Mixing
Integral of matrix exponential
$begingroup$
Let $A$ be an $n times n$ matrix. Then the solution of the initial value problem
begin{align*}
dot{x}(t) = A x(t), quad x(0) = x_0
end{align*}
is given by $x(t) = mathrm{e}^{At} x_0$.
I am interested in the following matrix
begin{align*}
int_{0}^T mathrm{e}^{At}, dt
end{align*}
for some $T>0$. Can one write down a general solution to this without distinguishing cases (e.g. $A$ nonsingular)?
Is this matrix always invertible?
calculus linear-algebra matrices exponential-function matrix-calculus
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $n times n$ matrix. Then the solution of the initial value problem
begin{align*}
dot{x}(t) = A x(t), quad x(0) = x_0
end{align*}
is given by $x(t) = mathrm{e}^{At} x_0$.
I am interested in the following matrix
begin{align*}
int_{0}^T mathrm{e}^{At}, dt
end{align*}
for some $T>0$. Can one write down a general solution to this without distinguishing cases (e.g. $A$ nonsingular)?
Is this matrix always invertible?
calculus linear-algebra matrices exponential-function matrix-calculus
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
$endgroup$
1
$begingroup$
you can represent matrix exponential by power series
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
1
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
$endgroup$
– TZakrevskiy
Jan 31 '14 at 10:05
add a comment |
$begingroup$
Let $A$ be an $n times n$ matrix. Then the solution of the initial value problem
begin{align*}
dot{x}(t) = A x(t), quad x(0) = x_0
end{align*}
is given by $x(t) = mathrm{e}^{At} x_0$.
I am interested in the following matrix
begin{align*}
int_{0}^T mathrm{e}^{At}, dt
end{align*}
for some $T>0$. Can one write down a general solution to this without distinguishing cases (e.g. $A$ nonsingular)?
Is this matrix always invertible?
calculus linear-algebra matrices exponential-function matrix-calculus
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
$endgroup$
Let $A$ be an $n times n$ matrix. Then the solution of the initial value problem
begin{align*}
dot{x}(t) = A x(t), quad x(0) = x_0
end{align*}
is given by $x(t) = mathrm{e}^{At} x_0$.
I am interested in the following matrix
begin{align*}
int_{0}^T mathrm{e}^{At}, dt
end{align*}
for some $T>0$. Can one write down a general solution to this without distinguishing cases (e.g. $A$ nonsingular)?
Is this matrix always invertible?
calculus linear-algebra matrices exponential-function matrix-calculus
calculus linear-algebra matrices exponential-function matrix-calculus
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
edited Oct 28 '14 at 22:48
Yiorgos S. Smyrlis
63.7k1385165
63.7k1385165
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
asked Jan 31 '14 at 9:45
samsa44samsa44
166124
166124
1
$begingroup$
you can represent matrix exponential by power series
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
1
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
$endgroup$
– TZakrevskiy
Jan 31 '14 at 10:05
add a comment |
1
$begingroup$
you can represent matrix exponential by power series
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
1
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
$endgroup$
– TZakrevskiy
Jan 31 '14 at 10:05
1
1
$begingroup$
you can represent matrix exponential by power series
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
you can represent matrix exponential by power series
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
1
1
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
$endgroup$
– TZakrevskiy
Jan 31 '14 at 10:05
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
$endgroup$
– TZakrevskiy
Jan 31 '14 at 10:05
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Case I. If $A$ is nonsingular, then
$$
int_0^Tmathrm{e}^{tA},dt=A^{-1}big(mathrm{e}^{TA}-Ibig),
$$
where $I$ is the identity matrix.
Case II. If $A$ is singular, then using the Jordan form we can write $A$ as
$$
A=U^{-1}left(begin{matrix}B&0\0&Cend{matrix}right)U,
$$
where $C$ is nonsingular, and $B$ is strictly upper triangular. Then
$$
mathrm{e}^{tA}=U^{-1}left(begin{matrix}mathrm{e}^{tB}&0\0&mathrm{e}^{tC}
end{matrix}right)U,
$$
and
$$
int_0^Tmathrm{e}^{tA},dt=U^{-1}left(begin{matrix}int_0^Tmathrm{e}^{tB}dt&0\0&C^{-1}big(mathrm{e}^{TC}-Ibig)
end{matrix}right)U
$$
But $int_0^Tmathrm{e}^{tB}dt$ may have different expressions. For example if
$$
B_1=left(begin{matrix}0&0\0&0end{matrix}right), quad
B_2=left(begin{matrix}0&1\0&0end{matrix}right),
$$
then
$$
int_0^Tmathrm{e}^{tB_1}dt=left(begin{matrix}T&0\0&Tend{matrix}right), quad
int_0^Tmathrm{e}^{tB_2}dt=left(begin{matrix}T&T^2/2\0&Tend{matrix}right).
