Thanks that’s better
              $endgroup$
              – Ben
              Dec 4 '18 at 1:59














            1












            1








            1





            $begingroup$

            The equation is equivalent to $e^{iA} = e^{iBt}$ where $t = x_1/x_0$.



            Since the exponential map is a group homomorphism, the preimage is the set of matrices of the form $Bt + X$ where $e^{iX} = 1$, i.e. $X$ is in the kernel. The kernel includes diagonal matrices with entries of the form $2pi n$ for $ninmathbb Z$ which are Hermitian.






            share|cite|improve this answer











            $endgroup$



            The equation is equivalent to $e^{iA} = e^{iBt}$ where $t = x_1/x_0$.



            Since the exponential map is a group homomorphism, the preimage is the set of matrices of the form $Bt + X$ where $e^{iX} = 1$, i.e. $X$ is in the kernel. The kernel includes diagonal matrices with entries of the form $2pi n$ for $ninmathbb Z$ which are Hermitian.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 1:59

























            answered Dec 4 '18 at 0:48









            BenBen

            3,218616




            3,218616








            • 1




              $begingroup$
              What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
              $endgroup$
              – jgon
              Dec 4 '18 at 1:19










            • $begingroup$
              Thanks that’s better
              $endgroup$
              – Ben
              Dec 4 '18 at 1:59














            • 1




              $begingroup$
              What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
              $endgroup$
              – jgon
              Dec 4 '18 at 1:19










            • $begingroup$
              Thanks that’s better
              $endgroup$
              – Ben
              Dec 4 '18 at 1:59








            1




            1




            $begingroup$
            What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
            $endgroup$
            – jgon
            Dec 4 '18 at 1:19




            $begingroup$
            What? I'm pretty sure the $i$ out front makes that not work. Like, the diagonal matrix with $2pi$ on the diagonal is Hermitian and in the kernel of $e^{iX}$.
            $endgroup$
            – jgon
            Dec 4 '18 at 1:19












            $begingroup$
            Thanks that’s better
            $endgroup$
            – Ben
            Dec 4 '18 at 1:59




            $begingroup$
            Thanks that’s better
            $endgroup$
            – Ben
            Dec 4 '18 at 1:59











            0












            $begingroup$

            Let $A=(Bx_1+2pi k)/x_0$ (which is Hermitian), for any $kinBbb{Z}$. Then
            $$e^{iAx_0}=e^{i(Bx_1+2pi k)}=e^{iBx_1+2pi i}=e^{iBx_1},$$
            so as long as $x_0ne 0$, then the answer is yes, you can always find $Ane B$ with the property you want.



            Also you lost the $i$ in the equation in the body, but its there in the title, so I assume I'm answering the right question.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $A=(Bx_1+2pi k)/x_0$ (which is Hermitian), for any $kinBbb{Z}$. Then
              $$e^{iAx_0}=e^{i(Bx_1+2pi k)}=e^{iBx_1+2pi i}=e^{iBx_1},$$
              so as long as $x_0ne 0$, then the answer is yes, you can always find $Ane B$ with the property you want.



              Also you lost the $i$ in the equation in the body, but its there in the title, so I assume I'm answering the right question.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $A=(Bx_1+2pi k)/x_0$ (which is Hermitian), for any $kinBbb{Z}$. Then
                $$e^{iAx_0}=e^{i(Bx_1+2pi k)}=e^{iBx_1+2pi i}=e^{iBx_1},$$
                so as long as $x_0ne 0$, then the answer is yes, you can always find $Ane B$ with the property you want.



                Also you lost the $i$ in the equation in the body, but its there in the title, so I assume I'm answering the right question.






                share|cite|improve this answer









                $endgroup$



                Let $A=(Bx_1+2pi k)/x_0$ (which is Hermitian), for any $kinBbb{Z}$. Then
                $$e^{iAx_0}=e^{i(Bx_1+2pi k)}=e^{iBx_1+2pi i}=e^{iBx_1},$$
                so as long as $x_0ne 0$, then the answer is yes, you can always find $Ane B$ with the property you want.



                Also you lost the $i$ in the equation in the body, but its there in the title, so I assume I'm answering the right question.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 1:30









                jgonjgon

                13.5k22041




                13.5k22041






























                    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%2f3024920%2fis-there-always-another-non-trivial-hermitian-matrix-a-for-b-that-eiax-0%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

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                    Read more

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Image
                    0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                    Read more