How to calculate the following $ fracpartialpartial x log (det X(x))$?

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How to calculate the following



$$ fracpartialpartial x log (det X(x))$$



where $X$ is a matrix in $mathbbR^ntimes n$ which is a function of $xin mathbbR^d$?




linear-algebra multivariable-calculus derivatives matrix-calculus




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  • 1




    Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
    – Michael Paris
    Jul 23 at 6:26











  • @MichaelParis, no, I don't know. Would you please explain a little?
    – user85361
    Jul 23 at 6:53






  • 1




    en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
    – Michael Paris
    Jul 23 at 6:57














up vote
2
down vote

favorite












How to calculate the following



$$ fracpartialpartial x log (det X(x))$$



where $X$ is a matrix in $mathbbR^ntimes n$ which is a function of $xin mathbbR^d$?




linear-algebra multivariable-calculus derivatives matrix-calculus




share|cite|improve this question

















  • 1




    Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
    – Michael Paris
    Jul 23 at 6:26











  • @MichaelParis, no, I don't know. Would you please explain a little?
    – user85361
    Jul 23 at 6:53






  • 1




    en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
    – Michael Paris
    Jul 23 at 6:57












up vote
2
down vote

favorite









up vote
2
down vote

favorite











How to calculate the following



$$ fracpartialpartial x log (det X(x))$$



where $X$ is a matrix in $mathbbR^ntimes n$ which is a function of $xin mathbbR^d$?




linear-algebra multivariable-calculus derivatives matrix-calculus




share|cite|improve this question













How to calculate the following



$$ fracpartialpartial x log (det X(x))$$



where $X$ is a matrix in $mathbbR^ntimes n$ which is a function of $xin mathbbR^d$?





linear-algebra multivariable-calculus derivatives matrix-calculus





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 7:17









user550103

502213




502213









asked Jul 23 at 6:16









user85361

331215




331215







  • 1




    Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
    – Michael Paris
    Jul 23 at 6:26











  • @MichaelParis, no, I don't know. Would you please explain a little?
    – user85361
    Jul 23 at 6:53






  • 1




    en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
    – Michael Paris
    Jul 23 at 6:57












  • 1




    Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
    – Michael Paris
    Jul 23 at 6:26











  • @MichaelParis, no, I don't know. Would you please explain a little?
    – user85361
    Jul 23 at 6:53






  • 1




    en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
    – Michael Paris
    Jul 23 at 6:57







1




1




Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
– Michael Paris
Jul 23 at 6:26





Do you know how to express a determinant as a sum? It will be a step in the right direction (Jacobi's formula). Also know that "inner times outer" applies here as well.
– Michael Paris
Jul 23 at 6:26













@MichaelParis, no, I don't know. Would you please explain a little?
– user85361
Jul 23 at 6:53




@MichaelParis, no, I don't know. Would you please explain a little?
– user85361
Jul 23 at 6:53




1




1




en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
– Michael Paris
Jul 23 at 6:57




en.m.wikipedia.org/wiki/Jacobi%27s_formula. The determinant of a matrix has a couple of useful representations as a sum, one of which can be used to simplify your expression
– Michael Paris
Jul 23 at 6:57










2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










beginequation
fracpartialpartial xlog(det(X(x))) = frac1det(X(x)) trleft(adj(X(x)) dX(x)right)
endequation
Edit:



"d" implies the gradient operator in the covariant form, which acts on the provided (1,1)-tensor in the usual way, making it a (2,1)-tensor.






share|cite|improve this answer























  • Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
    – user85361
    Jul 23 at 7:17










  • Is the result a vector or a matrix? What is the dimension of the result?
    – user85361
    Jul 23 at 7:29










  • It's just in mathrmR
    – Michael Paris
    Jul 23 at 7:34










  • $mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
    – user85361
    Jul 23 at 7:39











  • Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
    – Michael Paris
    Jul 23 at 7:41

















up vote
1
down vote













The gradient of $log det X$ with respect to the entries of $X$ is $X^-1$, so the total derivative with respect to $x$ is $sum_i sum_j (X^-1(x))_ij (X'_ij(x))$. Here, $X_ij(x)$ denotes the $i,j$-entry of $X(x)$, and $X'_ij$ denotes the derivative of this entry with respect to $x$. $X^-1(x)$ denotes the matrix inverse of $X(x)$.






share|cite|improve this answer





















  • Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
    – user85361
    Jul 23 at 6:52










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










beginequation
fracpartialpartial xlog(det(X(x))) = frac1det(X(x)) trleft(adj(X(x)) dX(x)right)
endequation
Edit:



"d" implies the gradient operator in the covariant form, which acts on the provided (1,1)-tensor in the usual way, making it a (2,1)-tensor.






share|cite|improve this answer























  • Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
    – user85361
    Jul 23 at 7:17










  • Is the result a vector or a matrix? What is the dimension of the result?
    – user85361
    Jul 23 at 7:29










  • It's just in mathrmR
    – Michael Paris
    Jul 23 at 7:34










  • $mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
    – user85361
    Jul 23 at 7:39











  • Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
    – Michael Paris
    Jul 23 at 7:41














up vote
1
down vote



accepted










beginequation
fracpartialpartial xlog(det(X(x))) = frac1det(X(x)) trleft(adj(X(x)) dX(x)right)
endequation
Edit:



