Suppose every eigenvalue $lambda$ of $X$ satisfies $|lambda|<2pi$. If $exp(X)v=v$ for some $vin mathbb...
$begingroup$
Let $Xin M_n(mathbb C)$. Suppose every eigenvalue $lambda$ of $X$ satisfies $|lambda|<2pi$. If $e^Xv=v$ for some $vin mathbb C^n$ then $Xv=0$.
I know that if I suppose $e^Xv=v$, then 1 is an eigenvalue of $e^X$, so then $log e^X=X$ has eigenvalue $log 1=0$. So $Xv=0$. However, I know there are some requirements to be able to use the matrix logarithm, and that I have not used all my hypothesis, so this is suspect, but I am not sure what else to do.
I have the hint to use the series of $frac{e^x-1}{x}$ but I'm not sure how to fit this in either.
matrices
asked Jan 23 at 20:07
AlexAlex
36429
$endgroup$
add a comment |
$begingroup$
Let $Xin M_n(mathbb C)$. Suppose every eigenvalue $lambda$ of $X$ satisfies $|lambda|<2pi$. If $e^Xv=v$ for some $vin mathbb C^n$ then $Xv=0$.
I know that if I suppose $e^Xv=v$, then 1 is an eigenvalue of $e^X$, so then $log e^X=X$ has eigenvalue $log 1=0$. So $Xv=0$. However, I know there are some requirements to be able to use the matrix logarithm, and that I have not used all my hypothesis, so this is suspect, but I am not sure what else to do.
I have the hint to use the series of $frac{e^x-1}{x}$ but I'm not sure how to fit this in either.
matrices
asked Jan 23 at 20:07
AlexAlex
36429
$endgroup$
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
$endgroup$
– Omnomnomnom
Jan 24 at 5:23
add a comment |
$begingroup$
Let $Xin M_n(mathbb C)$. Suppose every eigenvalue $lambda$ of $X$ satisfies $|lambda|<2pi$. If $e^Xv=v$ for some $vin mathbb C^n$ then $Xv=0$.
I know that if I suppose $e^Xv=v$, then 1 is an eigenvalue of $e^X$, so then $log e^X=X$ has eigenvalue $log 1=0$. So $Xv=0$. However, I know there are some requirements to be able to use the matrix logarithm, and that I have not used all my hypothesis, so this is suspect, but I am not sure what else to do.
I have the hint to use the series of $frac{e^x-1}{x}$ but I'm not sure how to fit this in either.
matrices
asked Jan 23 at 20:07
AlexAlex
36429
$endgroup$
Let $Xin M_n(mathbb C)$. Suppose every eigenvalue $lambda$ of $X$ satisfies $|lambda|<2pi$. If $e^Xv=v$ for some $vin mathbb C^n$ then $Xv=0$.
I know that if I suppose $e^Xv=v$, then 1 is an eigenvalue of $e^X$, so then $log e^X=X$ has eigenvalue $log 1=0$. So $Xv=0$. However, I know there are some requirements to be able to use the matrix logarithm, and that I have not used all my hypothesis, so this is suspect, but I am not sure what else to do.
I have the hint to use the series of $frac{e^x-1}{x}$ but I'm not sure how to fit this in either.
matrices
matrices
asked Jan 23 at 20:07
AlexAlex
36429
asked Jan 23 at 20:07
AlexAlex
36429
asked Jan 23 at 20:07
AlexAlex
36429
asked Jan 23 at 20:07
AlexAlex
36429
asked Jan 23 at 20:07
AlexAlex
36429
36429
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
$endgroup$
– Omnomnomnom
Jan 24 at 5:23
add a comment |
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
$endgroup$
– Omnomnomnom
Jan 24 at 5:23
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
$endgroup$
– Omnomnomnom
Jan 24 at 5:23
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
$endgroup$
– Omnomnomnom
Jan 24 at 5:23
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k ge 1$.
