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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
asked Jul 26 at 14:45
Simon Zhai
133
add a comment |Â
up vote
1
down vote
favorite
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
asked Jul 26 at 14:45
Simon Zhai
133
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
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Jul 26 at 14:45
Simon Zhai
133
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
asked Jul 26 at 14:45
Simon Zhai
133
edited Jul 26 at 18:56
asked Jul 26 at 14:45
Simon Zhai
133
asked Jul 26 at 14:45
Simon Zhai
133
asked Jul 26 at 14:45
Simon Zhai
133
133
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
add a comment |Â
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
add a comment |Â
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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