Complex scalar derivative of trace of complex matrices $frac{d}{d z} Tr[A U(z) B U^H(z)] $
$begingroup$
I'm trying to numerically find the maximum of
$$
f(z) = Tr[ A;U B; U^H], quad U=U(z,z^*),\
A,B,Uinmathbb{C}^{ntimes n},;;
A=A^H, quad B=B^H
$$
using gradient descent(ascent) w.r.t. $z in {C}$.
As I understand, I have to use gradient scheme
$z_{n+1} = z_n + eta;2 frac{partial f(z,z^*)}{partial z^*}$,
so I have to calculate the derivative $ frac{partial f(z,z^*)}{partial z^*}$. Here is calculation:
$$begin{align}
&dTr[ A; U B; U^H] \
&= Tr[ B; U^H A; color{red}{dU } ] + Tr[ B^T U^T A^T color{blue}{dU^*}] \
&left< X:= B; U^H A right>\
&= Tr[ X;; (U'_z;; color{red}{dz} + U'_{z^*};;color{blue}{dz^*}) ] \
&+ Tr[ X^* ({U^*}'_z color{red}{dz} + {U^*}'_{z^*} color{blue}{dz^*})] \
& left< text{using } {f^*}'_z = (f'_{z^*})^*text{ and } {f^*}'_{z^*} = (f'_z)^* right>
\
&= Tr[ X ;;; (U'_z quad color{red}{dz} + U'_{z^*}quad color{blue}{dz^*}) ] \
&+ Tr[ X^* ( (U'_{z^*})^* color{red}{dz} + (U'_z)^* color{blue}{dz^*})]
end{align}$$
Denote $U'_z = V,;; U'_{z^*} = W$. Then
$$begin{align}
dTr
&= Tr[ X ; V + X^* W^*] color{red}{dz} \
&+ Tr[ X ; W + X^* V^*] color{blue}{dz^*}
end{align}$$
Finally $ frac{partial f}{partial z^*} = Tr[ B; U^H A; W] + Tr[B; U^H A; V]^* $
Am I correct? I'm pretty unsure here, because I've never studied matrix or Wirtinger calculus.
(U(z) is squeeze operator here, if you are interested:)
complex-analysis matrix-calculus gradient-descent
asked Jan 8 at 23:19
Mike_Mike_
616
$endgroup$
add a comment |
$begingroup$
I'm trying to numerically find the maximum of
$$
f(z) = Tr[ A;U B; U^H], quad U=U(z,z^*),\
A,B,Uinmathbb{C}^{ntimes n},;;
A=A^H, quad B=B^H
$$
using gradient descent(ascent) w.r.t. $z in {C}$.
As I understand, I have to use gradient scheme
$z_{n+1} = z_n + eta;2 frac{partial f(z,z^*)}{partial z^*}$,
so I have to calculate the derivative $ frac{partial f(z,z^*)}{partial z^*}$. Here is calculation:
$$begin{align}
&dTr[ A; U B; U^H] \
&= Tr[ B; U^H A; color{red}{dU } ] + Tr[ B^T U^T A^T color{blue}{dU^*}] \
&left< X:= B; U^H A right>\
&= Tr[ X;; (U'_z;; color{red}{dz} + U'_{z^*};;color{blue}{dz^*}) ] \
&+ Tr[ X^* ({U^*}'_z color{red}{dz} + {U^*}'_{z^*} color{blue}{dz^*})] \
& left< text{using } {f^*}'_z = (f'_{z^*})^*text{ and } {f^*}'_{z^*} = (f'_z)^* right>
\
&= Tr[ X ;;; (U'_z quad color{red}{dz} + U'_{z^*}quad color{blue}{dz^*}) ] \
&+ Tr[ X^* ( (U'_{z^*})^* color{red}{dz} + (U'_z)^* color{blue}{dz^*})]
end{align}$$
Denote $U'_z = V,;; U'_{z^*} = W$. Then
$$begin{align}
dTr
&= Tr[ X ; V + X^* W^*] color{red}{dz} \
&+ Tr[ X ; W + X^* V^*] color{blue}{dz^*}
end{align}$$
Finally $ frac{partial f}{partial z^*} = Tr[ B; U^H A; W] + Tr[B; U^H A; V]^* $
Am I correct? I'm pretty unsure here, because I've never studied matrix or Wirtinger calculus.
(U(z) is squeeze operator here, if you are interested:)
complex-analysis matrix-calculus gradient-descent
asked Jan 8 at 23:19
Mike_Mike_
616
$endgroup$
1
$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
1
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
1
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06
add a comment |
$begingroup$
I'm trying to numerically find the maximum of
$$
f(z) = Tr[ A;U B; U^H], quad U=U(z,z^*),\
A,B,Uinmathbb{C}^{ntimes n},;;
A=A^H, quad B=B^H
$$
using gradient descent(ascent) w.r.t. $z in {C}$.