Suppose $J_{lambda,m}$ is a Jordan block of size $m$ with $lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{lambda,m}}-I = begin{bmatrix} e^lambda-1 & {1 over 2!} e^lambda &
{1 over 3!}e^lambda & cdots & {1 over (m-1)!} e^lambda \
0 & e^lambda-1 & {1 over 2!} e^lambda & cdots & {1 over (m-2)!} e^lambda \
vdots & vdots & ddots & & vdots\
0 & 0 & 0 & cdots & e^lambda-1
end{bmatrix}$.
From this we see that if $v neq 0$ and $(e^{J_{lambda,m}} -I)v = 0$ then $e^lambda =1$
and $v=e_k$.
Since $e^lambda =1 $, we must have $lambda in 2 pi i mathbb{Z}$ and so, by assumption, $lambda = 0$. Hence $J_{lambda,m} v = 0$. Hence if $(e^{J_{lambda,m}} -I)v = 0$ then $J_{lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = operatorname{diag}(e^{J_{lambda_1,m_1}},...,e^{J_{lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Suppose $Av = v$, we have
$$e^{2pi iA}v = e^{2pi i}v = v$$
Let $X = 2 pi inA, (n in mathbb Z)$, then
$$
e^{X}v = e^{2pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2pi inAv = 2pi i n v
$$
$X$ also have an eigenvalue of $lambda = 2pi in$ where $lvertlambdarvert = 2pi n$
From this example we can see that:
$$
e^Xv = lambda v Rightarrow Xv = (log lambda + 2 pi i n)v (n in mathbb Z)
$$
This is why we need $lvertlambdarvert < 2pi$
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+cdots)=e^D(I+M)$ where $M$ is nilpotent, $ker(M)=ker(N)$ and $MD=DM$.
$e^Xv=v$ $implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$.
We may sassume $D=diag(lambda_i)$ where $|lambda_i|<2pi$; then
$e^Dv=v$ IFF for every $i$, $e^{lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{lambda_i}=1$, that is $lambda_i=0$.
Thus, for every $i$, $lambda_iv_i=0$; therefore $Dv=0$ and we are done.
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
$endgroup$
add a comment |
Your Answer
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
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active
oldest
votes
$begingroup$
Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k ge 1$.
Suppose $J_{lambda,m}$ is a Jordan block of size $m$ with $lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{lambda,m}}-I = begin{bmatrix} e^lambda-1 & {1 over 2!} e^lambda &
{1 over 3!}e^lambda & cdots & {1 over (m-1)!} e^lambda \
0 & e^lambda-1 & {1 over 2!} e^lambda & cdots & {1 over (m-2)!} e^lambda \
vdots & vdots & ddots & & vdots\
0 & 0 & 0 & cdots & e^lambda-1
end{bmatrix}$.
From this we see that if $v neq 0$ and $(e^{J_{lambda,m}} -I)v = 0$ then $e^lambda =1$
and $v=e_k$.
Since $e^lambda =1 $, we must have $lambda in 2 pi i mathbb{Z}$ and so, by assumption, $lambda = 0$. Hence $J_{lambda,m} v = 0$. Hence if $(e^{J_{lambda,m}} -I)v = 0$ then $J_{lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = operatorname{diag}(e^{J_{lambda_1,m_1}},...,e^{J_{lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k ge 1$.
Suppose $J_{lambda,m}$ is a Jordan block of size $m$ with $lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{lambda,m}}-I = begin{bmatrix} e^lambda-1 & {1 over 2!} e^lambda &
{1 over 3!}e^lambda & cdots & {1 over (m-1)!} e^lambda \
0 & e^lambda-1 & {1 over 2!} e^lambda & cdots & {1 over (m-2)!} e^lambda \
vdots & vdots & ddots & & vdots\
0 & 0 & 0 & cdots & e^lambda-1
end{bmatrix}$.
From this we see that if $v neq 0$ and $(e^{J_{lambda,m}} -I)v = 0$ then $e^lambda =1$
and $v=e_k$.
Since $e^lambda =1 $, we must have $lambda in 2 pi i mathbb{Z}$ and so, by assumption, $lambda = 0$. Hence $J_{lambda,m} v = 0$. Hence if $(e^{J_{lambda,m}} -I)v = 0$ then $J_{lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = operatorname{diag}(e^{J_{lambda_1,m_1}},...,e^{J_{lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k ge 1$.