$$
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
Here A is not a constant matrix. It is a variable matrix
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
I am talking about $int e^{A(t)} dt$
$endgroup$
– Nirvana
Oct 6 '14 at 10:33
add a comment |
$begingroup$
The general formula is the power series
$$ int_0^T e^{At} dt = T left( I + frac{AT}{2!} + frac{(AT)^2}{3!} + dots + frac{(AT)^{n-1}}{n!} + dots right) $$
Note that also
$$ left(int_0^T e^{At} dt right) A + I = e^{AT} $$
is always satisfied.
A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $int_0^T e^{At} dt$ is invertible if
$$ T(mu - lambda) neq 2k pi i $$
for any nonzero integer $k$, where $lambda$ and $mu$ are any pair of eigenvalues of $A$.
Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
$endgroup$
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
add a comment |
$begingroup$
A Python numerical answer
It is surprisingly difficult to find a proper python package for numerical integration of matrix. I know it is not what the question want but I cannot find anywhere else to publish this.
Here, I provide a numerical solution to it. Just call the function
intergral_result = compute_exp_matrix_intergration(A,T)
will be enough
import numpy as np
def compute_exp_matrix_intergration(A,T,nbins=100):
f = lambda x: expm(A*x)
xv = np.linspace(0,T,nbins)
result = np.apply_along_axis(f,0,xv.reshape(1,-1))
return np.trapz(result,xv)
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
$endgroup$
$begingroup$
I get a shape error inresult = .... Also, to whoever it may help: theexpmis fromscipy.linalg.
$endgroup$
– anderstood
Jan 29 at 14:13
add a comment |
$begingroup$
Another Python implementation
Here is another implementation in Python, if it can be helpful to anyone... (and since ArtificiallyIntelligence's answer returns an error in my setup).
The value of the integral is integral and the last line verifies the equality $int_0^T e^{At}dt = A^{-1}(e^{tA}-I)$, provided $A$ is nonsingular (which is generically the case for randomly generated matrices).
import numpy as np
N = 5
t = 1
A = np.random.rand(N,N)
taylor = t*np.array([np.linalg.matrix_power(A*t,k)/np.math.factorial(k+1) for k in range(50)])
integral = taylor.sum(axis = 0)
print np.linalg.norm(integral - np.dot(np.linalg.inv(A),scipy.linalg.expm(t*A)-np.identity(N)))
Note that you should adjust the 50 in taylor = ... to check convergence.
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Case I. If $A$ is nonsingular, then
$$
int_0^Tmathrm{e}^{tA},dt=A^{-1}big(mathrm{e}^{TA}-Ibig),
$$
where $I$ is the identity matrix.
Case II. If $A$ is singular, then using the Jordan form we can write $A$ as
$$
A=U^{-1}left(begin{matrix}B&0\0&Cend{matrix}right)U,
$$
where $C$ is nonsingular, and $B$ is strictly upper triangular. Then
$$
mathrm{e}^{tA}=U^{-1}left(begin{matrix}mathrm{e}^{tB}&0\0&mathrm{e}^{tC}
end{matrix}right)U,
$$
and
$$
int_0^Tmathrm{e}^{tA},dt=U^{-1}left(begin{matrix}int_0^Tmathrm{e}^{tB}dt&0\0&C^{-1}big(mathrm{e}^{TC}-Ibig)
end{matrix}right)U
$$
But $int_0^Tmathrm{e}^{tB}dt$ may have different expressions. For example if
$$
B_1=left(begin{matrix}0&0\0&0end{matrix}right), quad
B_2=left(begin{matrix}0&1\0&0end{matrix}right),
$$
then
$$
int_0^Tmathrm{e}^{tB_1}dt=left(begin{matrix}T&0\0&Tend{matrix}right), quad
int_0^Tmathrm{e}^{tB_2}dt=left(begin{matrix}T&T^2/2\0&Tend{matrix}right).
$$
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
Here A is not a constant matrix. It is a variable matrix
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
I am talking about $int e^{A(t)} dt$
$endgroup$
– Nirvana
Oct 6 '14 at 10:33
add a comment |
$begingroup$
Case I. If $A$ is nonsingular, then
$$
int_0^Tmathrm{e}^{tA},dt=A^{-1}big(mathrm{e}^{TA}-Ibig),
$$
where $I$ is the identity matrix.
