Can I say that $frac{d}{dl}f=Tr((frac{partial}{partial K}f)^intercal frac{partial K}{partial l})$?
$begingroup$
Let's assume that I know $frac{partial}{partial K}f$, where f is scalar valued, and $K$ is a matrix-valued function of $lin mathbb{R}$, i.e. $K=K(l)$.
$df=Tr((frac{partial}{partial K}f)^intercal dK)$, with $dK=frac{partial K}{partial l} dl$.
Can I say that $frac{d}{dl}f=Tr((frac{partial}{partial K}f)^intercal frac{partial K}{partial l})$?
Since $dl$ is $1times 1$, I could take it out of the trace above...
Am I correct?
matrix-calculus
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
$endgroup$
add a comment |
$begingroup$
Let's assume that I know $frac{partial}{partial K}f$, where f is scalar valued, and $K$ is a matrix-valued function of $lin mathbb{R}$, i.e. $K=K(l)$.
$df=Tr((frac{partial}{partial K}f)^intercal dK)$, with $dK=frac{partial K}{partial l} dl$.
Can I say that $frac{d}{dl}f=Tr((frac{partial}{partial K}f)^intercal frac{partial K}{partial l})$?
Since $dl$ is $1times 1$, I could take it out of the trace above...
Am I correct?
matrix-calculus
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
$endgroup$
add a comment |
$begingroup$
Let's assume that I know $frac{partial}{partial K}f$, where f is scalar valued, and $K$ is a matrix-valued function of $lin mathbb{R}$, i.e. $K=K(l)$.
$df=Tr((frac{partial}{partial K}f)^intercal dK)$, with $dK=frac{partial K}{partial l} dl$.
Can I say that $frac{d}{dl}f=Tr((frac{partial}{partial K}f)^intercal frac{partial K}{partial l})$?
Since $dl$ is $1times 1$, I could take it out of the trace above...
Am I correct?
matrix-calculus
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
$endgroup$
Let's assume that I know $frac{partial}{partial K}f$, where f is scalar valued, and $K$ is a matrix-valued function of $lin mathbb{R}$, i.e. $K=K(l)$.
$df=Tr((frac{partial}{partial K}f)^intercal dK)$, with $dK=frac{partial K}{partial l} dl$.
Can I say that $frac{d}{dl}f=Tr((frac{partial}{partial K}f)^intercal frac{partial K}{partial l})$?
Since $dl$ is $1times 1$, I could take it out of the trace above...
Am I correct?
matrix-calculus
matrix-calculus
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
asked Dec 20 '18 at 19:24
An old man in the sea.An old man in the sea.
1,65711135
1,65711135
add a comment |
add a comment |
1 Answer
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$begingroup$
You are correct.
$$
frac{df}{dl} = sum_{i,j} frac{partial f}{partial K_{ij}}frac{d K_{ij}}{dl}
= sum_{i,j} left(frac{partial f}{partial K}right)^T_{ji}left(frac{d K}{dl}right)_{ij}
= sum_{j} left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right]_{jj}
=mathrm{tr}left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right] .
$$
See here for the formula $sum_{ij}X_{ij}Y_{ij}=mathrm{tr}(X^TY)$.
See also my comments to this other question for a similar but slightly more complex computation.
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
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1 Answer
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1 Answer
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$begingroup$
You are correct.
$$
frac{df}{dl} = sum_{i,j} frac{partial f}{partial K_{ij}}frac{d K_{ij}}{dl}
= sum_{i,j} left(frac{partial f}{partial K}right)^T_{ji}left(frac{d K}{dl}right)_{ij}
= sum_{j} left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right]_{jj}
=mathrm{tr}left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right] .
$$
See here for the formula $sum_{ij}X_{ij}Y_{ij}=mathrm{tr}(X^TY)$.
See also my comments to this other question for a similar but slightly more complex computation.
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
You are correct.
$$
frac{df}{dl} = sum_{i,j} frac{partial f}{partial K_{ij}}frac{d K_{ij}}{dl}
= sum_{i,j} left(frac{partial f}{partial K}right)^T_{ji}left(frac{d K}{dl}right)_{ij}
= sum_{j} left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right]_{jj}
=mathrm{tr}left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right] .
$$
See here for the formula $sum_{ij}X_{ij}Y_{ij}=mathrm{tr}(X^TY)$.
See also my comments to this other question for a similar but slightly more complex computation.
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
You are correct.
$$
frac{df}{dl} = sum_{i,j} frac{partial f}{partial K_{ij}}frac{d K_{ij}}{dl}
= sum_{i,j} left(frac{partial f}{partial K}right)^T_{ji}left(frac{d K}{dl}right)_{ij}
= sum_{j} left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right]_{jj}
=mathrm{tr}left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right] .
$$
See here for the formula $sum_{ij}X_{ij}Y_{ij}=mathrm{tr}(X^TY)$.
See also my comments to this other question for a similar but slightly more complex computation.
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
$endgroup$
You are correct.
$$
frac{df}{dl} = sum_{i,j} frac{partial f}{partial K_{ij}}frac{d K_{ij}}{dl}
= sum_{i,j} left(frac{partial f}{partial K}right)^T_{ji}left(frac{d K}{dl}right)_{ij}
= sum_{j} left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right]_{jj}
=mathrm{tr}left[left(frac{partial f}{partial K}right)^Tfrac{d K}{dl}right] .
$$
See here for the formula $sum_{ij}X_{ij}Y_{ij}=mathrm{tr}(X^TY)$.
See also my comments to this other question for a similar but slightly more complex computation.
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
answered Dec 20 '18 at 19:35
FedericoFederico
5,124514
5,124514
add a comment |
add a comment |
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