Suppose $J_{lambda,m}$ is a Jordan block of size $m$ with $lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{lambda,m}}-I = begin{bmatrix} e^lambda-1 & {1 over 2!} e^lambda &
{1 over 3!}e^lambda & cdots & {1 over (m-1)!} e^lambda \
0 & e^lambda-1 & {1 over 2!} e^lambda & cdots & {1 over (m-2)!} e^lambda \
vdots & vdots & ddots & & vdots\
0 & 0 & 0 & cdots & e^lambda-1
end{bmatrix}$.
From this we see that if $v neq 0$ and $(e^{J_{lambda,m}} -I)v = 0$ then $e^lambda =1$
and $v=e_k$.
Since $e^lambda =1 $, we must have $lambda in 2 pi i mathbb{Z}$ and so, by assumption, $lambda = 0$. Hence $J_{lambda,m} v = 0$. Hence if $(e^{J_{lambda,m}} -I)v = 0$ then $J_{lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = operatorname{diag}(e^{J_{lambda_1,m_1}},...,e^{J_{lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
$endgroup$
Here is a slightly unsatisfactory answer.
Let $f(x) = e^x-1$. Note that $f^{(k)}(x) = e^x$ for $k ge 1$.
Suppose $J_{lambda,m}$ is a Jordan block of size $m$ with $lambda$ on the diagonal.
Then (see, for example, https://en.wikipedia.org/wiki/Jordan_matrix#Functions_of_matrices),
$e^{J_{lambda,m}}-I = begin{bmatrix} e^lambda-1 & {1 over 2!} e^lambda &
{1 over 3!}e^lambda & cdots & {1 over (m-1)!} e^lambda \
0 & e^lambda-1 & {1 over 2!} e^lambda & cdots & {1 over (m-2)!} e^lambda \
vdots & vdots & ddots & & vdots\
0 & 0 & 0 & cdots & e^lambda-1
end{bmatrix}$.
From this we see that if $v neq 0$ and $(e^{J_{lambda,m}} -I)v = 0$ then $e^lambda =1$
and $v=e_k$.
Since $e^lambda =1 $, we must have $lambda in 2 pi i mathbb{Z}$ and so, by assumption, $lambda = 0$. Hence $J_{lambda,m} v = 0$. Hence if $(e^{J_{lambda,m}} -I)v = 0$ then $J_{lambda,m} v = 0$.
Let $J$ be a Jordan normal form of $X$. Then $e^J = operatorname{diag}(e^{J_{lambda_1,m_1}},...,e^{J_{lambda_k,m_k}})$ and from the above it follows
that if $(e^J-I)v=0$ then $Jv = 0$.
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
answered Jan 24 at 0:35
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
$begingroup$
Suppose $Av = v$, we have
$$e^{2pi iA}v = e^{2pi i}v = v$$
Let $X = 2 pi inA, (n in mathbb Z)$, then
$$
e^{X}v = e^{2pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2pi inAv = 2pi i n v
$$
$X$ also have an eigenvalue of $lambda = 2pi in$ where $lvertlambdarvert = 2pi n$
From this example we can see that:
$$
e^Xv = lambda v Rightarrow Xv = (log lambda + 2 pi i n)v (n in mathbb Z)
$$
This is why we need $lvertlambdarvert < 2pi$
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Suppose $Av = v$, we have
$$e^{2pi iA}v = e^{2pi i}v = v$$
Let $X = 2 pi inA, (n in mathbb Z)$, then
$$
e^{X}v = e^{2pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2pi inAv = 2pi i n v
$$
$X$ also have an eigenvalue of $lambda = 2pi in$ where $lvertlambdarvert = 2pi n$
From this example we can see that:
$$
e^Xv = lambda v Rightarrow Xv = (log lambda + 2 pi i n)v (n in mathbb Z)
$$
This is why we need $lvertlambdarvert < 2pi$
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
$endgroup$
add a comment |
$begingroup$
Suppose $Av = v$, we have
$$e^{2pi iA}v = e^{2pi i}v = v$$
Let $X = 2 pi inA, (n in mathbb Z)$, then
$$
e^{X}v = e^{2pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2pi inAv = 2pi i n v
$$
$X$ also have an eigenvalue of $lambda = 2pi in$ where $lvertlambdarvert = 2pi n$
From this example we can see that:
$$
e^Xv = lambda v Rightarrow Xv = (log lambda + 2 pi i n)v (n in mathbb Z)