Case II. If $A$ is singular, then using the Jordan form we can write $A$ as
$$
A=U^{-1}left(begin{matrix}B&0\0&Cend{matrix}right)U,
$$
where $C$ is nonsingular, and $B$ is strictly upper triangular. Then
$$
mathrm{e}^{tA}=U^{-1}left(begin{matrix}mathrm{e}^{tB}&0\0&mathrm{e}^{tC}
end{matrix}right)U,
$$
and
$$
int_0^Tmathrm{e}^{tA},dt=U^{-1}left(begin{matrix}int_0^Tmathrm{e}^{tB}dt&0\0&C^{-1}big(mathrm{e}^{TC}-Ibig)
end{matrix}right)U
$$
But $int_0^Tmathrm{e}^{tB}dt$ may have different expressions. For example if
$$
B_1=left(begin{matrix}0&0\0&0end{matrix}right), quad
B_2=left(begin{matrix}0&1\0&0end{matrix}right),
$$
then
$$
int_0^Tmathrm{e}^{tB_1}dt=left(begin{matrix}T&0\0&Tend{matrix}right), quad
int_0^Tmathrm{e}^{tB_2}dt=left(begin{matrix}T&T^2/2\0&Tend{matrix}right).
$$
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
Here A is not a constant matrix. It is a variable matrix
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
I am talking about $int e^{A(t)} dt$
$endgroup$
– Nirvana
Oct 6 '14 at 10:33
add a comment |
$begingroup$
Case I. If $A$ is nonsingular, then
$$
int_0^Tmathrm{e}^{tA},dt=A^{-1}big(mathrm{e}^{TA}-Ibig),
$$
where $I$ is the identity matrix.
Case II. If $A$ is singular, then using the Jordan form we can write $A$ as
$$
A=U^{-1}left(begin{matrix}B&0\0&Cend{matrix}right)U,
$$
where $C$ is nonsingular, and $B$ is strictly upper triangular. Then
$$
mathrm{e}^{tA}=U^{-1}left(begin{matrix}mathrm{e}^{tB}&0\0&mathrm{e}^{tC}
end{matrix}right)U,
$$
and
$$
int_0^Tmathrm{e}^{tA},dt=U^{-1}left(begin{matrix}int_0^Tmathrm{e}^{tB}dt&0\0&C^{-1}big(mathrm{e}^{TC}-Ibig)
end{matrix}right)U
$$
But $int_0^Tmathrm{e}^{tB}dt$ may have different expressions. For example if
$$
B_1=left(begin{matrix}0&0\0&0end{matrix}right), quad
B_2=left(begin{matrix}0&1\0&0end{matrix}right),
$$
then
$$
int_0^Tmathrm{e}^{tB_1}dt=left(begin{matrix}T&0\0&Tend{matrix}right), quad
int_0^Tmathrm{e}^{tB_2}dt=left(begin{matrix}T&T^2/2\0&Tend{matrix}right).
$$
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
$endgroup$
Case I. If $A$ is nonsingular, then
$$
int_0^Tmathrm{e}^{tA},dt=A^{-1}big(mathrm{e}^{TA}-Ibig),
$$
where $I$ is the identity matrix.
