Differentials to derivatives involving trace of matricesoptimization problem minimizing trace of a matrix with inversePartial derivative with matricesRemove the Kronecker operator in $mathrmtrace((Sigma^-1otimes S^-1)A)$Minimization with complex gradient descentMatrix Differentiation of Fraction PowerHow are the vector-valued trace and the unique linearization of $mathfrak L(X,Y):hatotimes_π:X→Y$ of $mathfrak L(X,Y)×X→Y,;(L,x)↦Lx$ related?Differentiation of expression involving matricesPerturbation Theory: Derivative of a trace.find the derivative of a function that involves complex-valued matricesGradient of a Scalar Function of a Matrix
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Differentials to derivatives involving trace of matrices
optimization problem minimizing trace of a matrix with inversePartial derivative with matricesRemove the Kronecker operator in $mathrmtrace((Sigma^-1otimes S^-1)A)$Minimization with complex gradient descentMatrix Differentiation of Fraction PowerHow are the vector-valued trace and the unique linearization of $mathfrak L(X,Y):hatotimes_π:X→Y$ of $mathfrak L(X,Y)×X→Y,;(L,x)↦Lx$ related?Differentiation of expression involving matricesPerturbation Theory: Derivative of a trace.find the derivative of a function that involves complex-valued matricesGradient of a Scalar Function of a Matrix
$begingroup$
Suppose $P$ is a real-valued function of the $ptimes m$ (real) matrix $mathbfQ$. After taking its differential, one arrives with the following:
$$
d(P(mathbfQ))
= operatornametraceleftmathbf1^top_pleft[ dmathbfQodot mathbfW right]left(mathbfwodot mathbfGright)right
$$
where $mathbf1_p$ is the $ptimes 1$ vector of $1$'s with $mathbfW$ is $ptimes m$ while $mathbfw$ and $mathbfG$ are both $m times 1$. $mathbfW$, $mathbfw$ and $mathbfG$ are matrices involving $mathbfQ$.
Question: What is
$
dfracdPdmathbfQ
$ ?
Attempt: $
dfracdPdmathbfQ =
mathbfW left(mathbfwodot mathbfGright)
mathbf1_p
$
But I think it's wrong. So my problem really is that Hadamard product of $dmathbfQ$ and $mathbfW $.
Some identities I have found online are these:
• $dfracd(mathbfa^topmathbfXmathbfb) d mathbfX = mathbfa mathbfb^top$
• $operatornametrace (mathbfAodot mathbfB)mathbfC = operatornametrace mathbfA (mathbfB^top odot mathbfC)$
UPDATE: To make it simpler, a general problem would be
$$
fracmathbfa^topleft[dmathbfQodot f(mathbfQ) right]g(mathbfQ)dmathbfQ
$$
where $mathbfain mathbbR^p$, $f:mathbbR^ptimes mrightarrow mathbbR^ptimes m$ and $g:mathbbR^ptimes mrightarrow mathbbR^m$.
The available identity I have encountered similar to this is
$$
fracoperatornametrace(mathbfAdmathbfX)dmathbfX = mathbfA
$$
from page 2 of this link.
derivatives matrix-calculus trace hadamard-product
asked Mar 22 at 14:13
venreyvenrey
16011
$endgroup$
add a comment |
$begingroup$
Suppose $P$ is a real-valued function of the $ptimes m$ (real) matrix $mathbfQ$. After taking its differential, one arrives with the following:
$$
d(P(mathbfQ))
= operatornametraceleftmathbf1^top_pleft[ dmathbfQodot mathbfW right]left(mathbfwodot mathbfGright)right
$$
where $mathbf1_p$ is the $ptimes 1$ vector of $1$'s with $mathbfW$ is $ptimes m$ while $mathbfw$ and $mathbfG$ are both $m times 1$. $mathbfW$, $mathbfw$ and $mathbfG$ are matrices involving $mathbfQ$.
Question: What is
$
dfracdPdmathbfQ
$ ?
