Evaluating matrix integral

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Context: This integral is part of the calculation of the step response to a linear system using the convolution equation, found in this book (pg. 150).



The part that is bugging me is the following integral:



$$
int_0^t e^A(t-tau) d tau,
$$
where $A$ is a matrix and $e^A(t-tau)$ is the matrix exponential.



My approach was:



$$
int_0^t e^A(t-tau) dtau = int_0^t e^Ate^-Atau dtau = e^Atint_0^t e^-Atau dtau = -e^AtA^-1(e^-At - I),
$$



where I used the property that $e^A+B = e^Ae^B$ when $AB=BA$.



However, the author does the following:
$$
int_0^t e^A(t-tau) d tau = int_0^t e^Asigma dsigma = A^-1(e^At - I).
$$



I do not understand why my approach is wrong and why the author is allowed to do what he did.




linear-algebra integration dynamical-systems matrix-calculus convolution




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  • $t$ and $tau$ are scalars.
    – pedroszattoni
    Jul 31 at 17:52














up vote
1
down vote

favorite
1












Context: This integral is part of the calculation of the step response to a linear system using the convolution equation, found in this book (pg. 150).



The part that is bugging me is the following integral:



$$
int_0^t e^A(t-tau) d tau,
$$
where $A$ is a matrix and $e^A(t-tau)$ is the matrix exponential.



My approach was:



$$
int_0^t e^A(t-tau) dtau = int_0^t e^Ate^-Atau dtau = e^Atint_0^t e^-Atau dtau = -e^AtA^-1(e^-At - I),
$$



where I used the property that $e^A+B = e^Ae^B$ when $AB=BA$.



However, the author does the following:
$$
int_0^t e^A(t-tau) d tau = int_0^t e^Asigma dsigma = A^-1(e^At - I).
$$



I do not understand why my approach is wrong and why the author is allowed to do what he did.




linear-algebra integration dynamical-systems matrix-calculus convolution




share|cite|improve this question





















  • $t$ and $tau$ are scalars.
    – pedroszattoni
    Jul 31 at 17:52












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Context: This integral is part of the calculation of the step response to a linear system using the convolution equation, found in this book (pg. 150).



The part that is bugging me is the following integral:



$$
int_0^t e^A(t-tau) d tau,
$$
where $A$ is a matrix and $e^A(t-tau)$ is the matrix exponential.



My approach was:



$$
int_0^t e^A(t-tau) dtau = int_0^t e^Ate^-Atau dtau = e^Atint_0^t e^-Atau dtau = -e^AtA^-1(e^-At - I),
$$



where I used the property that $e^A+B = e^Ae^B$ when $AB=BA$.



However, the author does the following:
$$
int_0^t e^A(t-tau) d tau = int_0^t e^Asigma dsigma = A^-1(e^At - I).
$$



I do not understand why my approach is wrong and why the author is allowed to do what he did.




linear-algebra integration dynamical-systems matrix-calculus convolution




share|cite|improve this question













Context: This integral is part of the calculation of the step response to a linear system using the convolution equation, found in this book (pg. 150).



The part that is bugging me is the following integral:



$$
int_0^t e^A(t-tau) d tau,
$$
where $A$ is a matrix and $e^A(t-tau)$ is the matrix exponential.



My approach was:



$$
int_0^t e^A(t-tau) dtau = int_0^t e^Ate^-Atau dtau = e^Atint_0^t e^-Atau dtau = -e^AtA^-1(e^-At - I),
$$



where I used the property that $e^A+B = e^Ae^B$ when $AB=BA$.



However, the author does the following:
$$
int_0^t e^A(t-tau) d tau = int_0^t e^Asigma dsigma = A^-1(e^At - I).
$$



I do not understand why my approach is wrong and why the author is allowed to do what he did.





linear-algebra integration dynamical-systems matrix-calculus convolution





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edited Jul 31 at 17:59
























asked Jul 31 at 17:45









pedroszattoni

14819




14819











  • $t$ and $tau$ are scalars.
    – pedroszattoni
    Jul 31 at 17:52
















  • $t$ and $tau$ are scalars.
    – pedroszattoni
    Jul 31 at 17:52















$t$ and $tau$ are scalars.
– pedroszattoni
Jul 31 at 17:52




$t$ and $tau$ are scalars.
– pedroszattoni
Jul 31 at 17:52










1 Answer
1






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up vote
3
down vote



accepted










We assume of course that $A$ is invertible. The author uses the substitution $tau = t - sigma$ , the identity $A^-1 A = I$, the linearity of the integral and the property $fracmathrmdmathrmd t mathrme^t A = A mathrme^t A$ of the matrix exponential to get
beginalign
int limits_0^t mathrme^(t-tau) A , mathrmd tau &= int limits_0^t mathrme^sigma A , mathrmd sigma = int limits_0^t I , mathrme^sigma A , mathrmd sigma = int limits_0^t A^-1 A
,mathrme^sigma A , mathrmd sigma \
&= A^-1 int limits_0^t A , mathrme^sigma A , mathrmd sigma = A^-1 left[mathrme^sigma A right]_sigma = 0^sigma=t = A^-1 left(mathrme^t A - Iright) , .
endalign
Note that since we have $ A^-1 mathrme^t A = mathrme^t A A^-1$, your answer is not wrong, but completely equivalent:
$$ - mathrme^t A A^-1 left(mathrme^-t A - Iright) = - A^-1 mathrme^t A left(mathrme^-t A - Iright)= A^-1 left(mathrme^t A - Iright) , . $$