$$
This is why we need $lvertlambdarvert < 2pi$
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
$endgroup$
Suppose $Av = v$, we have
$$e^{2pi iA}v = e^{2pi i}v = v$$
Let $X = 2 pi inA, (n in mathbb Z)$, then
$$
e^{X}v = e^{2pi i n A}v = v
$$
Which fit the requirements but
$$
Xv = 2pi inAv = 2pi i n v
$$
$X$ also have an eigenvalue of $lambda = 2pi in$ where $lvertlambdarvert = 2pi n$
From this example we can see that:
$$
e^Xv = lambda v Rightarrow Xv = (log lambda + 2 pi i n)v (n in mathbb Z)
$$
This is why we need $lvertlambdarvert < 2pi$
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
edited Jan 24 at 5:20
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
answered Jan 24 at 5:13
Zang MingJieZang MingJie
1356
1356
add a comment |
add a comment |
$begingroup$
Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+cdots)=e^D(I+M)$ where $M$ is nilpotent, $ker(M)=ker(N)$ and $MD=DM$.
$e^Xv=v$ $implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$.
We may sassume $D=diag(lambda_i)$ where $|lambda_i|<2pi$; then
$e^Dv=v$ IFF for every $i$, $e^{lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{lambda_i}=1$, that is $lambda_i=0$.
Thus, for every $i$, $lambda_iv_i=0$; therefore $Dv=0$ and we are done.
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
$endgroup$
add a comment |
$begingroup$
Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+cdots)=e^D(I+M)$ where $M$ is nilpotent, $ker(M)=ker(N)$ and $MD=DM$.
$e^Xv=v$ $implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$.
We may sassume $D=diag(lambda_i)$ where $|lambda_i|<2pi$; then
$e^Dv=v$ IFF for every $i$, $e^{lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{lambda_i}=1$, that is $lambda_i=0$.
Thus, for every $i$, $lambda_iv_i=0$; therefore $Dv=0$ and we are done.
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
$endgroup$
add a comment |
$begingroup$
Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+cdots)=e^D(I+M)$ where $M$ is nilpotent, $ker(M)=ker(N)$ and $MD=DM$.
$e^Xv=v$ $implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$.
We may sassume $D=diag(lambda_i)$ where $|lambda_i|<2pi$; then
$e^Dv=v$ IFF for every $i$, $e^{lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{lambda_i}=1$, that is $lambda_i=0$.
Thus, for every $i$, $lambda_iv_i=0$; therefore $Dv=0$ and we are done.
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
$endgroup$
Let $X=D+N$ where $D$ is diagonalizable, $N$ is nilpotent and $ND=DN$. Then $e^X=e^De^N=e^D(I+N+cdots)=e^D(I+M)$ where $M$ is nilpotent, $ker(M)=ker(N)$ and $MD=DM$.
$e^Xv=v$ $implies$ $e^Dv=v$, and $e^DMv=0$, that implies $Nv=0$ and finally $Xv=Dv$.
We may sassume $D=diag(lambda_i)$ where $|lambda_i|<2pi$; then
$e^Dv=v$ IFF for every $i$, $e^{lambda_i}v_i=v_i$ IFF for every $i$, $v_i=0$ or $e^{lambda_i}=1$, that is $lambda_i=0$.
Thus, for every $i$, $lambda_iv_i=0$; therefore $Dv=0$ and we are done.
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
answered Jan 26 at 23:05
loup blancloup blanc
23.5k21851
23.5k21851
add a comment |
add a comment |
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Required, but never shown
$begingroup$
Perhaps it's useful to note that if $e^X v = v$, then $$ (e^X - I)v = 0 $$ I still don't see how that series fits in though. Perhaps we could deduce that $frac{e^X - 1}{X} v = 0$.
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– Omnomnomnom
Jan 24 at 5:23