When $e^A = e^B$ for matrices $A,B$?
$begingroup$
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 $B$?
matrices matrix-calculus matrix-exponential
$endgroup$
add a comment |
$begingroup$
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 $B$?
matrices matrix-calculus matrix-exponential
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $B$?
matrices matrix-calculus matrix-exponential
$endgroup$
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
matrices matrix-calculus matrix-exponential
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $2pi i$).
$endgroup$
add a comment |
$begingroup$
$textbf{Definition.}$ 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 $A$.
$endgroup$
add a comment |
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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 $2pi i$).
$endgroup$
add a comment |
$begingroup$
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 $2pi i$).
$endgroup$
add a comment |
$begingroup$
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 $2pi i$).
$endgroup$
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 $2pi i$).
answered Dec 19 '18 at 4:36
Robert IsraelRobert Israel
323k23213467
323k23213467
add a comment |
add a comment |
$begingroup$
$textbf{Definition.}$ 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 $A$.
$endgroup$
add a comment |
$begingroup$
$textbf{Definition.}$ 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 $A$.
$endgroup$
add a comment |
$begingroup$
$textbf{Definition.}$ 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 $A$.
$endgroup$
$textbf{Definition.}$ 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 $A$.
answered Jan 18 at 22:36
loup blancloup blanc
23.3k21851
23.3k21851
add a comment |
add a comment |
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$begingroup$
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