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












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$begingroup$


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 $k in mathbb{Z}$?






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    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

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      284






















          2 Answers
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          $begingroup$

          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











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          • 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









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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            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






























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