Attempt: $
dfracdPdmathbfQ =
mathbfW left(mathbfwodot mathbfGright)
mathbf1_p
$
But I think it's wrong. So my problem really is that Hadamard product of $dmathbfQ$ and $mathbfW $.
Some identities I have found online are these:
• $dfracd(mathbfa^topmathbfXmathbfb) d mathbfX = mathbfa mathbfb^top$
• $operatornametrace (mathbfAodot mathbfB)mathbfC = operatornametrace mathbfA (mathbfB^top odot mathbfC)$
UPDATE: To make it simpler, a general problem would be
$$
fracmathbfa^topleft[dmathbfQodot f(mathbfQ) right]g(mathbfQ)dmathbfQ
$$
where $mathbfain mathbbR^p$, $f:mathbbR^ptimes mrightarrow mathbbR^ptimes m$ and $g:mathbbR^ptimes mrightarrow mathbbR^m$.
The available identity I have encountered similar to this is
$$
fracoperatornametrace(mathbfAdmathbfX)dmathbfX = mathbfA
$$
from page 2 of this link.
derivatives matrix-calculus trace hadamard-product
asked Mar 22 at 14:13
venreyvenrey
16011
$endgroup$
1
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28
add a comment |
$begingroup$
Suppose $P$ is a real-valued function of the $ptimes m$ (real) matrix $mathbfQ$. After taking its differential, one arrives with the following:
$$
d(P(mathbfQ))
= operatornametraceleftmathbf1^top_pleft[ dmathbfQodot mathbfW right]left(mathbfwodot mathbfGright)right
$$
where $mathbf1_p$ is the $ptimes 1$ vector of $1$'s with $mathbfW$ is $ptimes m$ while $mathbfw$ and $mathbfG$ are both $m times 1$. $mathbfW$, $mathbfw$ and $mathbfG$ are matrices involving $mathbfQ$.
Question: What is
$
dfracdPdmathbfQ
$ ?
Attempt: $
dfracdPdmathbfQ =
mathbfW left(mathbfwodot mathbfGright)
mathbf1_p
$
But I think it's wrong. So my problem really is that Hadamard product of $dmathbfQ$ and $mathbfW $.
Some identities I have found online are these:
• $dfracd(mathbfa^topmathbfXmathbfb) d mathbfX = mathbfa mathbfb^top$
• $operatornametrace (mathbfAodot mathbfB)mathbfC = operatornametrace mathbfA (mathbfB^top odot mathbfC)$
UPDATE: To make it simpler, a general problem would be
$$
fracmathbfa^topleft[dmathbfQodot f(mathbfQ) right]g(mathbfQ)dmathbfQ
$$
where $mathbfain mathbbR^p$, $f:mathbbR^ptimes mrightarrow mathbbR^ptimes m$ and $g:mathbbR^ptimes mrightarrow mathbbR^m$.
The available identity I have encountered similar to this is
$$
fracoperatornametrace(mathbfAdmathbfX)dmathbfX = mathbfA
$$
from page 2 of this link.
derivatives matrix-calculus trace hadamard-product
asked Mar 22 at 14:13
venreyvenrey
16011
$endgroup$
Suppose $P$ is a real-valued function of the $ptimes m$ (real) matrix $mathbfQ$. After taking its differential, one arrives with the following:
$$
d(P(mathbfQ))
= operatornametraceleftmathbf1^top_pleft[ dmathbfQodot mathbfW right]left(mathbfwodot mathbfGright)right
$$
where $mathbf1_p$ is the $ptimes 1$ vector of $1$'s with $mathbfW$ is $ptimes m$ while $mathbfw$ and $mathbfG$ are both $m times 1$. $mathbfW$, $mathbfw$ and $mathbfG$ are matrices involving $mathbfQ$.
Question: What is
$
dfracdPdmathbfQ
$ ?
Attempt: $
dfracdPdmathbfQ =
mathbfW left(mathbfwodot mathbfGright)
mathbf1_p
$
But I think it's wrong. So my problem really is that Hadamard product of $dmathbfQ$ and $mathbfW $.