Case II. If $A$ is singular, then using the Jordan form we can write $A$ as
$$
A=U^{-1}left(begin{matrix}B&0\0&Cend{matrix}right)U,
$$
where $C$ is nonsingular, and $B$ is strictly upper triangular. Then
$$
mathrm{e}^{tA}=U^{-1}left(begin{matrix}mathrm{e}^{tB}&0\0&mathrm{e}^{tC}
end{matrix}right)U,
$$
and
$$
int_0^Tmathrm{e}^{tA},dt=U^{-1}left(begin{matrix}int_0^Tmathrm{e}^{tB}dt&0\0&C^{-1}big(mathrm{e}^{TC}-Ibig)
end{matrix}right)U
$$
But $int_0^Tmathrm{e}^{tB}dt$ may have different expressions. For example if
$$
B_1=left(begin{matrix}0&0\0&0end{matrix}right), quad
B_2=left(begin{matrix}0&1\0&0end{matrix}right),
$$
then
$$
int_0^Tmathrm{e}^{tB_1}dt=left(begin{matrix}T&0\0&Tend{matrix}right), quad
int_0^Tmathrm{e}^{tB_2}dt=left(begin{matrix}T&T^2/2\0&Tend{matrix}right).
$$
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
edited Jan 15 '17 at 14:57
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
answered Jan 31 '14 at 10:06
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.7k1385165
63.7k1385165
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
Here A is not a constant matrix. It is a variable matrix
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
I am talking about $int e^{A(t)} dt$
$endgroup$
– Nirvana
Oct 6 '14 at 10:33
add a comment |
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
Here A is not a constant matrix. It is a variable matrix
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
$begingroup$
I am talking about $int e^{A(t)} dt$
$endgroup$
– Nirvana
Oct 6 '14 at 10:33
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
What we will do if we use $A(t)$ instead of $tA$
$endgroup$
– Nirvana
Oct 6 '14 at 8:06
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
There is nothing wrong with that.
$endgroup$
– Yiorgos S. Smyrlis
Oct 6 '14 at 10:22
$begingroup$
@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
$endgroup$
– Nirvana
Oct 6 '14 at 10:28
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@ Yiorgos S. Smyrlis So you are saying even if $A(t)$ and $A^{'}(t)$ are not commutative,this will work
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– Nirvana
Oct 6 '14 at 10:28
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Here A is not a constant matrix. It is a variable matrix
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– Nirvana
Oct 6 '14 at 10:28
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Here A is not a constant matrix. It is a variable matrix
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– Nirvana
Oct 6 '14 at 10:28
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I am talking about $int e^{A(t)} dt$
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– Nirvana
Oct 6 '14 at 10:33
$begingroup$
I am talking about $int e^{A(t)} dt$
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– Nirvana
Oct 6 '14 at 10:33
add a comment |
$begingroup$
The general formula is the power series
$$ int_0^T e^{At} dt = T left( I + frac{AT}{2!} + frac{(AT)^2}{3!} + dots + frac{(AT)^{n-1}}{n!} + dots right) $$
Note that also
$$ left(int_0^T e^{At} dt right) A + I = e^{AT} $$
is always satisfied.
A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $int_0^T e^{At} dt$ is invertible if
$$ T(mu - lambda) neq 2k pi i $$
for any nonzero integer $k$, where $lambda$ and $mu$ are any pair of eigenvalues of $A$.
Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
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$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
add a comment |
$begingroup$
The general formula is the power series
$$ int_0^T e^{At} dt = T left( I + frac{AT}{2!} + frac{(AT)^2}{3!} + dots + frac{(AT)^{n-1}}{n!} + dots right) $$
Note that also
$$ left(int_0^T e^{At} dt right) A + I = e^{AT} $$
is always satisfied.
A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $int_0^T e^{At} dt$ is invertible if
$$ T(mu - lambda) neq 2k pi i $$
for any nonzero integer $k$, where $lambda$ and $mu$ are any pair of eigenvalues of $A$.
Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
$endgroup$
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
add a comment |
$begingroup$
The general formula is the power series
$$ int_0^T e^{At} dt = T left( I + frac{AT}{2!} + frac{(AT)^2}{3!} + dots + frac{(AT)^{n-1}}{n!} + dots right) $$
Note that also
$$ left(int_0^T e^{At} dt right) A + I = e^{AT} $$
is always satisfied.
A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $int_0^T e^{At} dt$ is invertible if
$$ T(mu - lambda) neq 2k pi i $$
for any nonzero integer $k$, where $lambda$ and $mu$ are any pair of eigenvalues of $A$.
Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
$endgroup$
The general formula is the power series
$$ int_0^T e^{At} dt = T left( I + frac{AT}{2!} + frac{(AT)^2}{3!} + dots + frac{(AT)^{n-1}}{n!} + dots right) $$
Note that also
$$ left(int_0^T e^{At} dt right) A + I = e^{AT} $$
is always satisfied.
