Mixing
Derivative of a trace with Hadamard division
$begingroup$
I am trying to solve the derivative:
$$frac{partial Tr,[AX'(X oslash B)]}{partial X},$$
where $oslash$ is the symbol used for the Hadamard division (or element-wise division) and A is a square matrix. I think that the derivative should be:
$$(XA + XA') oslash B,$$
but I couldn't find a reference to double check it and I recently started doing similar calculations. Any hints?
linear-algebra derivatives self-learning
asked Jan 6 at 5:04
merchmerch
1082
$endgroup$
add a comment |
$begingroup$
I am trying to solve the derivative:
$$frac{partial Tr,[AX'(X oslash B)]}{partial X},$$
where $oslash$ is the symbol used for the Hadamard division (or element-wise division) and A is a square matrix. I think that the derivative should be:
$$(XA + XA') oslash B,$$
but I couldn't find a reference to double check it and I recently started doing similar calculations. Any hints?
linear-algebra derivatives self-learning
asked Jan 6 at 5:04
merchmerch
1082
$endgroup$
add a comment |
$begingroup$
I am trying to solve the derivative:
$$frac{partial Tr,[AX'(X oslash B)]}{partial X},$$
where $oslash$ is the symbol used for the Hadamard division (or element-wise division) and A is a square matrix. I think that the derivative should be:
$$(XA + XA') oslash B,$$
but I couldn't find a reference to double check it and I recently started doing similar calculations. Any hints?
linear-algebra derivatives self-learning
asked Jan 6 at 5:04
merchmerch
1082
$endgroup$
I am trying to solve the derivative:
$$frac{partial Tr,[AX'(X oslash B)]}{partial X},$$
where $oslash$ is the symbol used for the Hadamard division (or element-wise division) and A is a square matrix. I think that the derivative should be:
$$(XA + XA') oslash B,$$
but I couldn't find a reference to double check it and I recently started doing similar calculations. Any hints?
linear-algebra derivatives self-learning
linear-algebra derivatives self-learning
asked Jan 6 at 5:04
merchmerch
1082
asked Jan 6 at 5:04
merchmerch
1082
asked Jan 6 at 5:04
merchmerch
1082
asked Jan 6 at 5:04
merchmerch
1082
asked Jan 6 at 5:04
merchmerch
1082
1082
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's use a colon to denote the trace/Frobenius product, i.e.
$$eqalign{A:B={rm Tr}(A^TB)}$$
Properties of the trace give rise to rules for rearranging terms in a Frobenius product, such as
$$eqalign{
A:B &= B:A cr&= A^T!:!B^T cr
A:BC &= B^TA:C cr&= AC^T:B cr
}$$
Also note that the Hadamard product commutes with itself and the Frobenius product
$$eqalign{
Aodot B &= Bodot A cr
A:(Bodot C) &= (Aodot C):B cr&= (Bodot A):C cr
}$$
Instead of $B$ let's use its Hadamard inverse, and while we're at it, let's define a few other matrices.
$$eqalign{
C &= 1oslash B cr
Y &= X^TA cr
Z &= Xodot C = Xoslash B cr
}$$
to write the cost function. Then find its differential and gradient.
$$eqalign{
phi &= XA^T:Xodot C cr&= Y:Z cr
dphi
&= Z:dY + Y:dZ cr
&= Z:dX,A^T + Y:dXodot C cr
&= ZA:dX + Yodot C:dX cr
&= (ZA + Yodot C):dX cr
frac{partialphi}{partial X}
&= ZA + Yodot C cr
&= (Xoslash B)A + (XA^T)oslash B cr
}$$
answered Jan 6 at 16:26
greggreg
7,7951821
$endgroup$
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063510%2fderivative-of-a-trace-with-hadamard-division%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's use a colon to denote the trace/Frobenius product, i.e.
$$eqalign{A:B={rm Tr}(A^TB)}$$
Properties of the trace give rise to rules for rearranging terms in a Frobenius product, such as
$$eqalign{
A:B &= B:A cr&= A^T!:!B^T cr
A:BC &= B^TA:C cr&= AC^T:B cr
}$$
Also note that the Hadamard product commutes with itself and the Frobenius product
$$eqalign{
Aodot B &= Bodot A cr
A:(Bodot C) &= (Aodot C):B cr&= (Bodot A):C cr
}$$
Instead of $B$ let's use its Hadamard inverse, and while we're at it, let's define a few other matrices.
$$eqalign{
C &= 1oslash B cr
Y &= X^TA cr
Z &= Xodot C = Xoslash B cr
}$$
to write the cost function. Then find its differential and gradient.
$$eqalign{
phi &= XA^T:Xodot C cr&= Y:Z cr
dphi
&= Z:dY + Y:dZ cr
&= Z:dX,A^T + Y:dXodot C cr
&= ZA:dX + Yodot C:dX cr
&= (ZA + Yodot C):dX cr
frac{partialphi}{partial X}
&= ZA + Yodot C cr
&= (Xoslash B)A + (XA^T)oslash B cr
}$$
answered Jan 6 at 16:26
greggreg
7,7951821
$endgroup$
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
add a comment |
$begingroup$
Let's use a colon to denote the trace/Frobenius product, i.e.
