Given $S$ is a $n$-dimensional square symmetric matrix, how to calculate $fracpartial tr(sqrtSS^T)partial S$(perhaps by chain rule)?

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I am trying to go through this process using the chain rule:
$$fracpartial tr(sqrtSS^T)partial S=fracpartial tr(sqrtSS^T) partial sqrtSS^T cdotfracpartial sqrtSS^Tpartial S,$$



I am assuming the result $fracpartial tr(sqrtSS^T)partial S$ should also be an $n$-dimensional square matrix. If I am understanding correctly, according to the chain rule, the first part results:
$$fracpartial tr(sqrtSS^T) partial sqrtSS^T =mathbbI_ninmathbbR^ntimes n,$$ and the second part results(according to
https://math.stackexchange.com/q/1320527): $$fracpartial sqrtSS^Tpartial S = (sqrtS^ToplussqrtS)^-1in mathbbR^n^2times n^2,$$ and my question is how to understand/interpret the "$cdot$" in chain rule(Since we cannot just simply multiple a $mathbbR^ntimes n$ matrix with a $mathbbR^n^2times n^2$ matrix)? Or say how to use chain rule to calculate the result for this derivative properly?



Thanks!




multivariable-calculus derivatives




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  • It's a composition of functions – which corresponds to the product of matrices.
    – Bernard
    Jul 26 at 14:48










  • @Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
    – Simon Zhai
    Jul 26 at 14:52















up vote
1
down vote

favorite
1












I am trying to go through this process using the chain rule:
$$fracpartial tr(sqrtSS^T)partial S=fracpartial tr(sqrtSS^T) partial sqrtSS^T cdotfracpartial sqrtSS^Tpartial S,$$



I am assuming the result $fracpartial tr(sqrtSS^T)partial S$ should also be an $n$-dimensional square matrix. If I am understanding correctly, according to the chain rule, the first part results:
$$fracpartial tr(sqrtSS^T) partial sqrtSS^T =mathbbI_ninmathbbR^ntimes n,$$ and the second part results(according to
https://math.stackexchange.com/q/1320527): $$fracpartial sqrtSS^Tpartial S = (sqrtS^ToplussqrtS)^-1in mathbbR^n^2times n^2,$$ and my question is how to understand/interpret the "$cdot$" in chain rule(Since we cannot just simply multiple a $mathbbR^ntimes n$ matrix with a $mathbbR^n^2times n^2$ matrix)? Or say how to use chain rule to calculate the result for this derivative properly?



Thanks!




multivariable-calculus derivatives




share|cite|improve this question





















  • It's a composition of functions – which corresponds to the product of matrices.
    – Bernard
    Jul 26 at 14:48










  • @Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
    – Simon Zhai
    Jul 26 at 14:52













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I am trying to go through this process using the chain rule:
$$fracpartial tr(sqrtSS^T)partial S=fracpartial tr(sqrtSS^T) partial sqrtSS^T cdotfracpartial sqrtSS^Tpartial S,$$



I am assuming the result $fracpartial tr(sqrtSS^T)partial S$ should also be an $n$-dimensional square matrix. If I am understanding correctly, according to the chain rule, the first part results:
$$fracpartial tr(sqrtSS^T) partial sqrtSS^T =mathbbI_ninmathbbR^ntimes n,$$ and the second part results(according to
https://math.stackexchange.com/q/1320527): $$fracpartial sqrtSS^Tpartial S = (sqrtS^ToplussqrtS)^-1in mathbbR^n^2times n^2,$$ and my question is how to understand/interpret the "$cdot$" in chain rule(Since we cannot just simply multiple a $mathbbR^ntimes n$ matrix with a $mathbbR^n^2times n^2$ matrix)? Or say how to use chain rule to calculate the result for this derivative properly?



Thanks!




multivariable-calculus derivatives




share|cite|improve this question













I am trying to go through this process using the chain rule:
$$fracpartial tr(sqrtSS^T)partial S=fracpartial tr(sqrtSS^T) partial sqrtSS^T cdotfracpartial sqrtSS^Tpartial S,$$



I am assuming the result $fracpartial tr(sqrtSS^T)partial S$ should also be an $n$-dimensional square matrix. If I am understanding correctly, according to the chain rule, the first part results:
$$fracpartial tr(sqrtSS^T) partial sqrtSS^T =mathbbI_ninmathbbR^ntimes n,$$ and the second part results(according to
https://math.stackexchange.com/q/1320527): $$fracpartial sqrtSS^Tpartial S = (sqrtS^ToplussqrtS)^-1in mathbbR^n^2times n^2,$$ and my question is how to understand/interpret the "$cdot$" in chain rule(Since we cannot just simply multiple a $mathbbR^ntimes n$ matrix with a $mathbbR^n^2times n^2$ matrix)? Or say how to use chain rule to calculate the result for this derivative properly?



Thanks!





multivariable-calculus derivatives





share|cite|improve this question












share|cite|improve this question




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edited Jul 26 at 18:56
























asked Jul 26 at 14:45









Simon Zhai

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  • It's a composition of functions – which corresponds to the product of matrices.
    – Bernard
    Jul 26 at 14:48










  • @Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
    – Simon Zhai
    Jul 26 at 14:52

















  • It's a composition of functions – which corresponds to the product of matrices.
    – Bernard
    Jul 26 at 14:48










  • @Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
    – Simon Zhai
    Jul 26 at 14:52
















It's a composition of functions – which corresponds to the product of matrices.
– Bernard
Jul 26 at 14:48




It's a composition of functions – which corresponds to the product of matrices.
– Bernard
Jul 26 at 14:48












@Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
– Simon Zhai
Jul 26 at 14:52





@Bernard Thanks for your reply, but I am still confused about the "product of matrices", can you provide some more details(perhaps using this example) or provide some reference for me to check?
– Simon Zhai
Jul 26 at 14:52
















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