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  • I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
    – pedroszattoni
    Jul 31 at 19:06










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










We assume of course that $A$ is invertible. The author uses the substitution $tau = t - sigma$ , the identity $A^-1 A = I$, the linearity of the integral and the property $fracmathrmdmathrmd t mathrme^t A = A mathrme^t A$ of the matrix exponential to get
beginalign
int limits_0^t mathrme^(t-tau) A , mathrmd tau &= int limits_0^t mathrme^sigma A , mathrmd sigma = int limits_0^t I , mathrme^sigma A , mathrmd sigma = int limits_0^t A^-1 A
,mathrme^sigma A , mathrmd sigma \
&= A^-1 int limits_0^t A , mathrme^sigma A , mathrmd sigma = A^-1 left[mathrme^sigma A right]_sigma = 0^sigma=t = A^-1 left(mathrme^t A - Iright) , .
endalign
Note that since we have $ A^-1 mathrme^t A = mathrme^t A A^-1$, your answer is not wrong, but completely equivalent:
$$ - mathrme^t A A^-1 left(mathrme^-t A - Iright) = - A^-1 mathrme^t A left(mathrme^-t A - Iright)= A^-1 left(mathrme^t A - Iright) , . $$






share|cite|improve this answer





















  • I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
    – pedroszattoni
    Jul 31 at 19:06














up vote
3
down vote



accepted










We assume of course that $A$ is invertible. The author uses the substitution $tau = t - sigma$ , the identity $A^-1 A = I$, the linearity of the integral and the property $fracmathrmdmathrmd t mathrme^t A = A mathrme^t A$ of the matrix exponential to get
beginalign
int limits_0^t mathrme^(t-tau) A , mathrmd tau &= int limits_0^t mathrme^sigma A , mathrmd sigma = int limits_0^t I , mathrme^sigma A , mathrmd sigma = int limits_0^t A^-1 A
,mathrme^sigma A , mathrmd sigma \
&= A^-1 int limits_0^t A , mathrme^sigma A , mathrmd sigma = A^-1 left[mathrme^sigma A right]_sigma = 0^sigma=t = A^-1 left(mathrme^t A - Iright) , .
endalign
Note that since we have $ A^-1 mathrme^t A = mathrme^t A A^-1$, your answer is not wrong, but completely equivalent:
$$ - mathrme^t A A^-1 left(mathrme^-t A - Iright) = - A^-1 mathrme^t A left(mathrme^-t A - Iright)= A^-1 left(mathrme^t A - Iright) , . $$






share|cite|improve this answer





















  • I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
    – pedroszattoni
    Jul 31 at 19:06












up vote
3
down vote



accepted







up vote
3
down vote



accepted






We assume of course that $A$ is invertible. The author uses the substitution $tau = t - sigma$ , the identity $A^-1 A = I$, the linearity of the integral and the property $fracmathrmdmathrmd t mathrme^t A = A mathrme^t A$ of the matrix exponential to get
beginalign
int limits_0^t mathrme^(t-tau) A , mathrmd tau &= int limits_0^t mathrme^sigma A , mathrmd sigma = int limits_0^t I , mathrme^sigma A , mathrmd sigma = int limits_0^t A^-1 A
,mathrme^sigma A , mathrmd sigma \
&= A^-1 int limits_0^t A , mathrme^sigma A , mathrmd sigma = A^-1 left[mathrme^sigma A right]_sigma = 0^sigma=t = A^-1 left(mathrme^t A - Iright) , .
endalign
Note that since we have $ A^-1 mathrme^t A = mathrme^t A A^-1$, your answer is not wrong, but completely equivalent:
$$ - mathrme^t A A^-1 left(mathrme^-t A - Iright) = - A^-1 mathrme^t A left(mathrme^-t A - Iright)= A^-1 left(mathrme^t A - Iright) , . $$






share|cite|improve this answer













We assume of course that $A$ is invertible. The author uses the substitution $tau = t - sigma$ , the identity $A^-1 A = I$, the linearity of the integral and the property $fracmathrmdmathrmd t mathrme^t A = A mathrme^t A$ of the matrix exponential to get
beginalign
int limits_0^t mathrme^(t-tau) A , mathrmd tau &= int limits_0^t mathrme^sigma A , mathrmd sigma = int limits_0^t I , mathrme^sigma A , mathrmd sigma = int limits_0^t A^-1 A
,mathrme^sigma A , mathrmd sigma \
&= A^-1 int limits_0^t A , mathrme^sigma A , mathrmd sigma = A^-1 left[mathrme^sigma A right]_sigma = 0^sigma=t = A^-1 left(mathrme^t A - Iright) , .
endalign
Note that since we have $ A^-1 mathrme^t A = mathrme^t A A^-1$, your answer is not wrong, but completely equivalent:
$$ - mathrme^t A A^-1 left(mathrme^-t A - Iright) = - A^-1 mathrme^t A left(mathrme^-t A - Iright)= A^-1 left(mathrme^t A - Iright) , . $$







share|cite|improve this answer













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answered Jul 31 at 18:27









ComplexYetTrivial

2,602624




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  • I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
    – pedroszattoni
    Jul 31 at 19:06
















  • I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
    – pedroszattoni
    Jul 31 at 19:06















I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
– pedroszattoni
Jul 31 at 19:06




I failed to realize that $ A^-1 mathrme^t A = mathrme^t A A^-1$, which can easily be proved using the definition of the matrix exponential. Thanks!
– pedroszattoni
Jul 31 at 19:06












 

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