A sufficient condition for this matrix to be non-singular is the so-called Kalman-Ho-Narendra Theorem, which states that the matrix $int_0^T e^{At} dt$ is invertible if
$$ T(mu - lambda) neq 2k pi i $$
for any nonzero integer $k$, where $lambda$ and $mu$ are any pair of eigenvalues of $A$.
Note to the interested: This matrix also comes from the discretization of a continuous linear time invariant system. It can also be said that controllability is preserved under discretization if and only if this matrix has an inverse.
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
answered Jan 31 '14 at 18:40
obareeyobareey
3,06911128
3,06911128
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
add a comment |
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
$begingroup$
Thank you. This answer looks very promising. Btw, I am looking at this matrix exactly because of what you said about the discretization!
$endgroup$
– samsa44
Feb 24 '14 at 11:19
add a comment |
$begingroup$
A Python numerical answer
It is surprisingly difficult to find a proper python package for numerical integration of matrix. I know it is not what the question want but I cannot find anywhere else to publish this.
Here, I provide a numerical solution to it. Just call the function
intergral_result = compute_exp_matrix_intergration(A,T)
will be enough
import numpy as np
def compute_exp_matrix_intergration(A,T,nbins=100):
f = lambda x: expm(A*x)
xv = np.linspace(0,T,nbins)
result = np.apply_along_axis(f,0,xv.reshape(1,-1))
return np.trapz(result,xv)
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
$endgroup$
$begingroup$
I get a shape error inresult = .... Also, to whoever it may help: theexpmis fromscipy.linalg.
$endgroup$
– anderstood
Jan 29 at 14:13
add a comment |
$begingroup$
A Python numerical answer
It is surprisingly difficult to find a proper python package for numerical integration of matrix. I know it is not what the question want but I cannot find anywhere else to publish this.
Here, I provide a numerical solution to it. Just call the function
intergral_result = compute_exp_matrix_intergration(A,T)
will be enough
import numpy as np
def compute_exp_matrix_intergration(A,T,nbins=100):
f = lambda x: expm(A*x)
xv = np.linspace(0,T,nbins)
result = np.apply_along_axis(f,0,xv.reshape(1,-1))
return np.trapz(result,xv)
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
$endgroup$
$begingroup$
I get a shape error inresult = .... Also, to whoever it may help: theexpmis fromscipy.linalg.
$endgroup$
– anderstood
Jan 29 at 14:13
add a comment |
$begingroup$
A Python numerical answer
It is surprisingly difficult to find a proper python package for numerical integration of matrix. I know it is not what the question want but I cannot find anywhere else to publish this.
Here, I provide a numerical solution to it. Just call the function
intergral_result = compute_exp_matrix_intergration(A,T)
will be enough
import numpy as np
def compute_exp_matrix_intergration(A,T,nbins=100):
f = lambda x: expm(A*x)
xv = np.linspace(0,T,nbins)
result = np.apply_along_axis(f,0,xv.reshape(1,-1))
return np.trapz(result,xv)
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
$endgroup$
A Python numerical answer
It is surprisingly difficult to find a proper python package for numerical integration of matrix. I know it is not what the question want but I cannot find anywhere else to publish this.
Here, I provide a numerical solution to it. Just call the function
intergral_result = compute_exp_matrix_intergration(A,T)
will be enough
import numpy as np
def compute_exp_matrix_intergration(A,T,nbins=100):
f = lambda x: expm(A*x)
xv = np.linspace(0,T,nbins)
result = np.apply_along_axis(f,0,xv.reshape(1,-1))
return np.trapz(result,xv)
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
answered Jun 3 '18 at 21:13
ArtificiallyIntelligenceArtificiallyIntelligence
310112
310112
$begingroup$
I get a shape error inresult = .... Also, to whoever it may help: theexpmis fromscipy.linalg.
$endgroup$
– anderstood
Jan 29 at 14:13
add a comment |
$begingroup$
I get a shape error inresult = .... Also, to whoever it may help: theexpmis fromscipy.linalg.
