Mixing
Show that $exp(C^{-1} AC ) = C^{-1} exp(A C)$
$begingroup$
Show that $exp(C^{-1} AC) = C^{-1} exp(A C)$ for any matrices $A in L_{n}(mathbb{R})$ and $C in GL_{n}(mathbb{R})$.
The hint of the question is given below:
Consider the linear operator $alpha$ given in some basis (e) by the matrix $A$, and find the matrix of $e^{alpha}$ in the basis (e)$C$ in two ways.
But I just want any type of proof either by calculation or by the hint given above, because I will use this result in solving another problem, just to be convinced with it while using it.
Could anyone help me in doing so please?
Thank you!
linear-algebra matrices representation-theory
edited Jan 20 at 1:49
jordan_glen
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
$endgroup$
add a comment |
$begingroup$
Show that $exp(C^{-1} AC) = C^{-1} exp(A C)$ for any matrices $A in L_{n}(mathbb{R})$ and $C in GL_{n}(mathbb{R})$.
The hint of the question is given below:
Consider the linear operator $alpha$ given in some basis (e) by the matrix $A$, and find the matrix of $e^{alpha}$ in the basis (e)$C$ in two ways.
But I just want any type of proof either by calculation or by the hint given above, because I will use this result in solving another problem, just to be convinced with it while using it.
Could anyone help me in doing so please?
Thank you!
linear-algebra matrices representation-theory
edited Jan 20 at 1:49
jordan_glen
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
$endgroup$
2
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31
add a comment |
$begingroup$
Show that $exp(C^{-1} AC) = C^{-1} exp(A C)$ for any matrices $A in L_{n}(mathbb{R})$ and $C in GL_{n}(mathbb{R})$.
The hint of the question is given below:
Consider the linear operator $alpha$ given in some basis (e) by the matrix $A$, and find the matrix of $e^{alpha}$ in the basis (e)$C$ in two ways.
But I just want any type of proof either by calculation or by the hint given above, because I will use this result in solving another problem, just to be convinced with it while using it.
Could anyone help me in doing so please?
Thank you!
linear-algebra matrices representation-theory
edited Jan 20 at 1:49
jordan_glen
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
$endgroup$
Show that $exp(C^{-1} AC) = C^{-1} exp(A C)$ for any matrices $A in L_{n}(mathbb{R})$ and $C in GL_{n}(mathbb{R})$.
The hint of the question is given below:
Consider the linear operator $alpha$ given in some basis (e) by the matrix $A$, and find the matrix of $e^{alpha}$ in the basis (e)$C$ in two ways.
But I just want any type of proof either by calculation or by the hint given above, because I will use this result in solving another problem, just to be convinced with it while using it.
Could anyone help me in doing so please?
Thank you!
linear-algebra matrices representation-theory
linear-algebra matrices representation-theory
edited Jan 20 at 1:49
jordan_glen
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
edited Jan 20 at 1:49
jordan_glen
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
edited Jan 20 at 1:49
jordan_glen
1
edited Jan 20 at 1:49
jordan_glen
1
edited Jan 20 at 1:49
jordan_glen
1
1
asked Jan 20 at 0:00
hopefullyhopefully
237114
asked Jan 20 at 0:00
hopefullyhopefully
237114
asked Jan 20 at 0:00
hopefullyhopefully
237114
237114
2
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31
add a comment |
2
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31
2
2
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Look at the power series defining $e^B$ for any $B in L_n(Bbb R)$:
$e^B = displaystyle sum_0^infty dfrac{B^n}{n!}; tag 1$
now observe that for any $n ge 0$,
$C^{-1}A^n C = (C^{-1}AC)^n; tag 2$
then
$C^{-1}e^A C = C^{-1}left ( displaystyle sum_0^infty dfrac{A^n}{n!} right )C$
$= displaystyle sum_0^infty dfrac{C^{-1}A^nC}{n!} = displaystyle sum_0^infty dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. tag 3$
That's about the shortest derivation of
$C^{-1}e^A C = e^{C^{-1}AC} tag 4$
I know!