Some identities I have found online are these:
• $dfracd(mathbfa^topmathbfXmathbfb) d mathbfX = mathbfa mathbfb^top$
• $operatornametrace (mathbfAodot mathbfB)mathbfC = operatornametrace mathbfA (mathbfB^top odot mathbfC)$
UPDATE: To make it simpler, a general problem would be
$$
fracmathbfa^topleft[dmathbfQodot f(mathbfQ) right]g(mathbfQ)dmathbfQ
$$
where $mathbfain mathbbR^p$, $f:mathbbR^ptimes mrightarrow mathbbR^ptimes m$ and $g:mathbbR^ptimes mrightarrow mathbbR^m$.
The available identity I have encountered similar to this is
$$
fracoperatornametrace(mathbfAdmathbfX)dmathbfX = mathbfA
$$
from page 2 of this link.
derivatives matrix-calculus trace hadamard-product
derivatives matrix-calculus trace hadamard-product
asked Mar 22 at 14:13
venreyvenrey
16011
asked Mar 22 at 14:13
venreyvenrey
16011
edited Mar 23 at 1:26
venrey
asked Mar 22 at 14:13
venreyvenrey
16011
asked Mar 22 at 14:13
venreyvenrey
16011
asked Mar 22 at 14:13
venreyvenrey
16011
16011
1
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28
add a comment |
1
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28
1
1
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For convenience, let
$$eqalign
b &= wodot G, quad a &= 1_p cr
$$
Rearrange the given differential to isolate the gradient wrt $Q$.
$$eqalign
dP
&= rm TrBig(a^T,(dQodot W),bBig) cr
&= a^T,(dQodot W),b quad rm trace,does,not,affect,scalar,values cr
&= ab^T:(dQodot W) cr
&= (ab^Todot W):dQ cr
fracpartial Ppartial Q &= ab^Todot W cr
&= Big(1_p(wodot G)^TBig)odot W cr
$$
where a colon is used to write the trace in product form, i.e.
$$A:B = rm Trbig(A^TBbig)$$
UPDATE
The updated question uses $(f,g)$ in place of $(W,b),$ so the gradient becomes
$$eqalign
fracpartial Ppartial Q &= ag^Todot f cr
$$
answered Mar 22 at 23:46
greggreg
9,2531825
$endgroup$
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
add a comment |
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active
oldest
votes
$begingroup$
For convenience, let
$$eqalign
b &= wodot G, quad a &= 1_p cr
$$
Rearrange the given differential to isolate the gradient wrt $Q$.
$$eqalign
dP
&= rm TrBig(a^T,(dQodot W),bBig) cr
&= a^T,(dQodot W),b quad rm trace,does,not,affect,scalar,values cr
&= ab^T:(dQodot W) cr
&= (ab^Todot W):dQ cr
fracpartial Ppartial Q &= ab^Todot W cr
&= Big(1_p(wodot G)^TBig)odot W cr
$$
where a colon is used to write the trace in product form, i.e.
$$A:B = rm Trbig(A^TBbig)$$
UPDATE
The updated question uses $(f,g)$ in place of $(W,b),$ so the gradient becomes
$$eqalign
fracpartial Ppartial Q &= ag^Todot f cr
$$
answered Mar 22 at 23:46
greggreg
9,2531825
$endgroup$
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
add a comment |
$begingroup$
For convenience, let
$$eqalign
b &= wodot G, quad a &= 1_p cr
$$
Rearrange the given differential to isolate the gradient wrt $Q$.
$$eqalign
dP
&= rm TrBig(a^T,(dQodot W),bBig) cr
&= a^T,(dQodot W),b quad rm trace,does,not,affect,scalar,values cr
&= ab^T:(dQodot W) cr
&= (ab^Todot W):dQ cr
fracpartial Ppartial Q &= ab^Todot W cr
&= Big(1_p(wodot G)^TBig)odot W cr
$$
where a colon is used to write the trace in product form, i.e.