$$eqalign{A:B={rm Tr}(A^TB)}$$
Properties of the trace give rise to rules for rearranging terms in a Frobenius product, such as
$$eqalign{
A:B &= B:A cr&= A^T!:!B^T cr
A:BC &= B^TA:C cr&= AC^T:B cr
}$$
Also note that the Hadamard product commutes with itself and the Frobenius product
$$eqalign{
Aodot B &= Bodot A cr
A:(Bodot C) &= (Aodot C):B cr&= (Bodot A):C cr
}$$
Instead of $B$ let's use its Hadamard inverse, and while we're at it, let's define a few other matrices.
$$eqalign{
C &= 1oslash B cr
Y &= X^TA cr
Z &= Xodot C = Xoslash B cr
}$$
to write the cost function. Then find its differential and gradient.
$$eqalign{
phi &= XA^T:Xodot C cr&= Y:Z cr
dphi
&= Z:dY + Y:dZ cr
&= Z:dX,A^T + Y:dXodot C cr
&= ZA:dX + Yodot C:dX cr
&= (ZA + Yodot C):dX cr
frac{partialphi}{partial X}
&= ZA + Yodot C cr
&= (Xoslash B)A + (XA^T)oslash B cr
}$$
answered Jan 6 at 16:26
greggreg
7,7951821
$endgroup$
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
add a comment |
$begingroup$
Let's use a colon to denote the trace/Frobenius product, i.e.
$$eqalign{A:B={rm Tr}(A^TB)}$$
Properties of the trace give rise to rules for rearranging terms in a Frobenius product, such as
$$eqalign{
A:B &= B:A cr&= A^T!:!B^T cr
A:BC &= B^TA:C cr&= AC^T:B cr
}$$
Also note that the Hadamard product commutes with itself and the Frobenius product
$$eqalign{
Aodot B &= Bodot A cr
A:(Bodot C) &= (Aodot C):B cr&= (Bodot A):C cr
}$$
Instead of $B$ let's use its Hadamard inverse, and while we're at it, let's define a few other matrices.
$$eqalign{
C &= 1oslash B cr
Y &= X^TA cr
Z &= Xodot C = Xoslash B cr
}$$
to write the cost function. Then find its differential and gradient.
$$eqalign{
phi &= XA^T:Xodot C cr&= Y:Z cr
dphi
&= Z:dY + Y:dZ cr
&= Z:dX,A^T + Y:dXodot C cr
&= ZA:dX + Yodot C:dX cr
&= (ZA + Yodot C):dX cr
frac{partialphi}{partial X}
&= ZA + Yodot C cr
&= (Xoslash B)A + (XA^T)oslash B cr
}$$
answered Jan 6 at 16:26
greggreg
7,7951821
$endgroup$
Let's use a colon to denote the trace/Frobenius product, i.e.
$$eqalign{A:B={rm Tr}(A^TB)}$$
Properties of the trace give rise to rules for rearranging terms in a Frobenius product, such as
$$eqalign{
A:B &= B:A cr&= A^T!:!B^T cr
A:BC &= B^TA:C cr&= AC^T:B cr
}$$
Also note that the Hadamard product commutes with itself and the Frobenius product
$$eqalign{
Aodot B &= Bodot A cr
A:(Bodot C) &= (Aodot C):B cr&= (Bodot A):C cr
}$$
Instead of $B$ let's use its Hadamard inverse, and while we're at it, let's define a few other matrices.
$$eqalign{
C &= 1oslash B cr
Y &= X^TA cr
Z &= Xodot C = Xoslash B cr
}$$
to write the cost function. Then find its differential and gradient.
$$eqalign{
phi &= XA^T:Xodot C cr&= Y:Z cr
dphi
&= Z:dY + Y:dZ cr
&= Z:dX,A^T + Y:dXodot C cr
&= ZA:dX + Yodot C:dX cr
&= (ZA + Yodot C):dX cr
frac{partialphi}{partial X}
&= ZA + Yodot C cr
&= (Xoslash B)A + (XA^T)oslash B cr
}$$
answered Jan 6 at 16:26
greggreg
7,7951821
edited Jan 6 at 21:13
answered Jan 6 at 16:26
greggreg
7,7951821
answered Jan 6 at 16:26
greggreg
7,7951821
answered Jan 6 at 16:26
greggreg
7,7951821
7,7951821
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
add a comment |
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
Thanks. Would you please clarify what do you mean with two colons in a row? (e.g., the right-hand side for $dphi$)
$endgroup$
– merch
Jan 6 at 19:22
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
@merch Hopefully the introduction of the matrices $(Y, Z)$ have clarified things.
$endgroup$
– greg
Jan 6 at 21:15
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
Thank you, it simplifies it a little. I understand the final result. However, I am still struggling with part of your notation. How do you re-write $Z:dY+Y:dZ$ via the trace operator?
$endgroup$
– merch
Jan 6 at 23:47
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
$begingroup$
@merch From the 1st equation above, the Frobenius products correspond to $${Z:dY + Y:dZ = {rm Tr}(Z^TdY+Y^TdZ)}$$
$endgroup$
– greg
Jan 7 at 3:54
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063510%2fderivative-of-a-trace-with-hadamard-division%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
