Exponential of singular matrix times inverse of singular matrix












0












$begingroup$


I need to find a way to solve the following problem:



Suppose I have a singular matrix A. I need to find a solution to



$frac{e^{hA}-I}{A}$



However, note that the inverse of A is infinity. Can anyone help with this problem? Maybe l'hospital rule could help? I am not really sure. Thanks in advance.






matrices limits







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  • $begingroup$
    Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
    $endgroup$
    – James
    Dec 4 '18 at 14:12
















0












$begingroup$


I need to find a way to solve the following problem:



Suppose I have a singular matrix A. I need to find a solution to



$frac{e^{hA}-I}{A}$



However, note that the inverse of A is infinity. Can anyone help with this problem? Maybe l'hospital rule could help? I am not really sure. Thanks in advance.






matrices limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
    $endgroup$
    – James
    Dec 4 '18 at 14:12














0












0








0


0



$begingroup$


I need to find a way to solve the following problem:



Suppose I have a singular matrix A. I need to find a solution to



$frac{e^{hA}-I}{A}$



However, note that the inverse of A is infinity. Can anyone help with this problem? Maybe l'hospital rule could help? I am not really sure. Thanks in advance.






matrices limits







share|cite|improve this question









$endgroup$




I need to find a way to solve the following problem:



Suppose I have a singular matrix A. I need to find a solution to



$frac{e^{hA}-I}{A}$



However, note that the inverse of A is infinity. Can anyone help with this problem? Maybe l'hospital rule could help? I am not really sure. Thanks in advance.






matrices limits




matrices limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 14:06









Thiago LimaThiago Lima

31




31












  • $begingroup$
    Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
    $endgroup$
    – James
    Dec 4 '18 at 14:12


















  • $begingroup$
    Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
    $endgroup$
    – James
    Dec 4 '18 at 14:12
















$begingroup$
Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
$endgroup$
– James
Dec 4 '18 at 14:12




$begingroup$
Welcome to MSE. If you say that you want to find a solution, you should write down an equation. Moreover, let us know what you have tried already and where your interest in this question comes from.
$endgroup$
– James
Dec 4 '18 at 14:12










1 Answer
1






active

oldest

votes


















2












$begingroup$

What you wrote doesn't make sense, however the function $f(z) = dfrac{e^{hz}-1}{z}$
has a removable singularity at $0$, and after removing it you have an entire function,
so you can define $f(A)$ by the functional calculus. For example, you could compute
it using the power series
$$ f(A) = sum_{n=1}^infty frac{h^n}{n!} A^{n-1} $$
Or if $A$ is diagonalizable: $A = S^{-1} D S$ where $D$ is diagonal, then
$f(A) = S^{-1} f(D) S$ where $f(D)$ is diagonal with $(f(D))_{jj} = f(D_{jj})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 14:59










  • $begingroup$
    There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 16:53










  • $begingroup$
    See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 17:07










  • $begingroup$
    Ok, I will look into it. Thank you very much again!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 18:56











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

What you wrote doesn't make sense, however the function $f(z) = dfrac{e^{hz}-1}{z}$
has a removable singularity at $0$, and after removing it you have an entire function,
so you can define $f(A)$ by the functional calculus. For example, you could compute
it using the power series
$$ f(A) = sum_{n=1}^infty frac{h^n}{n!} A^{n-1} $$
Or if $A$ is diagonalizable: $A = S^{-1} D S$ where $D$ is diagonal, then
$f(A) = S^{-1} f(D) S$ where $f(D)$ is diagonal with $(f(D))_{jj} = f(D_{jj})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 14:59










  • $begingroup$
    There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 16:53










  • $begingroup$
    See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 17:07










  • $begingroup$
    Ok, I will look into it. Thank you very much again!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 18:56
















2












$begingroup$

What you wrote doesn't make sense, however the function $f(z) = dfrac{e^{hz}-1}{z}$
has a removable singularity at $0$, and after removing it you have an entire function,
so you can define $f(A)$ by the functional calculus. For example, you could compute
it using the power series
$$ f(A) = sum_{n=1}^infty frac{h^n}{n!} A^{n-1} $$
Or if $A$ is diagonalizable: $A = S^{-1} D S$ where $D$ is diagonal, then
$f(A) = S^{-1} f(D) S$ where $f(D)$ is diagonal with $(f(D))_{jj} = f(D_{jj})$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 14:59










  • $begingroup$
    There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 16:53










  • $begingroup$
    See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 17:07










  • $begingroup$
    Ok, I will look into it. Thank you very much again!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 18:56














2












2








2





$begingroup$

What you wrote doesn't make sense, however the function $f(z) = dfrac{e^{hz}-1}{z}$
has a removable singularity at $0$, and after removing it you have an entire function,
so you can define $f(A)$ by the functional calculus. For example, you could compute
it using the power series
$$ f(A) = sum_{n=1}^infty frac{h^n}{n!} A^{n-1} $$
Or if $A$ is diagonalizable: $A = S^{-1} D S$ where $D$ is diagonal, then
$f(A) = S^{-1} f(D) S$ where $f(D)$ is diagonal with $(f(D))_{jj} = f(D_{jj})$.






share|cite|improve this answer









$endgroup$



What you wrote doesn't make sense, however the function $f(z) = dfrac{e^{hz}-1}{z}$
has a removable singularity at $0$, and after removing it you have an entire function,
so you can define $f(A)$ by the functional calculus. For example, you could compute
it using the power series
$$ f(A) = sum_{n=1}^infty frac{h^n}{n!} A^{n-1} $$
Or if $A$ is diagonalizable: $A = S^{-1} D S$ where $D$ is diagonal, then
$f(A) = S^{-1} f(D) S$ where $f(D)$ is diagonal with $(f(D))_{jj} = f(D_{jj})$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 14:13









Robert IsraelRobert Israel

320k23210462




320k23210462












  • $begingroup$
    Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 14:59










  • $begingroup$
    There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 16:53










  • $begingroup$
    See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 17:07










  • $begingroup$
    Ok, I will look into it. Thank you very much again!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 18:56


















  • $begingroup$
    Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 14:59










  • $begingroup$
    There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 16:53










  • $begingroup$
    See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
    $endgroup$
    – Robert Israel
    Dec 4 '18 at 17:07










  • $begingroup$
    Ok, I will look into it. Thank you very much again!
    $endgroup$
    – Thiago Lima
    Dec 4 '18 at 18:56
















$begingroup$
Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
$endgroup$
– Thiago Lima
Dec 4 '18 at 14:59




$begingroup$
Thank you very much for your response. Sorry to bother again, but just to be clear, there is no way to get rid of the power series? I have used software to compute this series and it is a convergent series, therefore I would like to find a simple way to compute it which would not involve a summation of infinity terms!
$endgroup$
– Thiago Lima
Dec 4 '18 at 14:59












$begingroup$
There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
$endgroup$
– Robert Israel
Dec 4 '18 at 16:53




$begingroup$
There are many ways to compute it besides the series. I noted one in the case where $A$ is diagonalizable. More generally you could use Jordan canonical form.
$endgroup$
– Robert Israel
Dec 4 '18 at 16:53












$begingroup$
See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
$endgroup$
– Robert Israel
Dec 4 '18 at 17:07




$begingroup$
See also Moler and van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
$endgroup$
– Robert Israel
Dec 4 '18 at 17:07












$begingroup$
Ok, I will look into it. Thank you very much again!
$endgroup$
– Thiago Lima
Dec 4 '18 at 18:56




$begingroup$
Ok, I will look into it. Thank you very much again!
$endgroup$
– Thiago Lima
Dec 4 '18 at 18:56


















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