What is the scalar derivative?
I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.
What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.
But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$
$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere
$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$
What is the link with $A^T$?
matrices derivatives
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
add a comment |
I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.
What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.
But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$
$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere
$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$
What is the link with $A^T$?
matrices derivatives
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33
add a comment |
I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.
What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.
But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$
$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere
$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$
What is the link with $A^T$?
matrices derivatives
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
I quote paragraph 2.5 of The Matrix Cookbook document: Assume $F(X)$ to be a differentiable function of each of the elements of $X$... $f(cdot)$ is the scalar derivative of $F(cdot)$. $X$ is here a matrix.
What is the scalar derivative? It is not defined in this document and I have issues to find a definition using Mister Google.
But the way, I'm puzzled by formula (100) of that document:
$$frac{partial}{partial X} mathsf{Tr}(XA) = A^T$$
$X mapsto {Tr}(XA)$ is a linear form defined on the matrices vector space and therefore it's derivative is itself everywhere
$$frac{partial}{partial X} mathsf{Tr}(XA).H = mathsf{Tr}(HA)$$
What is the link with $A^T$?
matrices derivatives
matrices derivatives
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
asked Jan 9 at 18:35
mathcounterexamples.net
24.3k21753
24.3k21753
Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33
add a comment |
Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33
Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33
Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33
add a comment |
2 Answers
2
active
oldest
votes
For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.
@mathcounterexamples.net See:
- http://cs231n.stanford.edu/vecDerivs.pdf
- https://arxiv.org/pdf/1802.01528.pdf
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The simplest explanation is that the word $scalar$ is a typo.
The formula itself seem correct. For instance, let
$$eqalign{
F(x) &= sin(x) cr
f(x) &= frac{dF}{dx} = cos(x) cr
}$$
Then, for a matrix argument $A$, one has the result
$$eqalign{
frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
}$$
...or $cos(A)$ depending on which layout convention you prefer.
answered Jan 10 at 3:20
greg
7,5251821
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.
@mathcounterexamples.net See:
- http://cs231n.stanford.edu/vecDerivs.pdf
- https://arxiv.org/pdf/1802.01528.pdf
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.
@mathcounterexamples.net See:
- http://cs231n.stanford.edu/vecDerivs.pdf
- https://arxiv.org/pdf/1802.01528.pdf
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.
@mathcounterexamples.net See:
- http://cs231n.stanford.edu/vecDerivs.pdf
- https://arxiv.org/pdf/1802.01528.pdf
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For applied matrix calculus in deep learning the term 'scalar derivative' is used to explicitly confirm that the output of the partial derivative of the function with respect to a variable is a scalar and not a vector.
@mathcounterexamples.net See:
- http://cs231n.stanford.edu/vecDerivs.pdf
- https://arxiv.org/pdf/1802.01528.pdf
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 14:30
Matthew Arthur
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 at 14:30
Matthew Arthur
161
answered Dec 26 at 14:30
Matthew Arthur
161
161
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matthew Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
The simplest explanation is that the word $scalar$ is a typo.
The formula itself seem correct. For instance, let
$$eqalign{
F(x) &= sin(x) cr
f(x) &= frac{dF}{dx} = cos(x) cr
}$$
Then, for a matrix argument $A$, one has the result
$$eqalign{
frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
}$$
...or $cos(A)$ depending on which layout convention you prefer.
answered Jan 10 at 3:20
greg
7,5251821
add a comment |
The simplest explanation is that the word $scalar$ is a typo.
The formula itself seem correct. For instance, let
$$eqalign{
F(x) &= sin(x) cr
f(x) &= frac{dF}{dx} = cos(x) cr
}$$
Then, for a matrix argument $A$, one has the result
$$eqalign{
frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
}$$
...or $cos(A)$ depending on which layout convention you prefer.
answered Jan 10 at 3:20
greg
7,5251821
add a comment |
The simplest explanation is that the word $scalar$ is a typo.
The formula itself seem correct. For instance, let
$$eqalign{
F(x) &= sin(x) cr
f(x) &= frac{dF}{dx} = cos(x) cr
}$$
Then, for a matrix argument $A$, one has the result
$$eqalign{
frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
}$$
...or $cos(A)$ depending on which layout convention you prefer.
answered Jan 10 at 3:20
greg
7,5251821
The simplest explanation is that the word $scalar$ is a typo.
The formula itself seem correct. For instance, let
$$eqalign{
F(x) &= sin(x) cr
f(x) &= frac{dF}{dx} = cos(x) cr
}$$
Then, for a matrix argument $A$, one has the result
$$eqalign{
frac{partial,{rm Tr}(sin(A))}{partial A} &= cos(A)^T cr
}$$
...or $cos(A)$ depending on which layout convention you prefer.
answered Jan 10 at 3:20
greg
7,5251821
answered Jan 10 at 3:20
greg
7,5251821
answered Jan 10 at 3:20
greg
7,5251821
answered Jan 10 at 3:20
greg
7,5251821
7,5251821
add a comment |
add a comment |
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Mister Wikipedia provides some help.
– Paul Sinclair
Jan 10 at 0:33