And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
$endgroup$
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at the power series defining $e^B$ for any $B in L_n(Bbb R)$:
$e^B = displaystyle sum_0^infty dfrac{B^n}{n!}; tag 1$
now observe that for any $n ge 0$,
$C^{-1}A^n C = (C^{-1}AC)^n; tag 2$
then
$C^{-1}e^A C = C^{-1}left ( displaystyle sum_0^infty dfrac{A^n}{n!} right )C$
$= displaystyle sum_0^infty dfrac{C^{-1}A^nC}{n!} = displaystyle sum_0^infty dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. tag 3$
That's about the shortest derivation of
$C^{-1}e^A C = e^{C^{-1}AC} tag 4$
I know!
And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
$endgroup$
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
add a comment |
$begingroup$
Look at the power series defining $e^B$ for any $B in L_n(Bbb R)$:
$e^B = displaystyle sum_0^infty dfrac{B^n}{n!}; tag 1$
now observe that for any $n ge 0$,
$C^{-1}A^n C = (C^{-1}AC)^n; tag 2$
then
$C^{-1}e^A C = C^{-1}left ( displaystyle sum_0^infty dfrac{A^n}{n!} right )C$
$= displaystyle sum_0^infty dfrac{C^{-1}A^nC}{n!} = displaystyle sum_0^infty dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. tag 3$
That's about the shortest derivation of
$C^{-1}e^A C = e^{C^{-1}AC} tag 4$
I know!
And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
$endgroup$
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
add a comment |
$begingroup$
Look at the power series defining $e^B$ for any $B in L_n(Bbb R)$:
$e^B = displaystyle sum_0^infty dfrac{B^n}{n!}; tag 1$
now observe that for any $n ge 0$,
$C^{-1}A^n C = (C^{-1}AC)^n; tag 2$
then
$C^{-1}e^A C = C^{-1}left ( displaystyle sum_0^infty dfrac{A^n}{n!} right )C$
$= displaystyle sum_0^infty dfrac{C^{-1}A^nC}{n!} = displaystyle sum_0^infty dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. tag 3$
That's about the shortest derivation of
$C^{-1}e^A C = e^{C^{-1}AC} tag 4$
I know!
And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
$endgroup$
Look at the power series defining $e^B$ for any $B in L_n(Bbb R)$:
$e^B = displaystyle sum_0^infty dfrac{B^n}{n!}; tag 1$
now observe that for any $n ge 0$,
$C^{-1}A^n C = (C^{-1}AC)^n; tag 2$
then
$C^{-1}e^A C = C^{-1}left ( displaystyle sum_0^infty dfrac{A^n}{n!} right )C$
$= displaystyle sum_0^infty dfrac{C^{-1}A^nC}{n!} = displaystyle sum_0^infty dfrac{(C^{-1}AC)^n}{n!} = e^{C^{-1}AC}. tag 3$
That's about the shortest derivation of
$C^{-1}e^A C = e^{C^{-1}AC} tag 4$
I know!
And incidentally, the method works for any matrix analytic function defined by a convergent power series . . .
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
edited Jan 20 at 2:16
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
answered Jan 20 at 0:16
Robert LewisRobert Lewis
47.5k23067
47.5k23067
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
add a comment |
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
2
2
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
$begingroup$
This. I don't get the hint given at all.
$endgroup$
– Git Gud
Jan 20 at 0:17
2
2
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
$begingroup$
@GitGud: you are not alone in that my friend!
$endgroup$
– Robert Lewis
Jan 20 at 0:19
1
1
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
$begingroup$
I think there is a mistake in eq.(2) the r.h.s
$endgroup$
– hopefully
Jan 20 at 2:13
1
1
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
$begingroup$
@hopefully: of course, I switched up $A$ and $C$, will fix now. Thanks!
$endgroup$
– Robert Lewis
Jan 20 at 2:15
1
1
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
$begingroup$
@GitGud Perhaps the hint makes sense if $e^{tA}$ is defined by the differential equation $frac{dy}{dt} = Ay$
$endgroup$
– Omnomnomnom
Jan 20 at 5:04
add a comment |
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2
$begingroup$
do you mean $$e^{C^{-1}AC}$$ on the left hand side? That is equivalent to $exp(C^{-1}AC)$, whereas $exp^{C^{-1}AC}$ is rather meaningless. When using $exp(cdot)$, there needs to be something standing in for $cdot$. In general, $e^{A} = exp(A)$
$endgroup$
– jordan_glen
Jan 20 at 0:06
$begingroup$
yes I mean what you said @jordan_glen
$endgroup$
– hopefully
Jan 20 at 0:31