When $e^A = e^B$ for matrices $A,B$?












0












$begingroup$


Let $A$ and $B$ be $ntimes n$ complex matrices such that $e^A = e^B$.
I would like to know relations between $A$ and $B$.



When $A,Binmathbb{C}$, we have a simple relation $e^A = e^B Leftrightarrow A-Bin 2pi i mathbb{Z}$.
Is there such simple relation for general matrices $A$ and $B$?










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$endgroup$












  • $begingroup$
    See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
    $endgroup$
    – user1551
    Dec 19 '18 at 4:00












  • $begingroup$
    See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 10:28


















0












$begingroup$


Let $A$ and $B$ be $ntimes n$ complex matrices such that $e^A = e^B$.
I would like to know relations between $A$ and $B$.



When $A,Binmathbb{C}$, we have a simple relation $e^A = e^B Leftrightarrow A-Bin 2pi i mathbb{Z}$.
Is there such simple relation for general matrices $A$ and $B$?










share|cite|improve this question









$endgroup$












  • $begingroup$
    See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
    $endgroup$
    – user1551
    Dec 19 '18 at 4:00












  • $begingroup$
    See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 10:28
















0












0








0


1



$begingroup$


Let $A$ and $B$ be $ntimes n$ complex matrices such that $e^A = e^B$.
I would like to know relations between $A$ and $B$.



When $A,Binmathbb{C}$, we have a simple relation $e^A = e^B Leftrightarrow A-Bin 2pi i mathbb{Z}$.
Is there such simple relation for general matrices $A$ and $B$?










share|cite|improve this question









$endgroup$




Let $A$ and $B$ be $ntimes n$ complex matrices such that $e^A = e^B$.
I would like to know relations between $A$ and $B$.



When $A,Binmathbb{C}$, we have a simple relation $e^A = e^B Leftrightarrow A-Bin 2pi i mathbb{Z}$.
Is there such simple relation for general matrices $A$ and $B$?







matrices matrix-calculus matrix-exponential






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asked Dec 19 '18 at 1:47









user356126user356126

1306




1306












  • $begingroup$
    See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
    $endgroup$
    – user1551
    Dec 19 '18 at 4:00












  • $begingroup$
    See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 10:28




















  • $begingroup$
    See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
    $endgroup$
    – user1551
    Dec 19 '18 at 4:00












  • $begingroup$
    See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
    $endgroup$
    – Jyrki Lahtonen
    Dec 22 '18 at 10:28


















$begingroup$
See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
$endgroup$
– user1551
Dec 19 '18 at 4:00






$begingroup$
See https://math.stackexchange.com/q/660823 and in particular the answer by user loup blanc. Briefly speaking, the exponential map is "mostly" injective.
$endgroup$
– user1551
Dec 19 '18 at 4:00














$begingroup$
See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 10:28






$begingroup$
See also here. My answer is basically that for any real number $x$ the exponential of $$pmatrix{0&-xcr4pi^2/x&0cr}$$ is equal to $I_2$ because the eigenvalues, $pm2pi i$ become $1$ when the exponential function is applied.
$endgroup$
– Jyrki Lahtonen
Dec 22 '18 at 10:28












2 Answers
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$begingroup$

For any nonsingular $n times n$ matrix $C$, there are matrices $A$ such that $exp(A)=C$.



Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $lambda$,
$$pmatrix{ lambda & 1 & 0 & ldots & 0 & 0cr
0 & lambda & 1 & ldots & 0& 0cr
ldots & ldots & ldots & ldots & ldots & ldotscr
0 & 0 & 0 & ldots &lambda & 1cr
0 & 0 & 0 & ldots & 0 &lambda cr}$$

$A$ can have a corresponding block of size $m$ where the diagonal elements are some
branch of $log(lambda)$, and the elements in the $k$'th super-diagonal are $dfrac{d^n}{dlambda^n} log(lambda)/n!$.
This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.



