What is the scalar derivative?
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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 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
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add a comment |
$begingroup$
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 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33
add a comment |
$begingroup$
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 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
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 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
asked Jan 9 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
asked Jan 9 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
asked Jan 9 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
asked Jan 9 '18 at 18:35
mathcounterexamples.netmathcounterexamples.net
26.9k22157
26.9k22157
$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33
add a comment |
$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33
$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33
$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33
add a comment |
2 Answers
2
active
oldest
votes
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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 '18 at 14:30
Matthew ArthurMatthew Arthur
183
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add a comment |
$begingroup$
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 '18 at 3:20
greggreg
8,5851823
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 14:30
Matthew ArthurMatthew Arthur
183
$endgroup$
add a comment |
$begingroup$
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 '18 at 14:30
Matthew ArthurMatthew Arthur
183
$endgroup$
add a comment |
$begingroup$
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 '18 at 14:30
Matthew ArthurMatthew Arthur
183
$endgroup$
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 '18 at 14:30
Matthew ArthurMatthew Arthur
183
answered Dec 26 '18 at 14:30
Matthew ArthurMatthew Arthur
183
answered Dec 26 '18 at 14:30
Matthew ArthurMatthew Arthur
183
answered Dec 26 '18 at 14:30
Matthew ArthurMatthew Arthur
183
183
add a comment |
add a comment |
$begingroup$
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 '18 at 3:20
greggreg
8,5851823
$endgroup$
add a comment |
$begingroup$
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 '18 at 3:20
greggreg
8,5851823
$endgroup$
add a comment |
$begingroup$
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 '18 at 3:20
greggreg
8,5851823
$endgroup$
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 '18 at 3:20
greggreg
8,5851823
answered Jan 10 '18 at 3:20
greggreg
8,5851823
answered Jan 10 '18 at 3:20
greggreg
8,5851823
answered Jan 10 '18 at 3:20
greggreg
8,5851823
8,5851823
add a comment |
add a comment |
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$begingroup$
Mister Wikipedia provides some help.
$endgroup$
– Paul Sinclair
Jan 10 '18 at 0:33