As I understand, I have to use gradient scheme
$z_{n+1} = z_n + eta;2 frac{partial f(z,z^*)}{partial z^*}$,
so I have to calculate the derivative $ frac{partial f(z,z^*)}{partial z^*}$. Here is calculation:
$$begin{align}
&dTr[ A; U B; U^H] \
&= Tr[ B; U^H A; color{red}{dU } ] + Tr[ B^T U^T A^T color{blue}{dU^*}] \
&left< X:= B; U^H A right>\
&= Tr[ X;; (U'_z;; color{red}{dz} + U'_{z^*};;color{blue}{dz^*}) ] \
&+ Tr[ X^* ({U^*}'_z color{red}{dz} + {U^*}'_{z^*} color{blue}{dz^*})] \
& left< text{using } {f^*}'_z = (f'_{z^*})^*text{ and } {f^*}'_{z^*} = (f'_z)^* right>
\
&= Tr[ X ;;; (U'_z quad color{red}{dz} + U'_{z^*}quad color{blue}{dz^*}) ] \
&+ Tr[ X^* ( (U'_{z^*})^* color{red}{dz} + (U'_z)^* color{blue}{dz^*})]
end{align}$$
Denote $U'_z = V,;; U'_{z^*} = W$. Then
$$begin{align}
dTr
&= Tr[ X ; V + X^* W^*] color{red}{dz} \
&+ Tr[ X ; W + X^* V^*] color{blue}{dz^*}
end{align}$$
Finally $ frac{partial f}{partial z^*} = Tr[ B; U^H A; W] + Tr[B; U^H A; V]^* $
Am I correct? I'm pretty unsure here, because I've never studied matrix or Wirtinger calculus.
(U(z) is squeeze operator here, if you are interested:)
complex-analysis matrix-calculus gradient-descent
asked Jan 8 at 23:19
Mike_Mike_
616
$endgroup$
I'm trying to numerically find the maximum of
$$
f(z) = Tr[ A;U B; U^H], quad U=U(z,z^*),\
A,B,Uinmathbb{C}^{ntimes n},;;
A=A^H, quad B=B^H
$$
using gradient descent(ascent) w.r.t. $z in {C}$.
As I understand, I have to use gradient scheme
$z_{n+1} = z_n + eta;2 frac{partial f(z,z^*)}{partial z^*}$,
so I have to calculate the derivative $ frac{partial f(z,z^*)}{partial z^*}$. Here is calculation:
$$begin{align}
&dTr[ A; U B; U^H] \
&= Tr[ B; U^H A; color{red}{dU } ] + Tr[ B^T U^T A^T color{blue}{dU^*}] \
&left< X:= B; U^H A right>\
&= Tr[ X;; (U'_z;; color{red}{dz} + U'_{z^*};;color{blue}{dz^*}) ] \
&+ Tr[ X^* ({U^*}'_z color{red}{dz} + {U^*}'_{z^*} color{blue}{dz^*})] \
& left< text{using } {f^*}'_z = (f'_{z^*})^*text{ and } {f^*}'_{z^*} = (f'_z)^* right>
\
&= Tr[ X ;;; (U'_z quad color{red}{dz} + U'_{z^*}quad color{blue}{dz^*}) ] \
&+ Tr[ X^* ( (U'_{z^*})^* color{red}{dz} + (U'_z)^* color{blue}{dz^*})]
end{align}$$
Denote $U'_z = V,;; U'_{z^*} = W$. Then
$$begin{align}
dTr
&= Tr[ X ; V + X^* W^*] color{red}{dz} \
&+ Tr[ X ; W + X^* V^*] color{blue}{dz^*}
end{align}$$
Finally $ frac{partial f}{partial z^*} = Tr[ B; U^H A; W] + Tr[B; U^H A; V]^* $
Am I correct? I'm pretty unsure here, because I've never studied matrix or Wirtinger calculus.
(U(z) is squeeze operator here, if you are interested:)
complex-analysis matrix-calculus gradient-descent
complex-analysis matrix-calculus gradient-descent
asked Jan 8 at 23:19
Mike_Mike_
616
asked Jan 8 at 23:19
Mike_Mike_
616
edited Jan 9 at 15:35
Mike_
asked Jan 8 at 23:19
Mike_Mike_
616
asked Jan 8 at 23:19
Mike_Mike_
616
asked Jan 8 at 23:19
Mike_Mike_
616
616
1
$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
1
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
1
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06
add a comment |
1
$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
1
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
1
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06
1
1
$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
1
1
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
1
1
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06
add a comment |
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$begingroup$
You have changed the order of the arguments to the trace function in a non-cyclical manner. In your equation defining $f(z)$ the product order is $(AU^HBU)$, but in the differential expression the order is $(AUBU^H)$. The traces of these two products are not equal (unless $U$ is hermitian).
$endgroup$
– greg
Jan 9 at 6:55
$begingroup$
@greg, yes, thanks. This was a typo, my fucntion is $Tr[AUBU^H]$. Fixed. Now everything seems correct(?).
$endgroup$
– Mike_
Jan 9 at 14:59
1
$begingroup$
Okay, now everything looks right.
$endgroup$
– greg
Jan 9 at 17:31
$begingroup$
I have looked through you other answers and trust you. Unfortunately, I have no other sources of verification right now, so thank you with this "confidence-request" question! How do you think, should I answer to my question in order to mark it as solved somehow?
$endgroup$
– Mike_
Jan 10 at 4:38
1
$begingroup$
Your question is of a type that invites either a "yes" or "no" response, and such responses typically take the form of a comment. And that's okay. Had your solution been wrong then someone would write an answer containing the correct solution, and you could accept it.
$endgroup$
– greg
Jan 11 at 2:06