$endgroup$
– anderstood
Jan 29 at 14:13
$begingroup$
I get a shape error in
result = ... . Also, to whoever it may help: the expm is from scipy.linalg.$endgroup$
– anderstood
Jan 29 at 14:13
$begingroup$
I get a shape error in
result = ... . Also, to whoever it may help: the expm is from scipy.linalg.$endgroup$
– anderstood
Jan 29 at 14:13
add a comment |
$begingroup$
Another Python implementation
Here is another implementation in Python, if it can be helpful to anyone... (and since ArtificiallyIntelligence's answer returns an error in my setup).
The value of the integral is integral and the last line verifies the equality $int_0^T e^{At}dt = A^{-1}(e^{tA}-I)$, provided $A$ is nonsingular (which is generically the case for randomly generated matrices).
import numpy as np
N = 5
t = 1
A = np.random.rand(N,N)
taylor = t*np.array([np.linalg.matrix_power(A*t,k)/np.math.factorial(k+1) for k in range(50)])
integral = taylor.sum(axis = 0)
print np.linalg.norm(integral - np.dot(np.linalg.inv(A),scipy.linalg.expm(t*A)-np.identity(N)))
Note that you should adjust the 50 in taylor = ... to check convergence.
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
$begingroup$
Another Python implementation
Here is another implementation in Python, if it can be helpful to anyone... (and since ArtificiallyIntelligence's answer returns an error in my setup).
The value of the integral is integral and the last line verifies the equality $int_0^T e^{At}dt = A^{-1}(e^{tA}-I)$, provided $A$ is nonsingular (which is generically the case for randomly generated matrices).
import numpy as np
N = 5
t = 1
A = np.random.rand(N,N)
taylor = t*np.array([np.linalg.matrix_power(A*t,k)/np.math.factorial(k+1) for k in range(50)])
integral = taylor.sum(axis = 0)
print np.linalg.norm(integral - np.dot(np.linalg.inv(A),scipy.linalg.expm(t*A)-np.identity(N)))
Note that you should adjust the 50 in taylor = ... to check convergence.
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
$endgroup$
add a comment |
$begingroup$
Another Python implementation
Here is another implementation in Python, if it can be helpful to anyone... (and since ArtificiallyIntelligence's answer returns an error in my setup).
The value of the integral is integral and the last line verifies the equality $int_0^T e^{At}dt = A^{-1}(e^{tA}-I)$, provided $A$ is nonsingular (which is generically the case for randomly generated matrices).
import numpy as np
N = 5
t = 1
A = np.random.rand(N,N)
taylor = t*np.array([np.linalg.matrix_power(A*t,k)/np.math.factorial(k+1) for k in range(50)])
integral = taylor.sum(axis = 0)
print np.linalg.norm(integral - np.dot(np.linalg.inv(A),scipy.linalg.expm(t*A)-np.identity(N)))
Note that you should adjust the 50 in taylor = ... to check convergence.
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
$endgroup$
Another Python implementation
Here is another implementation in Python, if it can be helpful to anyone... (and since ArtificiallyIntelligence's answer returns an error in my setup).
The value of the integral is integral and the last line verifies the equality $int_0^T e^{At}dt = A^{-1}(e^{tA}-I)$, provided $A$ is nonsingular (which is generically the case for randomly generated matrices).
import numpy as np
N = 5
t = 1
A = np.random.rand(N,N)
taylor = t*np.array([np.linalg.matrix_power(A*t,k)/np.math.factorial(k+1) for k in range(50)])
integral = taylor.sum(axis = 0)
print np.linalg.norm(integral - np.dot(np.linalg.inv(A),scipy.linalg.expm(t*A)-np.identity(N)))
Note that you should adjust the 50 in taylor = ... to check convergence.
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
answered Jan 30 at 9:56
anderstoodanderstood
2,1791030
2,1791030
add a comment |
add a comment |
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1
$begingroup$
you can represent matrix exponential by power series
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– dato datuashvili
Jan 31 '14 at 9:52
1
$begingroup$
en.wikipedia.org/wiki/Matrix_exponential
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– dato datuashvili
Jan 31 '14 at 9:52
$begingroup$
Spectral theorem says that if you take analytical function $f$, apply it to the matrix $A$ (its eigenvaules has to be in the domain of analaticity of $f$), then eigenvalues of $f(A)$ are $f(text{eigenvalues of } A)$. Exponent is never zero, hence $exp(A)$ is always invertible, moreover, $exp(A)^{-1}=exp(-A)$.
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– TZakrevskiy
Jan 31 '14 at 10:05