If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = pmatrix{1 & 1 & 0 & 0cr 0 & 1 & 0 & 0cr 0 & 0 & 1 & 1cr 0 & 0 & 0 & 1cr}$$
commutes with $$ J = pmatrix{cos(t) & 0 & sin(t) & 0cr 0 & cos(t) & 0 & sin(t)cr -sin(t) & 0 & cos(t) & 0cr 0 & -sin(t) & 0 & cos(t)}$$
logarithms of $C$ will include
$$ eqalign{J &pmatrix{ w_1 & 1 & 0 & 0cr 0 & w_1 & 0 & 0cr 0 & 0 & w_2 & 1cr 0 & 0 & 0 & w_2cr} J^{-1} cr = &pmatrix{
(w_1-w_2)cos(t)^2+w_2 & 1 & -cos(t)sin(t)(w_1-w_2) & 0cr
0 & (w_1-w_2)cos(t)^2+w_2 & 0 & -cos(t)sin(t)(w_1-w_2)cr
-cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1 & 1cr
0 & -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1cr}}
$$

where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2pi i$).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $textbf{Definition.}$ We say that the spectrum of $Ain M_n(mathbb{C})$ is $2ipi$ congruence free (denoted by $2ipi$ CF) iff for every $u,vin spectrum(A)$, $u-vnotin 2ipimathbb{Z}^*$.



    $textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,Bin M_n(mathbb{C})$. If $spectrum(A)$ is $2ipi$ CF and $e^A=e^B$, then



    i) $AB=BA$ and, therefore,



    ii) $A-B$ is similar to $diag(2ipialpha_1,cdots,2ipialpha_n)$ where $alpha_iinmathbb{Z}$.



    $textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)subset 2ipimathbb{Z}$.



    $textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2ipi$ CF and don't pairwise commute.



    ii) $A$ is $2ipi$ CF iff $A$ is a polynomial in $e^A$ iff the function $Xmapsto e^X$ is one to one in a neighborhood of $A$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      2












      $begingroup$

      For any nonsingular $n times n$ matrix $C$, there are matrices $A$ such that $exp(A)=C$.



      Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $lambda$,
      $$pmatrix{ lambda & 1 & 0 & ldots & 0 & 0cr
      0 & lambda & 1 & ldots & 0& 0cr
      ldots & ldots & ldots & ldots & ldots & ldotscr
      0 & 0 & 0 & ldots &lambda & 1cr
      0 & 0 & 0 & ldots & 0 &lambda cr}$$

      $A$ can have a corresponding block of size $m$ where the diagonal elements are some
      branch of $log(lambda)$, and the elements in the $k$'th super-diagonal are $dfrac{d^n}{dlambda^n} log(lambda)/n!$.
      This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.



      If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = pmatrix{1 & 1 & 0 & 0cr 0 & 1 & 0 & 0cr 0 & 0 & 1 & 1cr 0 & 0 & 0 & 1cr}$$
      commutes with $$ J = pmatrix{cos(t) & 0 & sin(t) & 0cr 0 & cos(t) & 0 & sin(t)cr -sin(t) & 0 & cos(t) & 0cr 0 & -sin(t) & 0 & cos(t)}$$
      logarithms of $C$ will include
      $$ eqalign{J &pmatrix{ w_1 & 1 & 0 & 0cr 0 & w_1 & 0 & 0cr 0 & 0 & w_2 & 1cr 0 & 0 & 0 & w_2cr} J^{-1} cr = &pmatrix{
      (w_1-w_2)cos(t)^2+w_2 & 1 & -cos(t)sin(t)(w_1-w_2) & 0cr
      0 & (w_1-w_2)cos(t)^2+w_2 & 0 & -cos(t)sin(t)(w_1-w_2)cr
      -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1 & 1cr
      0 & -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1cr}}
      $$

      where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2pi i$).






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        For any nonsingular $n times n$ matrix $C$, there are matrices $A$ such that $exp(A)=C$.



        Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $lambda$,
        $$pmatrix{ lambda & 1 & 0 & ldots & 0 & 0cr
        0 & lambda & 1 & ldots & 0& 0cr
        ldots & ldots & ldots & ldots & ldots & ldotscr
        0 & 0 & 0 & ldots &lambda & 1cr
        0 & 0 & 0 & ldots & 0 &lambda cr}$$

        $A$ can have a corresponding block of size $m$ where the diagonal elements are some
        branch of $log(lambda)$, and the elements in the $k$'th super-diagonal are $dfrac{d^n}{dlambda^n} log(lambda)/n!$.
        This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.



        If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = pmatrix{1 & 1 & 0 & 0cr 0 & 1 & 0 & 0cr 0 & 0 & 1 & 1cr 0 & 0 & 0 & 1cr}$$
        commutes with $$ J = pmatrix{cos(t) & 0 & sin(t) & 0cr 0 & cos(t) & 0 & sin(t)cr -sin(t) & 0 & cos(t) & 0cr 0 & -sin(t) & 0 & cos(t)}$$
        logarithms of $C$ will include
        $$ eqalign{J &pmatrix{ w_1 & 1 & 0 & 0cr 0 & w_1 & 0 & 0cr 0 & 0 & w_2 & 1cr 0 & 0 & 0 & w_2cr} J^{-1} cr = &pmatrix{
        (w_1-w_2)cos(t)^2+w_2 & 1 & -cos(t)sin(t)(w_1-w_2) & 0cr
        0 & (w_1-w_2)cos(t)^2+w_2 & 0 & -cos(t)sin(t)(w_1-w_2)cr
        -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1 & 1cr
        0 & -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1cr}}
        $$

        where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2pi i$).