"d" implies the gradient operator in the covariant form, which acts on the provided (1,1)-tensor in the usual way, making it a (2,1)-tensor.






share|cite|improve this answer























  • Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
    – user85361
    Jul 23 at 7:17










  • Is the result a vector or a matrix? What is the dimension of the result?
    – user85361
    Jul 23 at 7:29










  • It's just in mathrmR
    – Michael Paris
    Jul 23 at 7:34










  • $mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
    – user85361
    Jul 23 at 7:39











  • Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
    – Michael Paris
    Jul 23 at 7:41












up vote
1
down vote



accepted







up vote
1
down vote



accepted






beginequation
fracpartialpartial xlog(det(X(x))) = frac1det(X(x)) trleft(adj(X(x)) dX(x)right)
endequation
Edit:



"d" implies the gradient operator in the covariant form, which acts on the provided (1,1)-tensor in the usual way, making it a (2,1)-tensor.






share|cite|improve this answer















beginequation
fracpartialpartial xlog(det(X(x))) = frac1det(X(x)) trleft(adj(X(x)) dX(x)right)
endequation
Edit:



"d" implies the gradient operator in the covariant form, which acts on the provided (1,1)-tensor in the usual way, making it a (2,1)-tensor.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 23 at 7:56


























answered Jul 23 at 7:01









Michael Paris

451111




451111











  • Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
    – user85361
    Jul 23 at 7:17










  • Is the result a vector or a matrix? What is the dimension of the result?
    – user85361
    Jul 23 at 7:29










  • It's just in mathrmR
    – Michael Paris
    Jul 23 at 7:34










  • $mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
    – user85361
    Jul 23 at 7:39











  • Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
    – Michael Paris
    Jul 23 at 7:41
















  • Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
    – user85361
    Jul 23 at 7:17










  • Is the result a vector or a matrix? What is the dimension of the result?
    – user85361
    Jul 23 at 7:29










  • It's just in mathrmR
    – Michael Paris
    Jul 23 at 7:34










  • $mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
    – user85361
    Jul 23 at 7:39











  • Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
    – Michael Paris
    Jul 23 at 7:41















Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
– user85361
Jul 23 at 7:17




Do we have $X(x) fracpartialpartial xlog(det(X(x)))= tr(fracpartialpartial xX(x))$
– user85361
Jul 23 at 7:17












Is the result a vector or a matrix? What is the dimension of the result?
– user85361
Jul 23 at 7:29




Is the result a vector or a matrix? What is the dimension of the result?
– user85361
Jul 23 at 7:29












It's just in mathrmR
– Michael Paris
Jul 23 at 7:34




It's just in mathrmR
– Michael Paris
Jul 23 at 7:34












$mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
– user85361
Jul 23 at 7:39





$mathrmR$? How it is possible that a derivative with respect to a vector becomes only in $mathrmR$.?
– user85361
Jul 23 at 7:39













Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
– Michael Paris
Jul 23 at 7:41




Oh, my bad. Should be in R^d then. Didn't notice that x in R^d.
– Michael Paris
Jul 23 at 7:41










up vote
1
down vote













The gradient of $log det X$ with respect to the entries of $X$ is $X^-1$, so the total derivative with respect to $x$ is $sum_i sum_j (X^-1(x))_ij (X'_ij(x))$. Here, $X_ij(x)$ denotes the $i,j$-entry of $X(x)$, and $X'_ij$ denotes the derivative of this entry with respect to $x$. $X^-1(x)$ denotes the matrix inverse of $X(x)$.






share|cite|improve this answer





















  • Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
    – user85361
    Jul 23 at 6:52














up vote
1
down vote













The gradient of $log det X$ with respect to the entries of $X$ is $X^-1$, so the total derivative with respect to $x$ is $sum_i sum_j (X^-1(x))_ij (X'_ij(x))$. Here, $X_ij(x)$ denotes the $i,j$-entry of $X(x)$, and $X'_ij$ denotes the derivative of this entry with respect to $x$. $X^-1(x)$ denotes the matrix inverse of $X(x)$.






share|cite|improve this answer





















  • Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
    – user85361
    Jul 23 at 6:52












up vote
1
down vote










up vote
1
down vote









The gradient of $log det X$ with respect to the entries of $X$ is $X^-1$, so the total derivative with respect to $x$ is $sum_i sum_j (X^-1(x))_ij (X'_ij(x))$. Here, $X_ij(x)$ denotes the $i,j$-entry of $X(x)$, and $X'_ij$ denotes the derivative of this entry with respect to $x$. $X^-1(x)$ denotes the matrix inverse of $X(x)$.






share|cite|improve this answer













The gradient of $log det X$ with respect to the entries of $X$ is $X^-1$, so the total derivative with respect to $x$ is $sum_i sum_j (X^-1(x))_ij (X'_ij(x))$. Here, $X_ij(x)$ denotes the $i,j$-entry of $X(x)$, and $X'_ij$ denotes the derivative of this entry with respect to $x$. $X^-1(x)$ denotes the matrix inverse of $X(x)$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 6:32









angryavian

34.6k12874




34.6k12874











  • Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
    – user85361
    Jul 23 at 6:52
















  • Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
    – user85361
    Jul 23 at 6:52















Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
– user85361
Jul 23 at 6:52




Thank you. Although it must be very easy, I'm a little confused. Could you please show it in matrix notation form?
– user85361
Jul 23 at 6:52












 

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