$$A:B = rm Trbig(A^TBbig)$$
UPDATE
The updated question uses $(f,g)$ in place of $(W,b),$ so the gradient becomes
$$eqalign
fracpartial Ppartial Q &= ag^Todot f cr
$$
answered Mar 22 at 23:46
greggreg
9,2531825
$endgroup$
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
add a comment |
$begingroup$
For convenience, let
$$eqalign
b &= wodot G, quad a &= 1_p cr
$$
Rearrange the given differential to isolate the gradient wrt $Q$.
$$eqalign
dP
&= rm TrBig(a^T,(dQodot W),bBig) cr
&= a^T,(dQodot W),b quad rm trace,does,not,affect,scalar,values cr
&= ab^T:(dQodot W) cr
&= (ab^Todot W):dQ cr
fracpartial Ppartial Q &= ab^Todot W cr
&= Big(1_p(wodot G)^TBig)odot W cr
$$
where a colon is used to write the trace in product form, i.e.
$$A:B = rm Trbig(A^TBbig)$$
UPDATE
The updated question uses $(f,g)$ in place of $(W,b),$ so the gradient becomes
$$eqalign
fracpartial Ppartial Q &= ag^Todot f cr
$$
answered Mar 22 at 23:46
greggreg
9,2531825
$endgroup$
For convenience, let
$$eqalign
b &= wodot G, quad a &= 1_p cr
$$
Rearrange the given differential to isolate the gradient wrt $Q$.
$$eqalign
dP
&= rm TrBig(a^T,(dQodot W),bBig) cr
&= a^T,(dQodot W),b quad rm trace,does,not,affect,scalar,values cr
&= ab^T:(dQodot W) cr
&= (ab^Todot W):dQ cr
fracpartial Ppartial Q &= ab^Todot W cr
&= Big(1_p(wodot G)^TBig)odot W cr
$$
where a colon is used to write the trace in product form, i.e.
$$A:B = rm Trbig(A^TBbig)$$
UPDATE
The updated question uses $(f,g)$ in place of $(W,b),$ so the gradient becomes
$$eqalign
fracpartial Ppartial Q &= ag^Todot f cr
$$
answered Mar 22 at 23:46
greggreg
9,2531825
edited Mar 23 at 18:55
answered Mar 22 at 23:46
greggreg
9,2531825
answered Mar 22 at 23:46
greggreg
9,2531825
answered Mar 22 at 23:46
greggreg
9,2531825
9,2531825
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
add a comment |
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
I apologize. Prior to my edit, I have written that $mathbfW$ also involves $mathbfQ$, i.e., it is not constant with respect to $mathbfQ$. I have edited it now.
$endgroup$
– venrey
Mar 23 at 1:10
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
Yes, but you also provided the differential $dP$ in which all those dependencies have been sorted out. The differential only contains the $dQ$ term. If that differential is correct, then the gradient above is correct.
$endgroup$
– greg
Mar 23 at 1:27
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
But I can't seem to get pass with the third line of yours : $X = Qodot W implies dX = dQodot W$ where $X$ has just been introduced while the equality involving $dP$ has already been established prior to it. Did I miss something? Shouldn't that be product rule since $W$ involves $Q$?
$endgroup$
– venrey
Mar 23 at 1:34
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
I changed the name of my variable from $P$ to $phi$. The important point is that their differentials are equal.
$endgroup$
– greg
Mar 23 at 1:37
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
$begingroup$
The $X$ variable was a distraction. It has been removed.
$endgroup$
– greg
Mar 23 at 18:58
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1
$begingroup$
Note that $1_p^T$ is a row vector and everything to the right of it is a matrix (lump it together and call it $M$). The quantity $r^T=1_p^TM$ is also a row vector. How do you define the trace of a row vector?
$endgroup$
– greg
Mar 22 at 14:41
$begingroup$
Thank you @greg for point out. I correted it already together with an attempt.
$endgroup$
– venrey
Mar 22 at 21:28