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          For any nonsingular $n times n$ matrix $C$, there are matrices $A$ such that $exp(A)=C$.



          Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $lambda$,
          $$pmatrix{ lambda & 1 & 0 & ldots & 0 & 0cr
          0 & lambda & 1 & ldots & 0& 0cr
          ldots & ldots & ldots & ldots & ldots & ldotscr
          0 & 0 & 0 & ldots &lambda & 1cr
          0 & 0 & 0 & ldots & 0 &lambda cr}$$

          $A$ can have a corresponding block of size $m$ where the diagonal elements are some
          branch of $log(lambda)$, and the elements in the $k$'th super-diagonal are $dfrac{d^n}{dlambda^n} log(lambda)/n!$.
          This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.



          If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = pmatrix{1 & 1 & 0 & 0cr 0 & 1 & 0 & 0cr 0 & 0 & 1 & 1cr 0 & 0 & 0 & 1cr}$$
          commutes with $$ J = pmatrix{cos(t) & 0 & sin(t) & 0cr 0 & cos(t) & 0 & sin(t)cr -sin(t) & 0 & cos(t) & 0cr 0 & -sin(t) & 0 & cos(t)}$$
          logarithms of $C$ will include
          $$ eqalign{J &pmatrix{ w_1 & 1 & 0 & 0cr 0 & w_1 & 0 & 0cr 0 & 0 & w_2 & 1cr 0 & 0 & 0 & w_2cr} J^{-1} cr = &pmatrix{
          (w_1-w_2)cos(t)^2+w_2 & 1 & -cos(t)sin(t)(w_1-w_2) & 0cr
          0 & (w_1-w_2)cos(t)^2+w_2 & 0 & -cos(t)sin(t)(w_1-w_2)cr
          -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1 & 1cr
          0 & -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1cr}}
          $$

          where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2pi i$).






          share|cite|improve this answer









          $endgroup$



          For any nonsingular $n times n$ matrix $C$, there are matrices $A$ such that $exp(A)=C$.



          Suppose $C$ is in Jordan form. For each Jordan block of size $m$ with eigenvalue $lambda$,
          $$pmatrix{ lambda & 1 & 0 & ldots & 0 & 0cr
          0 & lambda & 1 & ldots & 0& 0cr
          ldots & ldots & ldots & ldots & ldots & ldotscr
          0 & 0 & 0 & ldots &lambda & 1cr
          0 & 0 & 0 & ldots & 0 &lambda cr}$$

          $A$ can have a corresponding block of size $m$ where the diagonal elements are some
          branch of $log(lambda)$, and the elements in the $k$'th super-diagonal are $dfrac{d^n}{dlambda^n} log(lambda)/n!$.
          This is not a Jordan block, but its Jordan form will be a single Jordan block of size $m$.



          If the blocks of $C$ are all different, the decomposition is unique, and the only choice in $A$ is which branches of logarithm to use. However, if there are identical blocks, there will be a continuum of choices. Thus since $$ C = pmatrix{1 & 1 & 0 & 0cr 0 & 1 & 0 & 0cr 0 & 0 & 1 & 1cr 0 & 0 & 0 & 1cr}$$
          commutes with $$ J = pmatrix{cos(t) & 0 & sin(t) & 0cr 0 & cos(t) & 0 & sin(t)cr -sin(t) & 0 & cos(t) & 0cr 0 & -sin(t) & 0 & cos(t)}$$
          logarithms of $C$ will include
          $$ eqalign{J &pmatrix{ w_1 & 1 & 0 & 0cr 0 & w_1 & 0 & 0cr 0 & 0 & w_2 & 1cr 0 & 0 & 0 & w_2cr} J^{-1} cr = &pmatrix{
          (w_1-w_2)cos(t)^2+w_2 & 1 & -cos(t)sin(t)(w_1-w_2) & 0cr
          0 & (w_1-w_2)cos(t)^2+w_2 & 0 & -cos(t)sin(t)(w_1-w_2)cr
          -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1 & 1cr
          0 & -cos(t)sin(t)(w_1-w_2) & 0 & (-w_1+w_2)cos(t)^2+w_1cr}}
          $$

          where $w_1$ and $w_2$ are any logarithms of $1$ (e.g. $0$ and $2pi i$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 4:36









          Robert IsraelRobert Israel

          323k23213467




          323k23213467























              1












              $begingroup$

              $textbf{Definition.}$ We say that the spectrum of $Ain M_n(mathbb{C})$ is $2ipi$ congruence free (denoted by $2ipi$ CF) iff for every $u,vin spectrum(A)$, $u-vnotin 2ipimathbb{Z}^*$.



              $textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,Bin M_n(mathbb{C})$. If $spectrum(A)$ is $2ipi$ CF and $e^A=e^B$, then



              i) $AB=BA$ and, therefore,



              ii) $A-B$ is similar to $diag(2ipialpha_1,cdots,2ipialpha_n)$ where $alpha_iinmathbb{Z}$.



              $textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)subset 2ipimathbb{Z}$.



              $textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2ipi$ CF and don't pairwise commute.



              ii) $A$ is $2ipi$ CF iff $A$ is a polynomial in $e^A$ iff the function $Xmapsto e^X$ is one to one in a neighborhood of $A$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $textbf{Definition.}$ We say that the spectrum of $Ain M_n(mathbb{C})$ is $2ipi$ congruence free (denoted by $2ipi$ CF) iff for every $u,vin spectrum(A)$, $u-vnotin 2ipimathbb{Z}^*$.



                $textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,Bin M_n(mathbb{C})$. If $spectrum(A)$ is $2ipi$ CF and $e^A=e^B$, then



                i) $AB=BA$ and, therefore,



                ii) $A-B$ is similar to $diag(2ipialpha_1,cdots,2ipialpha_n)$ where $alpha_iinmathbb{Z}$.



                $textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)subset 2ipimathbb{Z}$.



                $textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2ipi$ CF and don't pairwise commute.



                ii) $A$ is $2ipi$ CF iff $A$ is a polynomial in $e^A$ iff the function $Xmapsto e^X$ is one to one in a neighborhood of $A$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $textbf{Definition.}$ We say that the spectrum of $Ain M_n(mathbb{C})$ is $2ipi$ congruence free (denoted by $2ipi$ CF) iff for every $u,vin spectrum(A)$, $u-vnotin 2ipimathbb{Z}^*$.



                  $textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,Bin M_n(mathbb{C})$. If $spectrum(A)$ is $2ipi$ CF and $e^A=e^B$, then



                  i) $AB=BA$ and, therefore,



                  ii) $A-B$ is similar to $diag(2ipialpha_1,cdots,2ipialpha_n)$ where $alpha_iinmathbb{Z}$.



                  $textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)subset 2ipimathbb{Z}$.



                  $textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2ipi$ CF and don't pairwise commute.



                  ii) $A$ is $2ipi$ CF iff $A$ is a polynomial in $e^A$ iff the function $Xmapsto e^X$ is one to one in a neighborhood of $A$.






                  share|cite|improve this answer









                  $endgroup$



                  $textbf{Definition.}$ We say that the spectrum of $Ain M_n(mathbb{C})$ is $2ipi$ congruence free (denoted by $2ipi$ CF) iff for every $u,vin spectrum(A)$, $u-vnotin 2ipimathbb{Z}^*$.



                  $textbf{Theorem}.$ (for i), Hille https://link.springer.com/article/10.1007%2FBF01350286 ). Let $A,Bin M_n(mathbb{C})$. If $spectrum(A)$ is $2ipi$ CF and $e^A=e^B$, then



                  i) $AB=BA$ and, therefore,



                  ii) $A-B$ is similar to $diag(2ipialpha_1,cdots,2ipialpha_n)$ where $alpha_iinmathbb{Z}$.



                  $textbf{Proof of ii)}$. From i), $e^{A-B}=I_n$ is diagonalizable and, therefore, $A-B$ is diagonalizable. Moreover $spectrum(A-B)subset 2ipimathbb{Z}$.



                  $textbf{Remark}.$ i) Note that the matrices of Jyrki Lahtonen are not $2ipi$ CF and don't pairwise commute.



                  ii) $A$ is $2ipi$ CF iff $A$ is a polynomial in $e^A$ iff the function $Xmapsto e^X$ is one to one in a neighborhood of $A$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 22:36









                  loup blancloup blanc

                  23.3k21851




                  23.3k21851






























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